The concentration of the acid decreases
Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = <span>0.02 M
n(</span>(NH4)2CO3) = ?
For the purpose, here we will use the next equation:
c=n/V ⇒ n=cxV
n((NH4)2CO3) = 0.02M x 0.001L
n((NH4)2CO3) = 2x10⁻⁵ mole of (NH4)2CO3 is presented in the solution
Answer:
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Explanation:
Answer:
Explanation:
Hello,
In this case, since 454 g are equivalent to 1 pound and 1000 millilitres are equivalent to 1 liter, the required density is computed below by applying the corresponding conversion factor:
Regards.
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
