2.5 moles H20 and H20g)H28 H21-0040 M. Which of the following is true? 100 g of
ID: 592882 • Letter: 2
Question
2.5 moles H20 and H20g)H28 H21-0040 M. Which of the following is true? 100 g of C are placed in a 50-L container. At equilibrium for the reaction as- 50-L.container. At 15) A) no carbon is left B) [H20 0.010 M C)[C01 = 0.020 M D) [H201-0.020 M E) none of these 16) For the reaction: CH4(8)+2H20(g) CO2(g)42)+190 k] add N21g) at constant volume 16) A) the reaction reacts to the left. B) the temperature increases. C) the AH increases. D) the reaction reacts to the right. E) there is no change. 17) According to the collision theory in gaseous molecules, collision frequency is and reacti n . is because A) high, high, each collision results in a reaction B) low, high, molecules are moving so fast that each reaction causes many others C) low, low, molecules are so far apart D) high, relatively low, only a fraction of the collisions lead to a reaction E) low, low, molecules must collide before they can react 18) 18) For the reaction 2HgCl2+C2042-products, data are [HgCl2l 0.0836 00836 00418 M IC2042-1 0.202 0404 0404 M 0.53 M/hr Init, rate 0.26 1.04 The rate law is Rate-[HgChMC2042-ly.Thus A)x-1,y-2 B) x- 1,y-1 C)x-2,y-1 E) x=2,y-2 19) 19) What is the rate law for the following reaction and its mechanism? (overall reaction) 203-302 O+03-202 (F A) Rate- kKO3Mo2] B) Rate O3P C) Rate kKO3] D) Rate HO3P1021 E) Rate = kfollo1 A-4Explanation / Answer
Q15
n = 2.5 mol of H2O,
mol of C = mass/MW = 100/12 = 8.33 mol of C
in equilbirium
mol of H2 formed = M*V = 0.04*50 = 2 mol of H2 formed
therefore
C + H2O = CO + H2
if H2 formed, then C = 8.33-2 = 6.33 and H2O = 2.5-2 = 0.5
a) is false, since C is left
b)
[H2O] = mol/V = 0.5/50 = 0.01 M, this is true
c)
false, incorrect stoichiometry
d)
false, incorrect stoichiometry
e) false, since b is correct
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