Answer:
The answer to your question is a) N₂ b) 3.04 g of NH₃
Explanation:
Data
mass of H₂ = 2.5 g
mass of N₂ = 2.5 g
molar mass H₂ = 2.02 g
molar mass of N₂ = 28.02 g
molar mass of NH₃ = 17.04 g
Balanced chemical reaction
3H₂ + 1 N₂ ⇒ 2NH₃
A)
Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22
Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1
Conclusion
The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.
B)
28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃
2.5 g of N₂ -------------------- x
x = (2.5 x 2 x 17.04) / 28.02
x = 85.2 / 28.02
x = 3.04 g of NH₃
Answer:
B. Equal to 7.
Explanation:
Hydrobromic acid is a strong acid that decreases pH and ammonia is a strong base that increases pH.
As the initial pH of water is 7,0 the addition of 35.0mL of 0.400M HBr will produce a pH less than 7,0. But, the same effect of decreasing pH is reverted for the addition of 35.0mL of 0.400M HNO3.
That means the net effect of the two addition is to have a pH:
B. Equal to 7.
I hope it helps!
First convert celcius to Kelvin.
20 + 273 = 293K
31 + 273 = 304K
Now we can set up an equation based on the information we have.
V1 = 5
P1 = 365
T1 = 293
V2 = 5
P1 = x
T2 = 304
The equation be: 
Now just solve.
1825/293 = 5x/304
Cross multiply.
554800 = 1465x
Divide both sides by 1465
x = 378.7030717 which can then be rounded to 378.7 mmHg
C. 28 KJ
AMU of H2O2 = 2(1) + 2(16) = 34 g/mol
10 g / 34 g/mol = 0.294 mol H2O2
0.294 mol / H = 2 mol / 190 KJ
H = 28.9 KJ
