Explanation:
The given data is as follows.
n = 2 mol, P = 1 atm, T = 300 K
Q = +34166 J, W= -1216 J (work done against surrounding)
= 
Relation between internal energy, work and heat is as follows.
Change in internal energy (
) = Q + W
= [34166 + (-1216)] J
= 32950 J
Also, 
= 
32950 J = 
1321.06 K + 300 K = 
= 1621.06 K
Thus, we can conclude that the final temperature of the gas is 1621.06 K.
Answer:
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
Explanation:
The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.
2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)
i mol 0.10 0.050 0.10
c mol -0.038 -0.038 +0019 +0.038
e mol 0.062 0.012 00.019 0.057
Since the volume of the vessel is 1.0 L, the concentrations in molarity are:
[NO] = 0.062 M
[H2] = 0.012 M
[N2] = 0.019 M
[H2O] = 0.057 M
<span>The ideal gas law.
PV=nRT
pressure x volume = moles x Faraday's constant x Temp Kelvin (C+273)
Original data
Pressure 1 atmosphere
Volume 1 liter
Temp 25C = 298K
New data
Volume 0.5 liter
pressure X
Temp 260C = 533K
P1v1T1 = P2v2T2
plug and chug.
(1)(1)(293) = (x)(0.5)(533)
Solve for X, which is the new pressure. </span>
228 grams
start with mass of Cr multiply by molar mass of Cr mole to mole ratio between Cr and Cr2O3 times molar mass of Cr2O3
The first blank is a compound, second is a mixture
