%yield = 88.5%
<h3>Further explanation</h3>
Given
Reaction
Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
Required
The percent yield
Solution
mol AgNO₃(MW=169,87 g/mol) :
= mass : MW
= 127 : 169.87
= 0.748
mol Ag from equation :
= 2/2 x mol AgNO₃
= 2/2 x 0.748
= 0.748
Mass Ag (theoretical) :
= mol x Ar Ag
= 0.748 x 108
= 80.784
% yield = (actual/theoretical) x 100%
%yield = 71.5/80.784 x 100%
<em>%yield = 88.5%</em>
Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)
v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol
n(Na₂S)=n{Cd(NO₃)₂}=cv
m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv
m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
Answer:
I think mixture
Explanation:
Mixture because it is a mixture of water and carbon
Answer:
1.428 moles
Explanation:
If 0.0714 moles of N2 gas occupies 1.25 L space,
how many moles of N2 have a volume of 25.0 L?
Assume temperature and pressure stayed constant.
we experience it 0.0714 moles: 1.25L space
x moles : 25L of space
to get the x moles, cross multiply
(0.0714 x 25)/1.25
1.785/1.25 = 1.428 moles
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
