9) A 1,000-L fask is charged with 920-10-3 mot of N204 Al equilibrium 5.98-10 3
ID: 593055 • Letter: 9
Question
9) A 1,000-L fask is charged with 920-10-3 mot of N204 Al equilibrium 5.98-10 3 mol of NyOy remains Keq for this reaction ls_ A) 0212 B) 0197 C694-10-3 D) 0.183 10) What volume of 0 716 M KBr solution is needed to provide 13.0g of KBr? D) 655 ml A) 931 mL B) 185 mL C) 153 mL 11) At high temperatures boron carbide vaporizes according to the equation Which equation describes the relationship between AG and &G; for this reaction? A) AG AG.RT in (ps ICVB4CD 12) G-AG" for a reaction B) at STP D) at the start of the reaction Ciro-1 3) it AG is negative for a reaction C) K is between O and 1 14) Which solution will be the most basic? A) 0 10M CH30H B) 0.10MH20 C) 0.10M Ba(OH)2 D) 0 10 M KOH E) All solutions have equal basicity 15) What is the pH of a solution prepared by diluting 25.00 ml of 0 10M HCI with enough water to produce a total volume of 100.00 mL? A) 1.00 B) 1.60 C)3.20 D) 200 6) What is the pH of a 0 020 M Ba(OH02 solution? A) 12.60 B) 12.30 C)1.70 D) 1.40Explanation / Answer
Q9
initially
[N2O4] = 9.2*10^-3
[NO2] = 0
in equilbirium
[N2O4] = 9.2*10^-3 - x
[NO2] = 0 + 2x
and we know
[N2O4] = 9.2*10^-3 - x = 5.98*10^-3
x = 9.2*10^-3- 5.98*10^-3 = 0.00322
[NO2] = 0 + 2x = 2*0.00322 = 0.00644
Keq = [NO2]^2/[N2O4]
Keq = (0.00644^2)/(5.98*10^-3) = 0.0069353
K = 6.9*10^-3
nearest answer is C
Q10
M = mol/V
V = mol/M
V = (mass/MW)/M
V = (13/119.002 )/0.716)
V = 0.152572 L
V = 152.5 mL
chooes c
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