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: 1. Two elements R and Q combine to form two binary compounds.
: In the first compound 14 g of R combines with 3 g of Q. In
: the second compound 7 g of R combine with 4.5 g of Q. show
: that these data are in accord with the law of multiple proportions.
: If the formula of the second compound is RD, what is the formula
: of the first compound.(10%)
(14/3):(7/4.5)=3:1
A ratio of small whole numbers.
The RD shall be RQ?
If it is, the first compoubd is R3Q.
------------------------------------------------------------------------
: 2. consider the following data for three binary compounds of
: hydrogen and nitrogen:
: %H(by mass) %N(by mass)
: I 17.75 82.25
: II 12.58 87.42
: III 2.34 97.66
: When 1.00 L of each gaseous compound is decomposed to its elements,
: the following volumes of H2(g) and N2(g)are obtained
: H2(L) N2(L)
: I 1.5 0.5
: II 2.0 1.0
: III 0.5 1.5
: Use these data to determine the molecular formulas of compounds,
: I,II,III and to determine the relative values for the atomic
: masses of hydrogen and nitrogen. (10%)
According to the volumes data, since under same condition, gas's volume
is in direct proportion to number of moles, compound I shall be NH3,
compound II shall be NH2 and compound 3 shall be N3H.
The relative values fore the atomic masses of hydrogen and nitrogen shall
be (17.75/1.5)/(82.25/0.5)=0.072
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: 3. A sample is a mixture of KCL and KBr. When 0.1024 g of the sample
: is dissolved in water and reacted with excess silver nitrate, 0.1889 g
: of solid is obtained. What is the composition by mass percent of the
: mixture. (10%)
K=39 Cl=35.5 Br=80 Ag=108
KCl=74.5 KBr=119 AgCl=143.5 AgBr=188
74.5x+119y=0.1024
143.5x+188y=0.1889
x=0.00105, y=0.00020
119y=0.0238
0.0238/0.1024=23.24%
1-23.24%=76.76%
The composition by mass percent of the mixture is KBr 23.24%, KCl 76.76%
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: 4. You made 100.0 ml of a lead(II) nitrate solution for lab but forget
: to cap it. The next lab session you noticed that there was only 80.0 ml
: left( the rest had evaporated). In addition, you forget the initial
: concentration of the solution. You decide to take 2.00 ml of the solution
: and add an excess of a concentrated sodium chloride solution. You obtain
: a solid with a mass of 3.407 g. What was the concentration of the original
: lead(II) nitrate solution. (10%)
Pb=207 Cl=35.5 PbCl2=278
3.407/278=0.0123
0.0123/0.002/(80/100)=7.69(M)
The concentration of the original lead(II) nitrate solution was 7.69 M.
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: 5. A compound Z is known to have a composition of 34.38% Ni, 28.13% C and
: 37.48% O. In an experiment 1.00l of gaseous Z is mixed with 1.00 L of argon
: , where each gas is at P=2.00 atm and T=25oC. When the mixture of gases
: in put in an effusion chamber, the ratio of Z molecules to Ar molecules
: in the effused mixture is 0.4837. Using these data, calculate the following:
: a. the empirical formula for Z
: b. the molar mass of Z
: c. the molecules formula for Z
: d. the mole ration of Z to Argon in a sample of gas obtained by five
: effusion steps(starting with the original mixture) (10%)
Ni=58.7 C=12 O=16
(34.38/58.7):(28.13/12):(37.48/16)=1:4:4
The empirical formula for Z shall be NiC4O4.
(1/0.4837)^2=4.27
Ar=40
40*4.27=170.8
The molar mass of Z shall be 170.8.
NiC4O4=170.7
170.8/170.7=1.00059
The molecules formula for Z shall be NiC4O4.
0.4837^5=0.0265
0.0265/(1+0.0265)=0.0258
the mole ration of Z to Argon in a sample of gas obtained by five
effusion steps(starting with the original mixture) shall be 0.0258.
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: 6. At 125oC, Kp=0.25atm2 for the reaction
: 2NaHCO3(s)== Na2CO3(s)+CO2(g)+H2O(g)
: A 1.00-L flask containing 10.0 g of NaHCO3 is evacuated and heated to
: 125oC
: a. calculate the partial pressures of CO2 and H2O after equilibrium is
: established.
: b. Calculate the masses of NaHCO3 and Na2CO3 present at equilibrium.
: c. Calculate the minimum container volume necessary for all the NaHCO3
: to decompose. (10%)
NaHCO3=84
10/84/2*(125+273)*0.082/1=1.94
Sqrt[0.25]=0.5<1.94
The partial pressures of CO2 and H2O after equilibrium is established shall
both be 0.5 atm.
0.5*1/0.082/(125+273)=0.01532
10-0.01532*84*2=7.426
Na2CO3=106
0.01532*106=1.624
The mass of NaHCO3 present at equilibrium shall be 7.426 g.
The mass of Na2CO3 present at equilibrium shall be 1.624 g.
1.94/0.5*1=3.88
The minimum container volume necessary for all the NaHCO3 to decompose shall
be 3.88 L.
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: 7. A certain acid ,HA , has a vapor density of 5.11 g/L when in the gas
: phase at a temperature of 25oC and a pressure of 1.00 atm. When 1.50 g
: of this acid is dissolved in enough water to make 100.0 ml of solution,
: the PH is found to be 1.80. Calculate Ka for HA. (10%)
1.5/5.11*1/0.082/(273+25)=0.012(mole)
Ka=(10^-1.8)^2/(0.012/0.1-10^-1.8)=2.41*10^-3
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: 8. Consider a weak acid , HA, with a Ka value of 1.6*10(-7)次方, Calculate
: the PH of a solution that is 5.0*10(-7) M HA and 5.0*10(-7) M NaA. (10%)
In this case, the Hydrogen ion that ionized from water can't be ignored.
(5.0*10^-7 + x)*(1.0*10^-7+x)/(5.0*10^-7-x)=1.6*10^-7
x=3.76*10^-8
Log10[1.0*10^-7+x]=-6.86
The PH of a solution that is 5.0*10(-7) M HA and 5.0*10(-7) M NaA shall be
6.86.
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: 9. A 50 ml ,0.5 M H2CO3 solution is titrated with 0.25M NaOH solution, find
: the PH values at all the inflection points along the titration curve.
: Ka1=4.3*10(-7), Ka2=5.6*10(-11) (10%)
When 100 ml NaOH solution is added, [H+]=Sqrt[Ka1*Ka2]=4.91*10^-9
pH=-Log10[H+]=8.31
when 200 ml NaOH solution is added,
[H+]=1/[OH-]=1/Sqrt[(10^-14/Ka2)*0.1]=2.417*10^-12
pH=-Log10[H+]=11.62
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: 10. Name each of the following compounds(10%)
: a. NaHCO3
sodium hydrogen carbonate
: b. SF6
sulfur hexafluoride
: c. PbO2
lead dioxide
: d. HClO
hypochlorous acid
: e. NH4NO2
ammonium nitrite
------------------------------------------------------------------------------
This is the end of the test paper, if there is any incorrect answers, please
reply on the board.
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