The following reaction is investigated 2N_2O(g) + N_2H_4(g) 3N_2(g) + 2H_2O(g) I
ID: 529688 • Letter: T
Question
The following reaction is investigated 2N_2O(g) + N_2H_4(g) 3N_2(g) + 2H_2O(g) Initially there are 0.100 mol of N_0O and 0.25 mol of N_2H_4, in a 10.0-L container, If there are 0.059 mol of N_2O at equilibrium, how many moles of N_2 are present at equilibrium? A) 4.10 times 10^-2 B0 1.2 times 10^-1 C) 6.20 times 10^-2 D) 2.1 times 10^-2 E) none of the above A solution containing 10.0 g of an unknown liquid and 90.0 water has a freezing point if -3.33 Given K_f = 1.86 C/m for water, the molar mass of the unknown liquid is g/mol. A) 69.0 B) 333 C) 619 D) 161 E)62.1 The enthalpy of vaporization (Delta H degree _vap) of benzene is 30.7 kJ/mol at its normal boiling point of 353.3 K. What is Delta S degree _vap this temperature? a. 86.9 J/(mol middot K) b. 0.087 J/(mol middot K) c. 11.5 J/(mol middot K) d. 0.0115 J/(mol middot K) e. 383 J/(mol middot K) Determine Delta G degree for the following reaction: CH_4(g) + 2O_2(g) rightarrow CO_2(g) + 2H_2O(I) CH_4(g) Delta G degree_f(kJ/mol) CH_2(g) -50.65 O_0(g) 0 CO_2(g) -394.4 H_2O(I) -237.4 A. -581.2 kJ B. -818.6kJ V. 131.1 kJ D. -682.5 kJ E. -919 A strip of iron is placed in a 1 M solution of iron(II) sulfate, and a strip of copper is solution of copper(II) chloride. The two solutions arc connected with a salt metals are connected by a wire. Reduction Half-Reaction E degree (V) Fe^2+(aq) + 2^- Fe(s) -0.41 Cu^2+(aq) + 2e^- Cu(s) 0.34 Which of the following takes place? A) Sulfur deposits at the iron electrode. B) The Fe(II) concentration of the iron half-cell decreases. C) Copper atoms deposit at the cathode. D) Chlorine is produced at the copper electrode.Explanation / Answer
2 N2O (g) + N2H4 (g) <-----> 3 N2 (g) + 3 H2O (g)
Given
Inital
No. of moles of N2O = 0.1 moles
No. of moles of N2H4 = 0.25 moles
Volume = 10 L
[N2O] = 0.1 moles / 10 L = 0.01 mol/L
[N2H4] = 0.25 mol / 10 L = 0.025 mol/L
N2O N2H4 N2 H2O
Initial 0.1 0.25 - -
Converted - 2x - x 3x 3x
Equilibrium 0.1-2x 0.25-x 3x 3x
Given
[N2O]eq = 0.1 -2x = 0.059 moles
2x = 0.041
x = 0.0205 moles
[N2]eq = 3x = 3 * 0.0205 = 0.0615 = 6.2 * 10-2 Answer (c)
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