0.5
Explanation:
Given parameters:
Mass of Ca²⁺ = 10g
unknown:
Equivalent weight = ?
Solution:
Equivalent weight that is the amount of electrons which a substance gains or loses per mole.
Ca²⁺ has +3 charge
It lost 2e⁻;
therefore;
In 1 mole of Ca²⁺, we have 2 equivalent weight
1 mol Ca²⁺ = 2eq. wts.
1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt.
Therefore;
10.0 g Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.
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Answer:
The partial pressure of helium gas in a gaseous mixture of helium and hydrogen is the pressure that the helium would exert in the absence of the hydrogen. equal to the total pressure divided by helium's molar mass. O equal to the total pressure divided by the number of helium atoms present.
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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
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