When NaCl dissolves in water
1 answer:
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Answer:
Kp = 1.41 x 10⁻⁶
Explanation:
We have the chemical equation:
2 A(g) + 3 B(g)⇌ C(g)
In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):
dn= (sum moles products - sum moles reactants)
= (moles C - (moles A + moles B))
= (1 - (2+3))
= 1 - 5
= -4
We have also the following data:
Kc = 63.2
T= 81∘C + 273 = 354 K
R = 0.082 L.atm/K.mol (it is a constant)
Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:
= (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶
Detonation of nitroglycerin:
<span>Mass nitroglycerin = 2.00 mL x 1.592 g/mL = 3.184 g
</span>
Moles = mass / molar mass = 3.184<span> g/ 227.0872 g/mol = 0.01402
</span>
the ratio between nitroglycerin and Carbon dioxide is 4 : 12
So, moles CO2 = 0.01402 x 12 / 4 =0.0420
the ratio between nitroglycerin and N2 is 4 : 6
moles N2 = 0.01402 x 6 / 4 =0.0841
<span>the ratio between nitroglycerin and O2 is 4 : 1 </span>
moles O2 = 0.01402 x 1 / 4 = 0.0035
<span>the ratio between nitroglycerin and water is 4 : 1 </span>
<span>in the same way moles water = 0.005258 </span>
total moles = 0.0420 + 0.0841 + 0.0035 + 0.005258 = 0.130758
0.130758<span> x 55 = 5.78 L </span>
Mass N2 = 0.0841 mol x 28.0134 g/mol = 2.3548 g
<span>Mass nitroglycerin = 2.00 mL x 1.592 g/mL = 3.184 g
</span>
Moles = mass / molar mass = 3.184<span> g/ 227.0872 g/mol = 0.01402
</span>
the ratio between nitroglycerin and Carbon dioxide is 4 : 12
So, moles CO2 = 0.01402 x 12 / 4 =0.0420
the ratio between nitroglycerin and N2 is 4 : 6
moles N2 = 0.01402 x 6 / 4 =0.0841
<span>the ratio between nitroglycerin and O2 is 4 : 1 </span>
moles O2 = 0.01402 x 1 / 4 = 0.0035
<span>the ratio between nitroglycerin and water is 4 : 1 </span>
<span>in the same way moles water = 0.005258 </span>
total moles = 0.0420 + 0.0841 + 0.0035 + 0.005258 = 0.130758
0.130758<span> x 55 = 5.78 L </span>
Mass N2 = 0.0841 mol x 28.0134 g/mol = 2.3548 g