Answer:
The answer is explained below:
Explanation:
Given the chemical equation:
N2H4(g)+H2(g)→2NH3(g)
The standard enthalpy of formation is given by the following formula:
ΔH^0 rxn = ∑ B reactants - ∑ product
-187.78 kJ/mol = [( 1x B N-N) + (1 x B N-H ) + (1 x b H-H)] - [ 6 x B N-H]
-187.78 kJ/mol = [( 1x B N-N) +(4 x 391 kJ/mol) + (1 x 436 kJ/mol)] - [ 6x 391 kJ/mol ]
-187.78 kJ/mol = B N-N + (1564 + 436 - 2346) kJ/mol
B = 158.22 kJ/mol
So, in this case the enthalpy of N-N bond is 158 kJ/mol
Answer:
Atomic number of this isotope = 77
Explanation:
Given that,
Mass number = 193
No of neutrons = 116
We need to find the atomic no of this isotope.
We know that,
Atomic mass = No of protons + No. of neutrons
Also, atomic no = no of protons
So,
Atomic mass = atomic no + No. of neutrons
⇒ Atomic no = Atomic mass - no of neutrons
Atomic no = 193 - 116
Atomic no = 77
Hence, 77 is the atomic no of the isotope.
It would be C
2 kg x 1000 g/kg x 1mol/18.02 x 6.03 kj/mol = 669kj
Explanation:
2H2 + O2 = 2H2O
2mol. 1mol. 2mol
2mol reacts with 1mol
13mol reacts with x
x=<u>13mol</u><u> </u><u>×</u><u> </u><u>1mol</u>
<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>2mol</u>
x= <u>13mol</u>
<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>2mol
x= 6.5mol of oxygen
Answer:1
Explanation:cause castle told me
