MCAT Subject Tests
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_________________________________________________________G E N E R A L C H E M I S T R Y S U B J E C T T E S T2 General Chemistry Subject Test 2
1. Which one of the following reactions will be
accompanied by an increase in entropy?
A. Na(s) + H2O(l) ? NaOH(aq) + H2(g)
B. I2(g) ? I2(s)
C. H2SO4(aq) + Ba(OH)2(aq) ? BaSO4(s) +
H2O(l)
D. H2(g) + 1/2O2(g) ? H2O(l)
E. None of the above
2. Which one of the following has the largest ionic
radius?
A. Na+
B. K+
C. Mg++
D. Al3+
E. Cl–
3. The heat of combustion of gaseous ammonia,
NH3(g), is 81 kcal/mole. How much heat is released in the reaction of 34 grams of ammonia with excess
oxygen?
A. 40.5 kcal
B. 60.3 kcal
C. 75.8 kcal
D. 81 kcal
E. 162 kcal
4. What is the percent composition by mass of Al in
Al2(SO4)3?
A. 7.9 %
B. 31.6 %
C. 15.8 %
D. 12.7 %
E. 22.3 % 5. Compound _H of formation
(kcal/mole, 25 °C, 1 atm) H2O(l) –68.4
H2O(g) –57.8
C2H6(g) –20.2
C2H4(g) +12.5
C2H2(g) +54.2
CO(g) –26.4 CO2(g) –94.1
Based on the values contained in the table, what is the
heat of combustion of one mole of ethylene at 298K
and 1 atm pressure?
A. 316.1 kcal
B. 12.5 kcal
C. –291.1 kcal
D. –316.1 kcal
E. –337.3 kcal
6. The reaction below is carried out at constant
temperature in a 2.0 liter vessel.
N2 + O2? 2NO
At equilibrium, the vessel is found to contain 8 moles
of NO, 2 moles of N2 and 4 moles of O2. What is the
value of K eq for this reaction under these conditions?
A. 1
B. 8
C. 0.25
D. 2
E. 0.5
7. I A
slow
→
fast
→
fast
→
B
II B + C BC* (activated complex)
III BC* D + E
Given the above series of reactions leading to the
formation of products D and E, which is the rate-
determining step?
A. I
B. II
C. III
D. I and II
E. II and III
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G E N E R A L C H E M I S T R Y S U B J E C T T E S T 2 8.
According to VSEPR Theory, the H-Te-H bond angle in H 2Te should be closest to A. 90°. B. 120°. C. 109.5°. D. 149.5°. E. 180°.
9.
A 300 mL flask containing nitrogen at a pressure of 100 torr and a 200 mL flask containing oxygen at a pressure of 200 torr are connected such that the gases are allowed to fill the combined volume of the two flasks. What is the partial pressure of nitrogen in the combined volume? A. 60 torr B. 80 torr C. 100 torr D. 150 torr E. 200 torr
10. Which of the following statements is true with regard to the diagram below?
U W X Y Z Reaction Coordinate
Free
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A. Going from U to W requires less energy than going
form X to Y.
B. Going from U to X occurs more readily than going
from X to Z.
C. Although in the forward direction the reaction X to
Z will occur more rapidly than from U to X, in the reverse reaction Z to X will occur more slowly than X to U.
D. The reaction U to X is exothermic while the
reaction from X to Z would be endothermic. E. The overall reaction U to Z is endothermic.
11. Which of the elements below is the least
electronegative?
A. F
B. Cl
C. Br
D. K
E. C
12. Which one of the electronic configurations below
represents the atom with the highest ionization energy?
A. 1s 22s 22p 63s 1
B. 1s 22s 22p 63s 23p 64s 23d 104p 65s 14d 10
C. 1s 22s 22p 63s 23p 64s 23d 104p 65s 24d 105p 5
D. 1s 22s 22p 63s 23p 63s 23p 5
E. 1s 22s 22p 63s 23p 6
13. Electron density studies have revealed that X and Y
have an equal number of electrons. Which of the following could be X and Y respectively?
A. Ca + and K
B. H + and He
C. Cl and F
D. O – and S +
E None of the above 14.
1st Ionization 2nd Ionization Energy (eV) Energy (eV)
U 5.6 5.6
V
1.7
2.9 X 1.1 1
3.6 Y 12.4 2.8 Z
2.9 1.7
From the information given in the table above, which of the following is most probably a Group IA metal? A. U B. V C. X D. Y E. Z
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15. 2 HCl + Na 2CO 3 ? 2 NaCl + H 2O + CO 2(g )
(Atomic weights: H = 1, Cl = 35, Na = 23, O=16, C=12)
36 g of HCl are mixed with excess Na 2CO 3. If the reaction is carried out at STP, how many liters of CO 2 are formed? A. 5.60 B. 11.2 C. 16.8 D. 22.4 E. 44.8
16. Which of the following elements would be the most
reactive with fluorine?
A. Ar
B. Ti
C. Cl
D. Li
E. I
Questions 17 -18 refer to the following kinetic rates:
T rial [A]o (M) [B]o (M) rate (M/min) 1 1 x 10–2 3 x 10–3 2 x 10–3 2 1 x 10–2 6 x 10–3 8 x 10–3 3 2 x 10–2 1.2 x 10–2
3.2 x 10–2
17. What is the reaction order with respect to A?
A. 1
B. 1.5
C. 2
D. 3
E. 0
18. What would be the initial rate of the reaction in a
fourth trial if [A]o = 4 x 10–2 M and [B]o = 3 x 10–3 M?
A. 1 x 10–3 M/min
B. 2 x 10–3 M/min
C. 8 x 10–3 M/min
D. 8 x 10 –2 M/min
E. Cannot be determined from the information given
19. A spontaneous reaction is always accompanied by
A. a negative change in enthalpy.
B. a positive change in enthalpy.
C. a negative change in enthalpy and a positive
change in entropy.
D. a positive change in enthalpy and a negative
change in entropy.
E. a negative change in Gibbs free energy.
20. Combustion of a 50 gram sample of an unknown
hydrocarbon yields 132 grams of CO 2 and 126 grams of H 2O. Approximately how many grams of carbon were contained in the original sample?
A. 14 g
B. 28 g
C. 36 g
D. 63 g
E. 66 g
21. Under which of the following sets of conditions (temperature and pressure) will the ideal gas law be least likely to apply?
A. 250 K, 4 atm
B. 250 °C, 400 atm
C. 25 °C, 4 torr
D. 25 K, 400 torr
E. 25 K, 400 atm
22. The following reaction is first order with respect to W
and first order with respect to X. If the concentration of W and X are doubled what happens to the rate of the reaction?
W + X ? Y + Z A. It is doubled. B. It is quadrupled. C. It is squared.
D. It is increased but it cannot be determined how much from the information given.
E. It remains the same.
23. If atom A has a valence of 7 and atom B has a
valence of 2, which compound consisting of atoms A and B would be the most likely to form?
A. BA
B. B 2A
C. BA 2
D. B 2A 2
E. B 3A 2
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24. In moving down column II A of the Periodic Table
A. electron affinity increases.
B. ionization energy increases.
C. electronegativity remains constant.
D. the oxidation state remains constant.
E. the shielding effect decreases.
25. Which of the following choices lists elements, from
left to right, with decreasing atomic radii?
A. Li, Be, B, C, N
B. Be, Mg, Ca, Sr, Ba
C. Li, Mg, Sc, Zn, Ta
D. Au, Ag, Cu
E. Two of the above
26. Under which of the following conditions will a
reaction always be in thermodynamic equilibrium?
A. ? H
? S
> 1
B. ? H
? S
< 1
C. ? H
? S
= 1
D. ? H
T ?S
= 1
E. ? H
? S
< 0
27. An ideal gas is contained in a one-liter vessel at 350
K. What additional information is sufficient to
determine the pressure of the gas?
A. the molecular weight of the gas
B. the number of gas molecules
C. the molecular weight and the mass of the gas
D. B or C
E. A, B, or C 28. An ideal gas is placed in a 10 liter container that is
free to expand. If the temperature (K) is doubled and
the pressure is changed from 60 mmHg to 20 mmHg,
what is the new volume in liters?
A. 20
60
x 1
2
x 10
B. 20
60
x 2 x 10
C. 60
20
x 1
2
D. 60
20
x 1
( 2 x 10)
E. 60
20
x 2 x 10
29. The concentration of which of the following
substances would NOT decrease as a result of an
increase in the temperature?
X2 3AB + P _H = -10 kcal/mole
A. AB only
B. both AB and P
C. X2Y only
D. both X2Y and BD
E. Cannot be determined without a value for the
change in entropy
30. Boron found in nature has an atomic weight of
10.811 and is made up of the isotopes 10B (mass
10.013 amu) and 11B (mass 11.0093 amu). What
percentage of naturally occurring boron is made up of 10B and 11B respectively?
A. 30; 70
B. 25; 75
C. 20; 80
D. 15; 85
E. 10; 90
STOP! END OF TEST.
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THE ANSWER KEY AND EXPLANATIONS BEGIN ON THE FOLLOWING PAGE.
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GENERAL CHEMISTRY SUBJECT TEST 2
ANSWER KEY
1. A
2. E
3. E
4. C
5. E
6. B
7. A 8. C 9. A 10. C 11. D 12. E
13. A 14. C 15. B 16. D 17. E 18. B
19. E 20. C 21. E 22. B 23. C 24. D
25. E 26. D 27. D 28. E 29. D 30. C
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EXPLANATIONS
1. A
Qualitative thermodynamics questions can be answered by applying the definitions of the various state functions. This question asks about entropy, which can be defined as the energetic equivalent of randomness, or disorder. The question thus translates into “Which of the following will have an increase in disorder as the reaction proceeds from left to right?” Recall that the general order of increasing entropy goes: solid Precipitation takes aqueous ions to ionic solid and thus represents a decrease in entropy. Neutralization takes aqueous ions to molecules of liquid and is accompanied by a net decrease in disorder, therefore C is accompanied by a decrease in entropy. Choice D shows the formation of liquid water or, alternatively, the combustion of gaseous hydrogen. By either description, this reaction proceeds with a large decrease in entropy as the mole and a half of mixed gas phase molecules transform into one mole of pure liquid. 2. E In general, a cation is smaller than a neutral atom of the same element since the removal of an electron or electrons to form the cation leaves fewer electrons to shield one another from the attractive force of the nucleus and to repel one another. The remaining electrons on the cation are drawn closer to the nucleus, resulting in a reduction in size. By this same reasoning, it follows that the greater the positive charge, the smaller the ion. Conversely, anions are generally larger than the corresponding neutral atoms due to an increased number of electrons. A size trend is also observed in ions of equal charge: as one moves down a group, or column of the Periodic Table, the atomic and ionic radii increase due to the larger size of the highest occupied orbital. Therefore sodium atoms are smaller than potassium atoms, and Na + ions are smaller than K + ions. If we apply these facts and place the ions into a partial order by size, we have: K + > Na + > Mg 2+ > Al 3+ Thus choices A, C, and D can be eliminated, and we must now compare the remaining species, K + and Cl -. When potassium, atomic number 19, gives up an electron to form the cation K +, it is left with 18 electrons and is thus isoelectronic to argon, atomic number 18. Chlorine, atomic number 17, takes on an electron to form Cl -, and thus it likewise becomes isoelectronic to argon. Each species is isoelectronic to the other, and the relative radii will be affected most by the attractive force exerted on the outer electrons by the positive charge of the nucleus. It follows that the more protons the nucleus contains, i.e., the larger the atomic number is, the smaller the ionic radius will be. The final comparison is thus Cl - > K +, verifying choice E as the credited choice. 3. E This question combines some basic stoichiometry with thermodynamic values for a reaction. We are given the ?H comb per mole of ammonia and the mass of ammonia in grams. Applying dimensional analysis, we must convert the mass of ammonia into moles, then multiply by kcal/mol to find the total number of kcal. Ammonia has a formula weight of 14 + 3(1) = 17 g/mol, therefore: (34 g)/(17 g/mol) = 2 mol (2 mol).(81 kcal/mol) = 162 kcal. Note that it is not necessary to come up with a balanced equation for the combustion reaction to do this problem. _________________________________________________________ G E N E R A L C H E M I S T R Y S U B J E C T T E S T 2 K A P L A N ________________________________________________________________________________________9 4. C The percent by mass of an element in a compound is calculated by multiplying the atomic mass of that element by the number of atoms of that element in one formula of the compound, then dividing by the total formula weight of the compound and, finally, converting the result to a percentage. In this example there are two atoms of aluminum (atomic weight 27) and three sulfate ions (formula weight = 32 + 4(16) = 96) per formula of aluminum sulfate, so the calculation goes: 2(27)/(2(27) + 3(96)) = 54/342. which rounds off and converts to 15.8%. Calculation can, as usual, be avoided by noticing that 54/342 will be approximately equal to 50/350, or roughly 15%. Choice A is the result of inserting one aluminum atomic weight into the top of the calculation, while choices B, D, and E are the results of other random errors in calculation. 5. E This question simultaneously tests the basic definition of combustion and thermodynamic calculations involving heats of formation, along with a bit of basic organic nomenclature. To calculate thermodynamic values for a reaction or process from given heats of formation, we must set up a correctly balanced equation for the process and then subtract the sum of the heats of formation, ?H f , of the reactants from the sum of the heats of formation of the products. To do so for this problem, we must first identify ethylene as the C 2H 4(g ) entry in the table and then write a balanced equation for its combustion: C 2H 4(g ) + 3 O 2(g ) ? 2 CO 2(g ) + 2 H 2O (l ) Adding the values for the ?H f (products) = 2 ?H f (CO 2(g )) + 2 ?H f (H 2O (l )) and subtracting the sum of the ?H f (reactants) = ?H f (C 2H 4(g )) + 3 ?H f (O 2(g )), the result is : ?H comb (C 2H 4(g )) = ?H f (products) - ?H f (reactants) = [2 ?H f (CO 2(g )) + 2 ?H f (H 2O (l ))]- [?H f (C 2H 4(g )) + 3 ?H f (O 2(g )) = 2(-94) + 2(-68.4) -(12.5) -3(0) = -337.3 kcal/mol (Recall that the heat of formation of any standard state element, like O 2(g ) here, is zero.) As for the wrong choices, notice first that choices A and B could have been immediately eliminated since combustion, by nature, is exothermic; a correct answer to this question must therefore be negative. Choice D results from using the heat of formation of gaseous water rather than that of the liquid water that should be expected at 25o C, and choice C is the result obtained by adding 12.5 (when it should have been subtracted) in addition to using the value for gaseous rather than liquid water. 6. B We can calculate the equilibrium constant for this reaction by setting up the reaction quotient in terms of the concentrations of the reactants and products, then solving based on the numbers given in the problem. Since we are told that the vessel is at equilibrium the reaction quotient is the equilibrium constant. For the given reaction, the reaction quotient is: Q c = K eq = [NO]2 [N 2][O 2 ] Since the reaction takes place in a 2.0 liter container, we can convert the given numbers of moles into moles per liter by dividing each by two, then plug the results into the expression for Q c : [NO] = 8 mol/2 L = 4 mol/L [N 2] = 2 mol/2 L = 1 mol/L [O 2] = 4 mol/2 L = 2 mol/l G E N E R A L C H E M I S T R Y S U B J E C T T E S T 2 ________________________________________________________________ 10 _______________________________________________________________________________________K A P L A N ∴ K eq = [NO]2[N 2][O 2] =42 (1)(2) = 8 7. A This question can be answered based on fundamental definitions; the rate-determining step is simply the slowest step in a proposed mechanism. Of the three steps shown, only step I is labeled as 'slow' and must therefore be the bottleneck in the overall process. As such, it must be the rate-determining step. 8. C Tellurium, Te, is a Group VIA element, found below oxygen and sulfur in the Periodic Table. It should thus have six valence electrons available for bonding. If the central Te atom uses two of its valence electrons to form sigma bonds to the two hydrogens in H 2Te, it will still have two pairs of nonbonding electrons remaining. According to VSEPR theory H 2Te fits the general formula AB 2U 2 and will have tetrahedral electronic geometry and angular molecular geometry, with bond angles of 109.5_. VSEPR theory makes the same prediction for H 2O and H 2S, and similar predictions for other tetrahedral electronic species such as methane and ammonia. (Note: In reality, the bond angles generally vary from those predicted by the theory because of differences in repulsion between bonded and nonbonded electron pairs and because of orientational differences in the orbitals involved in the bonding.) 9. A This question about gas volumes and pressures can best be answered by applying Dalton's law of partial pressures, which states that the pressure exerted by any one gas in a mixture of ideal gases is the same as it would be if the gas were to occupy the same volume alone. If we focus on the nitrogen we can approach this problem as if the volume increases from the initial 300 mL to the combined 500 mL and apply Boyle's law: P 1V 1 = P 2V 2 (at constant temperature ). Since an increase in volume will lead to a decrease in pressure, the final pressure must be less than 100 torr. Thus we can automatically eliminate choices C, D, and E before completing the calculation as follows: P 2 = P 1V 1/V 2 = (100 torr)(300 mL)/(500 mL) = 3/5(100 torr) = 60 torr. 10. C This question is probably best approached by examining the answer choices in search of the one true statement required. Choice A is false; it appears from the reaction coordinate diagram provided that W and Y are at approximately the same level while U is lower in energy than is X. It will thus take more energy to go from the lower energy state U to the W/Y level than it will to go from the higher X energy level to that of W/Y. In other words, the activation energy for the reaction U ? X is larger than the activation energy for the reaction X ? Z. Choice B is essentially equivalent to choice A, since a reaction "occurs more readily" if and only if its activation energy is smaller. Choice C is true. In the forward direction, X ? Z has a smaller energy of activation (and is therefore a faster reaction) than does U ? X. Meanwhile, the free energy of activation for Z ? X, that is, the energy required to go from Z to Y, is larger than the energy required to move from X to W. Therefore, in the reverse direction, the reaction Z ? X will be slower than the reaction X ? U. Choices D and E are both wrong for two reasons. First, the diagram shown is a free energy diagram (?G on the vertical axis) rather than an enthalpy diagram (?H on the vertical axis); as such we cannot directly determine changes in enthalpy to which the terms exothermic and endothermic apply. What we can determine from a free energy diagram is whether a given reaction is spontaneous or not. Secondly, even if we did attempt to infer that enthalpy and free energy were following parallel trends, the reaction U ? X is an uphill move on the diagram and would thus be _________________________________________________________ G E N E R A L C H E M I S T R Y S U B J E C T T E S T 2 K A P L A N _______________________________________________________________________________________11 endothermic (making choice D wrong) while the reaction U ? Z results in a net decrease in energy and would thus be exothermic (disqualifying choice E). 11. D Electronegativity increases toward the upper right-hand corner of the Periodic Table, with fluorine, F, the most electronegative element and cesium, Cs, the least. Electronegativity is a property of atoms in bonds, i.e., it is actually a molecular property, therefore rare gases like helium, He, which do not form compounds, are not included in the electronegativity scale. (Heavier elements like francium, Fr, are not included because their half-lives are so short that their compounds can not be isolated.) The question can therefore be translated as "Which of the following is closest to the lower left-hand corner of the Periodic Table?" Choice D, potassium, lies far to the left of the other entries and as such is the least electronegative. 12. E Ionization energy is defined as the amount of energy required to remove an electron from a gas phase atom or ion, and increases toward the upper right-hand corner of the Periodic Table, with helium, He, having the largest such energy requirement of any neutral atom. This question can therefore be reworded as "Which of the following configurations corresponds to the element which is closest to the upper right-hand corner of the Periodic Table?" We can count the total number of electrons and, given that these are atoms rather than ions, thereby translate the answer choices into their corresponding elemental symbols: A. total number of electrons = atomic number = 11 _ Na B. total number of electrons = atomic number = 47 _ Ag C. total number of electrons = atomic number = 53 _ I D. total number of electrons = atomic number = 17 _ Cl E. total number of electrons = atomic number = 18 _ Ar We can then see that choice E, argon, lies furthest to the upper right and therefore has the greatest ionization energy. It is not necessary to convert the electron configurations into the atoms which they represent if one notices that choice E is the only one with all its subshells filled and would therefore be the hardest to remove an electron from. 13. A Two species with the same number of electrons are said to be isoelectronic ; this question can thus be reworded to read "Which of the following is an isoelectronic pair?" Calcium has atomic number 20. This means that a neutral atom of calcium has twenty protons in its nucleus and twenty electrons around its nucleus. Positive charges result from the removal of outer electrons, thus to acquire a charge of positive one, calcium would need to lose one electron from its outermost, or 4s , orbital, leaving Ca + with a total of 19 electrons. Potassium, meanwhile, has atomic number 19 and thus also has 19 electrons as a neutral atom. These two species are then seen to be isoelectronic, verifying choice A as the correct choice. The species in each of the other choices do not have the same number of electrons; H + has no electrons while He has two, Cl has 17 while F has 9, and O - has 9 electrons while S + has 15. These choices, along with choice E, can thus be eliminated. 14. C Ionization energy is defined as the amount of energy required to remove an electron from a given species. Ionization energy is usually expressed in energy per particle (as it is here) or in energy per mole, and energy, in turn, is usually expressed in electron volts (eV), joules, or kilojoules. The first ionization energy of an element is the energy required to remove an electron from a neutral atom of that element, while the second ionization energy is the energy required to remove a second electron, i.e., to remove an electron from the +1 cation. A particularly useful piece of logic/knowledge G E N E R A L C H E M I S T R Y S U B J E C T T E S T 2 ________________________________________________________________ 12 _______________________________________________________________________________________K A P L A N here is that it will be more difficult to remove an electron from a positively-charged species than it will be to remove an electron from a neutral version of the same species. In other words, the second ionization energy of an element is always greater than its first ionization energy. Based on this fact alone, choices A, D, and E can be eliminated. To choose between the remaining two choices, we must reason out that a small value for ionization energy corresponds to relative ease of removal of the electron. A Group IA metal atom will lose its first electron with relative ease, but after that it will possess an electronic configuration similar to that of a noble gas. It will therefore be difficult to remove another electron, implying a high second ionization energy. Looking at the numbers in the table, we see that both remaining choices have relatively small values for first ionization energy, but that X has a much higher second ionization energy. It follows that element V is most likely a Group IIA element (e.g., Mg, Ca) while element X is most likely a member of Group IA (e.g., Na, K). Choice C, as indicated above, is thus the best choice. 15. B According to the balanced equation, the complete reaction of two moles of HCl should result in the formation of one mole of CO 2. We are given that 36 grams, or one mole, of HCl reacts; there should thus be half a mole of CO 2 formed. Assuming that the gas formed behaves ideally, and recalling that one mole of any ideal gas occupies 22.4 liters at STP, this half a mole of CO 2 should occupy a volume of 11.2 liters as stated in choice B. 16. D Reading from left to right across the Periodic Table, metallic character decreases as electronegativity increases. The most reactive combinations of elements are those with the greatest differences in electronegativity, and this difference can be most easily predicted by observing the distance between the elements on the Periodic Table. This question can thus be translated to read: "Which of the following lies furthest from F on the Periodic Table?" Lithium, in Group IA, is therefore the correct answer. Titanium, choice B, would be the second most reactive of the options offered, while choices C and E, chlorine and iodine, are Group VIIA elements like fluorine, and thus would be only slightly reactive with it. Choice A, argon, is an inert gas and would not be reactive at all. 17. E The rate law for any reaction, A + B ? C, is given by: rate = k[A]x [B]y where k is the rate constant at the temperature at which the reaction is carried out, [A] and [B] are the concentrations of the two reactants, and x and y are the reaction orders with respect to A and B respectively. This question asks us to find x from the experimental rate data provided. To simplify the tabulated rate data, we can rewrite the table as small whole number multiples of the smallest entry in each column: line # [A] o [B] o rate 1 1 1 1 2 1 2 4 3 2 4 16 Looking at lines 1 and 2, where the concentration of reactant A is held constant, we see that a doubling in the concentration of B results in a quadrupling of the rate, therefore the reaction is second order with respect to B. We can now look at lines 2 and 3 (or 1 and 3) to determine the required reaction order. From line 2 to line 3, the concentration of B is doubled; the reaction rate should thus quadruple due to the influence of the concentration increase of second order reactant B. We can divide out this quadrupling effect and create a hypothetical line 4: _________________________________________________________G E N E R A L C H E M I S T R Y S U B J E C T T E S T2 line # [A]o [B]o rate 1 1 1 1 2 1 2 4 3 2 4 16 4 2 2 4 From this revision of the experimental data it can be seen (lines 2 and 4) that a doubling of [A] results in no change in the rate, i.e., 2x = 1. Therefore the reaction is zero order with respect to A. 18. B This question is based on the same reaction, and can be answered quickly based on the results outlined above. As determined in the explanation above, the reaction is zero order in A; as a result, changing the concentration of A will have no effect on the reaction rate. The original entry labeled as 'trial 1' has the same initial concentration of B, 3 x 10-3 M, as that offered in this question. The answer will thus be that the rate is the same as that in trial 1, or 2 x 10-3. 19. E This is a question on fundamental concepts of thermodynamics. To be classified as spontaneous, in thermodynamic terms, a reaction or process must be accompanied by a decrease in Gibbs free energy, i.e., ?G < 0. Recall that ?G = ?H - T?S; depending on the sign and magnitude of ?S, ?H can be either positive or negative and still result in a negative ?G, and choices A and B can be eliminated. Choice C is a bit tricky: while it is true that the conditions listed in choice C will always result in a negative value for ?G, these conditions are not, conversely, required for a reaction to be spontaneous and thus they will not always be true for any spontaneous reaction. In other words, while the statement in C guarantees a spontaneous reaction, a spontaneous reaction does not necessarily guarantee the statement in choice C. Finally, the conditions provided in choice D will never allow a process to proceed spontaneously; since T is the absolute temperature and hence always positive, ?H - T?S will always be positive for positive ?H and negative ?S. 20. C This question of stoichiometry can be easier than it appears if we notice that we are only asked for the carbon content of the original sample, rather than for its complete formula. We can therefore ignore the sample size and the water produced, focusing instead on the carbon containing product CO2. Since each formula of CO2 contains one carbon atom, it follows that each formula weight of CO2 will contain 12 grams of carbon. The calculation can then be completed as follows: (132 g CO2)(12 g C/44 g CO2) = (132)(12)/(44) g C = 36 g C The reactants in a combustion reaction are the substance being combusted and oxygen. All the carbon content in the form of carbon dioxide in the product thus has to come from the substance itself originally. Therefore 36 g of carbon in the product means 36 g of carbon in the original sample. 21. E Despite the numerical nature of the answer choices, this is a qualitative question on fundamental concepts regarding ideal gases. Ideal gases are those which obey the ideal gas law, PV = nRT. This obedience is most likely to occur for relatively inert, nonpolar gases under conditions of high temperature and low pressure. Conversely, it follows that the ideal gas law will be least likely to apply for highly reactive, polar gases at low temperature and high pressure. Since the K A P L A N_______________________________________________________________________________________13 G E N E R A L C H E M I S T R Y S U B J E C T T E S T2 ________________________________________________________________ question does not identify the chemical nature of a specific gas, we can thus translate the given question as “Which of the following combinations has the lowest temperature and the highest pressure?” Choice E does. Keep in mind that a torr is a much smaller unit of pressure than is an atmosphere. (1 atm = 760 torr) 22. B The rate law for any reaction of the type W + X ? products is given by: rate = k[W]x[X]y where k is the rate constant at the temperature at which the reaction is carried out, [W] and [X] are the concentrations of the two reactants, and x and y are the reaction orders with respect to W and X respectively. This question gives us x and y and asks us to predict the effect on the overall rate resulting from the given changes in concentration. Since x and y are each equal to one, as implied by the fact that the reaction is first order with respect to each reactant, we may rewrite the rate law as: rate = k[W][X]. If we substitute double concentrations for each of W and X, the new rate becomes k[2W][2X] = 4(k[W][X]), i.e. it has quadrupled. Note that if we were not given the information about the orders of the reaction, then the equation by itself is insufficient for us to predict the rate. 23. C This question is best answered based on the chemical properties of elements as read from the Periodic Table. A valence of 7 puts element A in column VIIA along with fluorine, chlorine, etc., while a valence of 2 puts element B in group IIA along with calcium and the rest of the alkaline earth metals. Group IIA metals most readily form 2+ cations while the halogens prefer a 1- charge as anions; charge balance then requires that two A combine with each B, giving the empirical formula BA2. (This question could have been answered most conveniently, and quickly, by comparison to a reference compound such as CaCl2.) 24. D Group IIA is the group commonly referred to as the 'alkaline earth metals'. Metallic character is greatest toward the lower left-hand corner of the Periodic Table. Metallic character includes physical properties such as malleability, ductility, and luster, as well as chemical properties such as low ionization energy, electron affinity, and electronegativity. Choices A and B are thus in complete opposition to the properties of metals, and should therefore be disposed of. Choice C is also untrue, as electronegativity decreases as you move down the group. On the other hand, the low ionization energy of metals allows them to easily give up electrons to a willing acceptor; the metals of Group IIA are easily oxidized to 2+ cations, making choice D a true statement. Finally, choice E is incorrect since the effective nuclear charge, Z eff, remains roughly constant as one moves down a column of the Periodic Table: all the elements in column IIA have the electron configuration [G]n s2, where G is the symbol for the previous rare gas and n is the principal quantum number of the valence shell of the metal in question. 25. E This is yet another question on atomic properties and periodic trends. Atomic radius is greatest toward the lower left-hand corner of the Periodic Table; it generally decreases from left to right across a period, and increases from top to bottom down a group. Choice A lists elements in order from left to right across period 2 and thus represents elements in order of decreasing atomic radius. Choice D lists three elements in order moving up a group and thus likewise in order of decreasing atomic radius. Choice E, two of the above, is therefore the credited response. Choice B lists the Group IIA elements in order of increasing atomic radius, while choice C is a somewhat randomized list of elements in (mostly) increasing order by size; choices B and C can thus be eliminated. 14 _______________________________________________________________________________________K A P L A N _________________________________________________________G E N E R A L C H E M I S T R Y S U B J E C T T E S T2 26. D To be classified as spontaneous, in thermodynamic terms, a reaction or process must be accompanied by a decrease in Gibbs free energy, i.e., ?G < 0. For a system to be at equilibrium, the forward and reverse processes must be equally spontaneous, therefore the net change in Gibbs free energy for the system must be equal to zero, i.e., ?G = 0. Recall that ?G = ?H - T?S; it follows then that at equilibrium ?H - T?S = 0, or, that ?H = T?S. An algebraic rearrangement of this last expression then gives choice D as the credited response. 27. D This question is asking us to determine the minimum additional information necessary to find an ideal gas's volume. If we begin with PV = nRT, we can focus on what is given, and figure out what is missing. Volume is given as 1 liter and temperature as 350K; we can thus rearrange the ideal gas law as: P = n(RT/V), where all the values inside the parentheses are known. To complete the determination of pressure, we need only to find n, the number of moles of gas in the container. Molecular weight alone does not tell us how many moles we have, so choice A can be eliminated. If we know the number of molecules, as given in choice B, we can divide by Avogadro's number to find the number of moles present, so choice B will be sufficient. Likewise, if we know the mass and molecular weight of the gas, as choice C offers, we can calculate the number of moles present and thereby complete the pressure calculation. Since choices B and C are each sufficient while choice A is not, choice D is the credited response. 28. E This question on gas volumes could be answered quantitatively by applying the ideal gas law, PV = nRT, but the format of the choices indicates that calculation is unnecessary. One approach is to use the combined gas law: P1V1/T1 = P2V2/T2, and rearrange it to solve for V2 = T2P1V1/P2T1 = (T2/T1)(P1/P2)V1 = (2)(60/20)(10) = 60/20 x 2 x 10, as written in choice E. An alternate, more logical approach is to use the choices and think: if temperature is increased, the gas will expand and the volume therefore increases; we thus need to multiply by 2 rather than divide by 2, and choices A, C, and D can be eliminated. Similarly, if the pressure is decreased, the gas will expand, and we must multiply the original volume, 10 liters, by 60/20 rather than by 20/60; this observation alone allows us to eliminate choices A and B. Choice E is then the only surviving choice after both of these observation/elimination sequences are completed. 29. D This question is on equilibrium and le Chatelier's principle. The reaction shown is exothermic, as evidenced by the negative value provided for ?H. Heat can thus be thought of as a product of this reaction, and raising the temperature should favor reactants, i.e., le Chatelier's principle predicts that the equilibrium will shift to the left. A shift toward the formation of the reactants would increase the concentrations of X2Y and BD, making choice D the credited choice for this 'NOT' question, while decreasing the concentrations of AB and P. 30. C The atomic weight of an element is calculated by multiplying the atomic mass of each isotope of that element by the relative abundance of that isotope. In this example, there are only two isotopes of boron so the calculation set-up goes: (x)(10.013) + (1–x)(11.0093) = 10.811, where x = the relative abundance of 10B, and thus 1–x = the relative abundance of 11B. To simplify the arithmetic, we can approximate and perform the claculation to just one decimal place: x(10.0) + (1–x)(11.0) = 10.8 10.0x + 11.0 – 11.0x = 10.8 –1.0x = –0.2 x = 0.2 = 20% K A P L A N_______________________________________________________________________________________15 虽然剑桥雅思已经出到了剑11，但是剑桥雅思4还是很有参考价值，今天雅思给大家带来了：剑桥雅思4Test2passage2阅读答案解析。相应的原文及翻译，请点击：剑桥雅思4Test2阅读passage2原文+译文：澳大利亚的另类疗法。更多解析，请点击：剑桥雅思4解析。剑4下载，请点击： passage1部分查看，请点击： 剑桥雅思4Test2阅读passage1原文+译文：语言的消失 剑桥雅思4Test2passage1阅读答案解析 剑桥雅思4Test2passage2阅读答案解析 Question 14 答案：C 关键词：Western 定位原文：第1段第1句“Australia has been unusual in the Western world in having a very conservative attitude to natural or alternative therapies, according to Dr Paul Laver, a lecturer in Public Health at the University of Sydney.” 解题思路：A答案说澳大利亚医生与制药公司关系紧密，属于完全未提及型答案。B答案认为澳大利亚医生总是和其他医师一同工作，与文中所说的事实恰好相反。D答案说澳大利亚医生会开出另类处方，这也是不正确的。只有C答案与文章叙述相符。 Question 15 答案：B 关键词：Americans 定位原文：第1段倒数第1句“Americans made more visits to alternative therapists than to orthodox doctors in 1990, and each year they spend about $US 12 billion on therapies that have not been scientifically tested.” 解题思路：文中这句话说1990年美国人去看另类疗法医师的次数比去看传统医生的次数还多。所以答案B是正确的。而A、C和D答案中提到的比较关系并不存在。 Question 16 答案：YES 关键词：20 years 定位原文：第2段第1句“Disenchantment with orthodox medicine has seen the 初三（下册）数学各章节重要知识点总结 二次函数 1. 二次函数的一般形式：y=ax2+bx+c.(a≠0) 2. 关于二次函数的几个概念：二次函数的图象是抛物线，所以也叫抛物线y=ax2+bx+c；抛物线关于对称轴对称且以对称轴为界，一半图象上坡，另一半图象下坡；其中c叫二次函数在y轴上的截距, 即二次函数图象必过（0，c）点. 3. y=ax2(a≠0)的特性：当y=ax2+bx+c (a≠0)中的b=0且c=0时二次函数为y=ax2(a≠0); 这个二次函数是一个特殊的二次函数，有下列特性： （1）图象关于y轴对称；（2）顶点（0，0）；（3）y=ax2 (a≠0)可以经过补0看做二次函数的一般式，顶点式和双根式，即： y=ax2+0x+0, y=a(x-0)2+0, y=a(x-0)(x-0). 4. 二次函数y=ax2+bx+c (a≠0)的图象及几个重要点的公式： 5. 二次函数y=ax2+bx+c (a≠0)中，a、b、c与Δ的符号与图象的关系： (1) a＞0 <=> 抛物线开口向上； a＜0 <=> 抛物线开口向下； (2) c＞0 <=> 抛物线从原点上方通过； c=0 <=> 抛物线从原点通过； c＜0 <=> 抛物线从原点下方通过； (3) a, b异号 <=> 对称轴在y轴的右侧； a, b同号 <=> 对称轴在y轴的左侧； b=0 <=> 对称轴是y轴； (4) Δ＞0 <=> 抛物线与x轴有两个交点； Δ=0 <=> 抛物线与x轴有一个交点（即相切）； Δ＜0 <=> 抛物线与x轴无交点. 6．求二次函数的解析式：已知二次函数图象上三点的坐标，可设解析式y=ax2+bx+c，并把这三点的坐标代入，解关于a、b、c的三元一次方程组，求出a、b、c的值, 从而求出解析式-------待定系数法. 8．二次函数的顶点式： y=a(x-h)2+k (a≠0)；由顶点式可直接得出二次函数的顶点坐标（h, k），对称轴方程 x=h 和函数的最值 y最值= k. 9．求二次函数的解析式：已知二次函数的顶点坐标（x0,y0）和图象上的另一点的坐标，可设解析式为y=a(x -x0)2+ y0，再代入另一点的坐标求a，从而求出解析式.（注意：习题无特殊说明，最后结果要求化为一般式） 10. 二次函数图象的平行移动：二次函数一般应先化为顶点式，然后才好判断图象的平行移 小学数学创新能力竞赛（预赛）试题 一、填空题（每空3分，共60分） 1．20500321000≈（）亿37094000＝（）万 2．甲比乙多20%，乙比甲少（）。 3．用2、0、0、6可以组成（）个不同的四位数。 4．能被2、3、5、7整除的三位数中，最大的是（）。 5．用2、6、8和4个零组成的7位数中，只读出一个零的最大的数是（）。 6．同学们排队从学校出发去看电影，队伍全长200米，从排头出校门到排尾进入电影院共用35分钟，如果步行的平均速度是每分钟50米，学校到电影院共（）米。 7．找规律填得数：2.5 1.250.625（）0.15625。 8．2006年世界杯足球赛分为8个小组，每组4支球队，每组进行循环赛（即：每支球队都与其它球队进行一场比赛），循环赛后每组选2支球队进行淘汰赛（即：每支球队进行一场比赛，赢的进入下一轮，输的淘汰），最后决出冠军。这次世界杯一共举行（）场足球赛。 9．在1～2006这2006个自然数中，不能同时被7和13整除的数共 有（）个。 10．如图1，大圆内画一个最大的正方形，正方形内画一个最大的 圆……，如此画下去，共画了4个圆，那么最大的圆的面积是最 小圆的（）倍。 11．学校门口到公路边有一条100米的路，如果在这条路的两边栽树，离校门口10米处栽一棵，然后每隔10米栽一棵，一共需要栽（）棵。 12．在方框里填上适当的数：50.15×［72.05－ －17.95)］＝2006 13．用88个小正方体表面积之和的比是（）。 14．请你用1～9这九个数字，写出五个平方数（某个数的平方），每个数字最多用一次，这五个平方数分别是（）。 15．2005年12月8日是星期四，推算一下，2006年5月1日是星期（）。16．一个池塘里的睡莲，每天增长一倍，到第5天已长满了整个池塘，第二天长到这个池塘的（）。 17．五年级参加植树活动，人数在30与50人之间，如果分3人一组，4人一组，6人一组或8人一组，都恰好分完。五年级参加植树的学生有（）人。 图1 投影与视图 ?解读考点 ?2年中考 1．（北海）一个几何体的三视图如图所示，则这个几何体是（） A．圆柱 B．圆锥 C．球 D．以上都不正确 【答案】A． 【解析】 试题分析：由于主视图和左视图为长方形可得此几何体为柱体，由俯视图为圆可得为圆柱体．故选A． 考点：由三视图判断几何体． 2．（南宁）如图是由四个大小相同的正方体组成的几何体，那么它的主视图是（） A． B． C． D． 【答案】B． 考点：简单组合体的三视图． 3．（柳州）如图是小李书桌上放的一本书，则这本书的俯视图是（） A． B． C． D．【答案】A． 【解析】 试题分析：根据俯视图的概念可知，几何体的俯视图是A图形，故选A．考点：简单几何体的三视图． 4．（桂林）下列四个物体的俯视图与右边给出视图一致的是（） A．B．C． D． 【答案】C． 【解析】 试题分析：几何体的俯视图为， 故选C． 考点：由三视图判断几何体． 5．（梧州）如图是一个圆锥，下列平面图形既不是它的三视图，也不是它的侧面展开图的是（） A．B．C． D． 【答案】D． 考点：1．几何体的展开图；2．简单几何体的三视图． 6．（扬州）如图所示的物体的左视图为（） A． B． C． D． 【答案】A． 【解析】 试题分析：从左面看易得第一层有1个矩形，第二层最左边有一个正方形．故选A． 考点：简单组合体的三视图． 7．（攀枝花）如图所示的几何体为圆台，其俯视图正确的是（） A．B．C． D． 【答案】C． 考点：简单几何体的三视图． 8．（达州）一个几何体由大小相同的小方块搭成，从上面看到的几何体的形状图如图所示，其中小正方形中的数字表示在该位置的小立方块的个数，则从正面看到几何体的形状图是（） 英语 TEST 1 Part I 听力题没有题所以没有答案 Part II 16-25 DAAACBBACD 26 yourself 27 sending 28 traditional 29 solution 30 analysis 31 Personally 32 told 33 able 34 stores 35 be sure Part III 36-40 ADACB 41-45 CACAD 46 The two styles of job interviewing 47 The traditional job interview 48 the behavioral interview 49 Ability to communicate than on the truthfulness or content of his answer. 50 Past performance is the best indicator of future behavior. 51 贸易逆差（M-trade gap）分期付款（I-installment） 52 商标（N-trade mark ）交易会（G-fair，书上印刷错误有2个G） 53 关税（C-customs duty）合同（A-contract） 54 申报价值（E-declared value）免税入境（F-duty-free entry） 55 还盘（B-counter-offer）装运（K-shipment） 56 Hiram B.Rogan 57 The Chief Executive of Bubble Recordings 58 He is one of our best customers in the states. 59 He is coming over to see our working methods. 60 He has to arrange a top singer or group to make records when Mr.Rogan comes and sees 61-64 CAAD 65 尊敬的怀特先生, 非常感谢你上周五精彩的演讲。我代表中心的全体师生写信给你。我们很喜 欢你上周的演讲。我们最近也了解了一下近年来国际IT市场发展的趋势，十 分感谢你能来到我们学校，给我们带来了如此生动和有意义的演讲。再次感 谢，希望期待你的下次演讲。 真诚地 李明 Part V Dear Mr. Jackson, I hope you are still keeping well and that business is prospering. I'm writing to see if you can help me. Mr.Zhang has just taken the place of Mr. Wang to be the international sale's manager in your company.He is familar with American market，and will visit America lately,mainly stay in Washington，Please take care of him. Thank you . I hope to hear from you soon. Yours Sincerely Li Ming 九年级讲义目录 专题01 二次根式的化简与求值 阅读与思考 二次根式的化简与求值问题常涉及最简根式、同类根式，分母有理化等概念，常用到分解、分拆、换元等技巧. 有条件的二次根式的化简与求值问题是代数变形的重点，也是难点，这类问题包含了整式、分式、二次根式等众多知识，又联系着分解变形、整体代换、一般化等重要的思想方法，解题的基本思路是： 1、直接代入 直接将已知条件代入待化简求值的式子. 2、变形代入 适当地变条件、适当地变结论，同时变条件与结论，再代入求值. 数学思想： 数学中充满了矛盾，如正与负，加与减，乘与除，数与形，有理数与无理数，常量与变量、有理式与无理式，相等与不等，正面与反面、有限与无限，分解与合并，特殊与一般，存在与不存在等，数学就是在矛盾中产生，又在矛盾中发展. =x , y , n 都是正整数） 例题与求解 【例1】 当x = 时，代数式32003 (420052001)x x --的值是（ ） A 、0 B 、－1 C 、1 D 、2003 2- （绍兴市竞赛试题） 【例2】 化简 （1(b a b ab b -÷-- （黄冈市中考试题） （2 （五城市联赛试题） （3 （北京市竞赛试题） （4 （陕西省竞赛试题） 解题思路：若一开始把分母有理化，则计算必定繁难，仔细观察每题中分子与分母的数字特点，通过分解、分析等方法寻找它们的联系，问题便迎刃而解. 思想精髓：因式分解是针对多项式而言的，在整式，分母中应用非常广泛，但是因式分解的思想也广泛应用于解二次根式的问题中，恰当地作类似于因式分解的变形，可降低一些二次根式问题的难度. 【例3】比6大的最小整数是多少？ （西安交大少年班入学试题） 解题思路：直接展开，计算较繁，可引入有理化因式辅助解题，即设x y == 想一想：设x=求 432 32 621823 7515 x x x x x x x --++ -++ 的值. （“祖冲之杯”邀请赛试题） 的根式为复合二次根式，常用配方，引入参数等方法来化简复合二次根式. ---------------------------------------------------------------最新资料推荐------------------------------------------------------ 小学数学创新案例 小学数学创新案例【活动内容】社会调查与简单统计【活动目的】让学生通过小组活动进行社会调查，并将调查的所得的数据进行简单的统计与分析，从而培养和提高学生的应用数学的知识的综合能力。 【活动目标】 1、在组织学生进行社会调查的过程中，让学生掌握搞社会调查的一般方法和技巧，明白搞社会调查的意义。 2、让学生在调查中学会应用学过的数学知识的作简单的统计和分析，并根据所需，将统计结果制成统计图，从中获取有关的信息。 3、培养学生的探究、合作的能力，提高学生的应用数学知识解决实际为题的能力【活动重点】运用学过的数学知识对学校所在的地区进行社会调查，并对统计的数据作以总结与分析，让学生明白数学与生活的密切关系，从而帮助学生树立为了生活而学好数学的远大志向。 【活动难点】让学生掌握社会调查的方法和技巧，并能根据需要灵活运用。 【活动准备】 1、抽调积极分子，组成社会调查领导小组，进行调查活动内容制定，并挑选组员，教师提供可供选择的调查项目供各小组选用。 2、进行小组社会调查项目的指导，教给学生调查的方法和 1 / 6 技巧，以及将如何获取数据填在统计表中。 3、根据各小组去确立的调查的项目，指导学生制作统计表。 附： 以下社会调查项目，供各小组选择： （1）社区外出务工人员的统计（2）农民收入情况统计（3）社区机动车数据统计（4）村区房屋情况统计（5）农民家庭经济收入统计（6）各村民小组辖区劳动力数据统计（7）家长给学生零花钱数额统计也可以根据小组学生提供的项目制定统计表。 【课时安排】 2 课时第一课时【活动内容】进行社会调查指导，并组织学生开展社会调查。 【活动指导】一、组织学生，谈话导入我们已经学过了哪几种简单的统计？根据不同的统计我们学会了哪几种统计图？（学生进行交流，说硕学过的简单的统计的内容和各种统计图表的名称和制作方法）今天，我们将针对社区的各种情况，进行社会调查，并将获取的有关数据加以统计和分析。 从中获取有关的信息，来提高学生的社会调查实践能力和应用知识的能力。 二、练习，加深印象 1、根据下列的情况及数据绘制统计表，并进行统计图的绘制。 红星机械厂人员的机构组成情况： 工人，男 89 人，女 65 人；技术人员，男 40 人，女 25 2021年中考数学专题复习：投影 一、选择题 1.关于盲区的说法正确的有（） （1）我们把视线看不到的地方称为盲区 （2）我们上山与下山时视野盲区是相同的 （3）我们坐车向前行驶，有时会发现一些高大的建筑物会被比矮的建筑物挡住（4）人们常说“站得高，看得远”，说明在高处视野盲区要小，视野范围大 A.1 个 B.2个 C.3个 D.4个 2.如图，在一间黑屋子里用一盏白炽灯照一个球，球在地面上的阴影的形状是一个 圆，当把白炽灯向上远移时，圆形阴影的大小的变化情况是（） A.越来越小 B.越来越大 C.大小不变 D.不能确定 3.如下图所示的四幅图中，灯光与影子的位置最合理的是( ) 4.如图，一个斜插吸管的盒装饮料的正投影是图中的( ) 5.如图所示，晚上小亮在路灯下散步，在小亮由A处走向B处的过程中，他在地上 的影子（） A.逐渐变短 B.逐渐变长 C.先变短后再变长 D.先变长后再变短 6.如图是一根电线杆在一天中不同时刻的影长图,试按其一天中发生的先后顺序排 列,正确的是 ( ) (A)①②③④. (B)④①③②. (C)④②③①. (D)④③②①. 7.下列各种现象属于中心投影现象的是( ) A.上午10点时，走在路上的人的影子 C.中午用来乘凉的树影 D.升国旗时，地上旗杆的影子 8.如图，晚上小亮在路灯下散步，在小亮由A处走到B处这一过程中，他在地上的 影子（） A.逐渐变短 B.逐渐变长 C.先变短后变长 D.先变长后变短 9.在同一时刻的阳光下，小明的影子比小强的影子长，那么在同一路灯下( ) A.小明的影子比小强的影子长 B.小明的影子比小强的影子短 C.小明的影子和小强的影子一样长 D.无法判断谁的影子长 10.下列说法正确的是（） A.物体在阳光下的投影只与物体的高度有关 B.小明的个子比小亮高，我们可以肯定，不论什么情况，小明的影子一定比小 亮的影子长. C.物体在阳光照射下，不同时刻，影长可能发生变化，方向也可能发生变化. D.物体在阳光照射下，影子的长度和方向都是固定不变的. 11.四个直立在地面上的字母广告牌在不同情况下，在地面上的投影(阴影部分)效果 如图.则在字母L、K、C的投影中，与字母N属于同一种投影的有( ) A.L、K B.C； C.K D.L、K、C 12.这是圆桌正上方的灯泡（看作一个点）发出的光线照射桌面后，在地面上形成阴影（圆形）的示意图，已知桌面的直径为1.2米，桌面距离地面1米，若灯泡距离地面3米，则地面上阴影部分的面积为（） A.0.36π平方米 B.0.81π平方米 C.2π平方米 D.3.24π平方米 二、填空题 13.有下列投影： ①阳光下遮阳伞的影子； ②探照灯光下小明读书的影子； ③阳光下大树的影子； ④阳光下农民锄地的影子； 综合B3Test2答案( 60minutes ) Part I Reading Comprehension ( 25minutes ) Passage One There was one shop in the town of Mufulira that was notorious (臭名昭著的；臭名远扬的) for its color bar. It was a drugstore. While Europeans were served at the counter, a long line of Africans queued at the window and often not only were kept waiting but, when their turn came to be served, were rudely treated by the shop assistants. One day I was determined to make a public protest against this kind of thing, and many of the schoolboys in my class followed me to the store and waited outside to see what would happen when I went in. I simply went into the shop and asked the manager politely for some medicine. As soon as he saw me standing in the place where only European customers were allowed to stand he shouted at me in a bastard (杂种的，不纯粹的；伪的，假的；奇形怪状的，异常的) language that is only used by a boss when speaking to his servants. I stood at the counter and politely requested in English that I should be served. The manager became exasperated (愤怒的，恼火的) and said to me in English, "Even if you stand there till Christmas I will never serve you." I went to the District Commissioner’s office. Fortunately the District Commissioner was out, for he was one of the old school; however, I saw a young District Officer who was a friend of mine. He was very concerned to hear my story and told me that if ever I wanted anything more from the drugstore all I had to do was come to him personally and he would buy my medicine for me. I protested that that was not good enough. I asked him to accompany me back to the store and to make a protest to the manager. This he did, and I well remember him saying to the manager, "Here is Mr. Kaunda who is a responsible member of the Urban Advisory Council, and you treat him like a common servant." The manager of the drugstore apologized and said, "If only he had introduced himself and explained who he was, then, of course I should have given him proper service." I had to explain once again that he had missed my point. Why should I have to introduce myself every time I went into a store … any more than I should have to buy my medicine by going to a European friend? I want to prove that any man of any color, whatever his position, should have the right to go into any shop and buy what he wants. 1. Color bar in the first paragraph comes closest in meaning to _____. A) a bar which is painted in different colors B) the fact that white and black customers are served separately C) a bar of chocolate having different colors D) a counter where people of different colors are served with beer 2. The writer was, at the time of the story, _____. A) a black, but a member of the Urban Advisory Council B) an African servant C) a black, but a friend of the drugstore manager D) a rich black 3. The manager of the drugstore shouted at the writer in a bastard language because _____. A) he hadn’t learned to speak polite English B) he thought the writer wouldn’t understand English C) that was the usual language when speaking to Africans D) that was the only language he could speak when he was angry 4. In the third paragraph, "he was one of the old school" means _____. A) he believed in the age-old practice of racial discrimination B) he was a very old man C) he graduated from an old, conservative school D) he was in charge of an old school 5. Why didn’t the writer wait at the window of the drugstore like other black Africans? A) Because he thought he was educated and should be treated differently. B) Because he thought, being an important person, he should not be kept waiting. 初三下册数学知识点 1 二次函数及其图像 二次函数(quadratic function)是指未知数的次数为二次的多项式函数。二次函数可以表示为f(x)=ax bx c(a不为0)。其图像是一条主轴平行于y轴的抛物线。 一般的，自变量x和因变量y之间存在如下关系： 一般式 y=ax∧2; bx c(a≠0,a、b、c为常数)，顶点坐标为(-b/2a，-(4ac-b∧2)/4a) ; 顶点式 y=a(x m)∧2 k(a≠0,a、m、k为常数)或y=a(x-h)∧2 k(a≠0,a、h、k为常数)，顶点坐标为(-m，k)对称轴为x=-m，顶点的位置特征和图像的开口方向与函数y=ax ∧2的图像相同，有时题目会指出让你用配方法把一般式化成顶点式; 交点式 y=a(x-x1)(x-x2) [仅限于与x轴有交点A(x1，0)和B(x2，0)的抛物线] ; 重要概念：a，b，c为常数，a≠0，且a决定函数的开口方向，a>0时，开口方向向上，a牛顿插值公式(已知三点求函数解析式) y=(y3(x-x1)(x-x2))/((x3-x1)(x3-x2) (y2(x-x1)(x-x3))/((x2-x1)(x2-x3) (y1(x-x2)(x-x3))/((x1-x2)(x1-x3) 。由此可引导出交点式的系数a=y1/(x1*x2) (y1为截距) 求根公式 二次函数表达式的右边通常为二次三项式。 求根公式 x是自变量，y是x的二次函数 x1,x2=[-b±(√(b -4ac))]/2a (即一元二次方程求根公式) 求根的方法还有因式分解法和配方法 在平面直角坐标系中作出二次函数y=2x的平方的图像， 可以看出，二次函数的图像是一条永无止境的抛物线。 不同的二次函数图像 如果所画图形准确无误，那么二次函数将是由一般式平移得到的。 注意：草图要有1本身图像，旁边注明函数。 2画出对称轴，并注明X=什么 3与X轴交点坐标，与Y轴交点坐标，顶点坐标。抛物线的性质 轴对称 1.抛物线是轴对称图形。对称轴为直线x = -b/2a。 对称轴与抛物线的交点为抛物线的顶点P。 特别地，当b=0时，抛物线的对称轴是y轴(即直线x=0) 顶点 2.抛物线有一个顶点P，坐标为P ( -b/2a ，4ac-b ;)/4a ) 当-b/2a=0时，P在y轴上;当Δ= b ;-4ac=0时，P在x轴上。 开口 3.二次项系数a决定抛物线的开口方向和大小。 当a>0时，抛物线向上开口;当a|a|越大，则抛物线的开口越小。 决定对称轴位置的因素 4.一次项系数b和二次项系数a共同决定对称轴的位置。 当a与b同号时(即ab>0)，对称轴在y轴左; 因为若对称轴在左边则对称轴小于0，也就是- b/2a当a与b异号时(即ab0, 所以b/2a要小于0，所以a、b要异号 可简单记忆为左同右异，即当a与b同号时(即ab>0)，对称轴在y轴左;当a 小学五年级数学创新思维竞赛试题 学校：班级：姓名： 一、学海无涯任我翱翔——把你认为最准确的作案写在括号里。 1、小林家住在三楼，他每上一层楼要走14级台阶，小林从一楼走到三楼要走（）级台阶。 2、请你在算式: 1+2×3+4×5+6 中添上适当的一个小括号，使算式的得数最大，最大的得数是（）。 3、一件毛衣102元，比一副手套的5倍还多12元，一副手套（）元。 4、简算: 7.29×4.6+46×1.271 =（） 5、小张有2元和5元的人民币共34张，总值110元，问2元的人民币有（）张；5元的人民币有（）张。 6、在爷爷是父亲现在的年龄时候，父亲才12岁。等父亲到爷爷现在这么大的年龄时，爷爷84岁。爷爷现在（）岁；父亲现在（）岁。 7、幼儿园的老师给小朋友发苹果，每位小朋友4个，就多出12个，每个小朋友6个，就少12个，共有苹果( )个. 8、小英4次语文测验的平均成绩是89分，第5次测验得了94分。她5次测验的平均成绩是( )分。 9、用5、5、5、1四个数字组成一个算式，使其结果为24。算式是（）。 10、已知三个连续偶数的和比其中最大的一个偶数的2倍还多2，这三个偶数分别是（）、（）、（）。 11、在1、2、3……499、500中，数字2在一共出现了（）次。 12、一座铁路桥全长1200米，一列火车开过大桥需要75秒，火车开过路旁的电线杆只需15秒，那么火车全长是（）米。 13、蜗牛从一个枯井网上爬，白天向上爬110厘米，夜里向下滑40厘米，若要第五天的白天爬到井口，这口井至少深（）厘米。 14、王飞以每小时40千米的速度行了240千米，按原路返回时每小时行60千米，王飞往返 22、在任三个自然数中，其中必然有两个的数和为（）数。 中考数学 投影与三视图练习题 1填空题 （1） 俯视图为圆的几何体是 ________ ， _______ 。 （2） 画视图时，看得见的轮廓线通常画成 _________ ， 看不见的部分通常画成 __________ 。 （3） 举两个左视图是三角形的物体例子： __________ ， ________ 。 （4） 如图所示是一个立体图形的三视图，请根据视图说出立体图形的名称 请将六棱柱的三视图名称填在相应的横线上 （6）、如图所示是一个立体图形的三视图，请根据视图说出立体图形的名称 （7）、一张桌子摆放若干碟子，从三个方向上看，三种视图如下图所示，则这 张桌子上共有 ________________ 碟子。 主视EJ （8） 、某几何体的三种视图分别如下图所示，那么这个几何体可能是 __________ 。 （9） 人在观察目标时，从眼睛到目标的 _______ 叫做视线。 _____ 所在的位置叫做 视点，有公共 _____ 的两条 ________________ 所成的角叫做视角。 视线不能到达的区域叫做 __________ 。 （10） 物体在光线的照射下，在某个 ______ 内形成的影子叫做 _________ ，这时光 线叫做 _____ ，投影所在的 ___________ 叫做投影面。 由 _________ 的投射线所形成的投影叫做平行投影。 由 的投射线所形成的投影叫做中心投影。 （11） 在平行投影中，如果投射线 ______ 垂直于投影面，那么这种投影就称为正 投影。 （12） 物体的三视图是物体在三个不同方向的 __________________ 。 i 厂 — 王视S 左 视 團 测试二 一、判断题（每小题1分，本题共10分。选A表示正确，选B表示错误。按要求答 题否则，不给分） 1.同一软盘上不允许出现同名文件。 2.对计算机RAM中的信息进行读、写操作时，主机必须通电。 3.半导体存储器ROM是一种易失性存储器件，电源关掉后，存储在其中的信息便 会丢失。 4.PrintScreen和Alt+PrintScreen的功能分别是把当前窗口容和当前桌面容送到剪贴 板上。 5.在windows98中，删除桌面的快捷方式，它所指向的项目也同时被删除。 6.在Windows98环境下，系统工具中的磁盘扫描程序主要用于清理磁盘，把不需要 的垃圾文件从磁盘中删掉。 7.在word中，“自动图文集”具有在录入过程中对某些缩写文字进行扩展的功能。 8.在excel的数据清单中，“数据”菜单的“记录单”命令可以方便地插入、删除或 修改记录数据及公式字段的值。 9.在powerpoint的幻灯片视图，在任一时刻，主窗口只能查看或编辑一幻灯片。 10.域名地址https://www.doczj.com/doc/0b18316347.html,中，www称为主机名。 二、单选题（每小题1分，本题共30分） 1.计算机系统加电时，应先给____________加电，后给主机加电。 A.主机 B.显示器 C.打印机 D 外部设备 2.ASCII是__________。 A 条件码 B 二~十进制编码 C 二进制码 D 美国信息交换标准代码 3.计算机所有的信息都是以______数码形式表示的。 A.八进制 B.十六进制 C.十进制 D.二进制 4.按存取速度来划分，下列哪类存储器的速度最快_____。 A.主()存储器 B. 硬盘 C.Cache D.优盘 5.计算机性能指标包括多项，下列项目中____________不属于性能指标。 A．主频 B.字长 C. 运算速度 D .是否带光驱 6.计算机硬件系统由_______组成。 A.控制器、显示器、打印机、主机和键盘 B.控制器、运算器、存储器、输入输出设备 C.CPU、主机、显示器、硬盘、电源 D.主机箱、集成块、显示器、电源 7. CAM是计算机应用领域中的一种，其含义是________。 A.计算机辅助设计 B.计算机辅助制造 C.计算机辅助教学 D.计算机辅助测试 8. 在windows中，实现窗口移动的操作是______。 A.将鼠标指针指向菜单栏，拖动鼠标 B.将鼠标指针指向边框，拖动鼠标 C.将鼠标指针指向标题栏，拖动鼠标 D. 将鼠标指针指向任何位置，拖动鼠标 9. 在windows环境下，使用资源管理器中的“复制软盘”可进行软盘整盘复制，对该命令的用法说明中，应选择_______。 A.需先对新目标盘进行格式化命令操作，才能使用 B.应先删除目标盘中所有容，否则有可能复制不下全部文件 C.应先检查是否有两个软盘驱动器 D.以上3种说法都不正确 10. 要在windows中修改日期或时间，则应运行____程序的“日期或时间”选项。 ----- 初三数学知识整理与重点难点总结 第21 章二次根式 知识框图 理解并掌握下列结论： （1）是非负数；（2）；（3）； I.二次根式的定义和概念： 1、定义：一般地，形如√ā（a≥0）的代数式叫做二次根式。当a＞0 时，√ a 表示 a 的算数平方根 ,√0=0 2、概念：式子√ā（a≥0）叫二次根式。√ā（a≥ 0）是一个非负数。 II.二次根式√ā 的简单性质和几何意义 1）a≥0; √ā≥0[双重非负性] 2）（√ā）^2=a（a≥0）[任何一个非负数都可以写成一个数的平方的形式] 3)√(a^2+b^2) 表示平面间两点之间的距离，即勾股定理推论。 IV. 二次根式的乘法和除法 1运算法则 √a·√ b= √ab（ a≥ 0，b≥0） -1- ---- √a/b= √a/√ b（a≥ 0，b>0 ） 二数二次根之积，等于二数之积的二次根。 2共轭因式 如果两个含有根式的代数式的积不再含有根式，那么这两个代数式叫做共轭因式，也称互为有理化根式。 V.二次根式的加法和减法 1同类二次根式 一般地，把几个二次根式化为最简二次根式后，如果它们的被开方数相同，就把这几个二次根式叫做同类二次根式。 2合并同类二次根式 把几个同类二次根式合并为一个二次根式就叫做合并同类二次根式。 3二次根式加减时，可以先将二次根式化为最简二次根式，再将被开方数相同的进行合并Ⅵ.二次根式的混合运算 1确定运算顺序 2灵活运用运算定律 3正确使用乘法公式 4大多数分母有理化要及时 5在有些简便运算中也许可以约分，不要盲目有理化 VII.分母有理化 分母有理化有两种方法 I.分母是单项式 如:√a/ √b= √a×√ b/√b×√ b=√ ab/b 思考题训练 第一组： 1、已知A≠0,且A×5/3＝B×9/10=C÷3/4=D×4/5＝E÷6/5,把A、B、C、D、E从小到大排列起来。 2、甲数的5/6等于乙数的2/3,甲乙两个数的和是90,求甲乙各是多少？ 3、实验小学六年级一班女生人数的3/4等于男生人数的2/3,男生比女生多3人,男生有多少人？ 4、甲、乙两个仓库，甲仓存粮的75%是乙仓存粮的2/3，甲仓的存粮比乙仓少40吨。甲、乙两个仓库各有粮食多少吨？ 5、甲乙丙三人合买一台电视机，甲付钱数的50%等于乙付钱数的1/3,等于丙付钱数的3/7,已知丙比甲多120元，这台电视机多少？ 6、有两支蜡烛，当第一支燃去4/5，第二支燃去2/3，它们剩下的部分一样长。第一支与第二支蜡烛原来长度的比是多少？ 7、甲、乙、丙三根木棒插入同一个水池中，三根木棒的长度和为270厘米，甲有3/4在水面外，乙有7/10在水面外，丙有2/5在水面外，求水有多深? 第二组： 1、六（2）班参加书法兴趣小组的同学，男女生人数比是4：3。后来，有一名女生不参加了，这是，男女生人数比是3：2。你知道，六（2）班原来参加书法兴趣小组的同学中，男女生各有多少人吗？ 2、学校田径组原来女生人数占1/3，后来又有6名女生参加进来，这样女生就占田经组总人数的4/9。现在田径组有女生多少人？ 3、有一群羊正在山坡吃草。其中白羊占45﹪，这时又来16只黑羊后，白羊就只占25﹪，问：这群羊中白羊有多少只？ 4、六（1）班有女生24人，占全班人数的4/9。今年转出了若干名女生，这是女生占全班人数的2/5。求今年转出多少名女生？ 5、某班女生人数是男生人数的4/5,后来又转进1名女生，这时女生是男生的5/6,现在班上有多少女生？ 6、学校棋类组原来女生人数占3/7，后来又有30名女生参加进来，这样女生就占棋类总人数的2/3。现在棋类组有女生多少人？ 7、育才小学，男教师占全体教师总数的1/3，如果调出4个女教师，调进4个男教师，则男女教师人数之比是4：5，育才小学有教师多少人？ 第三组： 1、生产一批零件，甲独做要6小时，乙每小时可以做36个。现在甲、乙两人合做，完成任务时，甲、乙两人生产零件数量的比是5︰3。这批零件一共有多少个。 2、修路队修一条公路，如果由甲队单独修要15天，而乙队每天可以修44米，当两队共同修完这段公路时，甲队修了全长的60%，这段公路全长多少米？ 3、两辆车同时从甲乙两地相对开出，客车行完全路程要10小时，货车每小时行40千米，当两车相遇时，客车行了全路程的3/5，甲乙两地相距多少千米？ 第四组： 1、甲、乙、丙三人共植树120棵,甲植的是乙丙的1/2,乙植的是甲丙的1/3,丙植多少棵数? 2、甲乙丙三人共同加工零件180 初三数学下册重要知识点总结 第25章概率 1、必然事件、不可能事件、随机事件的区别 2、概率 注意：（1）概率是随机事件发生的可能性的大小的数量反映. （2）概率是事件在大量重复试验中频率逐渐稳定到的值，即可以用大量重复试验中事件发生的频率去估计得到事件发生的概率，但二者不能简单地等同. 3、求概率的方法 （1）用列举法求概率（列表法、画树形图法） （2）用频率估计概率：一方面，可用大量重复试验中事件发生频率来估计事件发生的概率.另一方面,大量重复试验中事件发生的频率稳定在某个常数(事件发生的概率)附近，说明概率是个定值,而频率随不同试验次数而有所不同,是概率的近似值,二者不能简单地等同. 第26章二次函数 1. 二次函数的一般形式：y=ax2+bx+c.(a≠0) 4．求二次函数的解析式：已知二次函数图象上三点的坐标，可设解析式y=ax2+bx+c，并把这三点的坐标代入，解关于a、b、c的三元一次方程组，求出a、b、c的值, 从而求出解析式-------待定系数法. 5．二次函数的顶点式： y=a(x-h)2+k (a≠0)；由顶点式可直接得出二次函数的顶点坐标（h, k），对称轴方程 x=h 和函数的最值 y最值= k. 6．求二次函数的解析式：已知二次函数的顶点坐标（h,k）和图象上的另一点的坐标，可设解析式为y=a(x -h)2+ k，再代入另一点的坐标求a，从而求出解析式. 8. 二次函数y=ax2+bx+c (a≠0)的图象及几个重要点的公式： 9. 二次函数y=ax2+bx+c (a≠0)中，a、b、c与Δ的符号与图象的关系： (1) a＞0 <=> 抛物线开口向上； a＜0 <=> 抛物线开口向下； (2) c＞0 <=> 抛物线从原点上方通过； c=0 <=> 抛物线从原点通过； c＜0 <=> 抛物线从原点下方通过； (3) a, b异号 <=> 对称轴在y轴的右侧； a, b同号 <=> 对称轴在y轴的左侧； b=0 <=> 对称轴是y轴； (4) b2－4ac＞0 <=> 抛物线与x轴有两个交点； b2－4ac =0 <=> 抛物线与x 轴有一个交点（即相切）； b2－4ac＜0 <=> 抛物线与x轴无交点.剑桥雅思4Test2passage2阅读答案解析
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