3.456 x 10^2 is the answer you're looking for.
Hi!
• For this you want to convert moles to grams
• To do this you would simply multiple your moles times the atomic mass of potassium
3.25 x 39.098= 127.0685
• So your answer for this should be about 127.07 depending on how you need to round!
The answer is 2300 mg which is equal to one teaspoon of salt
Answer:
The enthalpy of the nitrogen-nitrogen bond in N2H4 is 162.6 kJ
Explanation:
For the reaction: N2H4(g)+H2(g)→2NH3(g), the enthalpy change of reaction is
ΔH rxn = 2 ΔHºf NH3 - ΔHºf N2H4
but we also know that the ΔH rxn is calculated by accounting the sum of number of bonds formed and bonds broken as follows:
ΔH rxn = 6H (N-H) + 4 (N-H) + 2H (H-H)
where H is the bond enthalpy .When bonds are broken H is positive, and negative when formed, in the product there are 6 N-H bonds , and in the reactants 4 N-H and 1 H-H bonds).
Consulting an appropiate reference handbook or table the following values are used:
ΔHºf (NH3) = -46 kJ/mol
ΔHºf (N2H4) = 95.94 kJ/mol
(The enthalpy of fomation of hydrogen in its standard state is zero)
H (N-H) = 391 kJ
H (H-H) = 432 kJ
H (N-N) = ?
So plugging our values:
ΔH rxn = 2mol ( -46.0 kJ/mol) - 1mol(95.4 kJ/mol) = -187.40 kJ
-187.40 kJ = 6(-391 kJ) + 4 (391 kJ) + 432 + H(N-N)
-187.40 kJ = -350 kJ + H(N-N)
H(N-N) = 162.6 kJ
