Missing question: what is the density of 53.4 wt% aqueous NaOH if 16.7 mL of the solution diluted to 2.00L gives 0.169 M NaOH?
Answer is: density is 1.52 g/mL.
c₁(NaOH) = ?; molarity of concentrated sodium hydroxide.
V₁(NaOH) = 16.7 mL; volume of concentrated sodium hydroxide.
c₂(NaOH) = 0.169 M; molarity of diluted sodium hydroxide.
V₂(NaOH) = 2.00 L · 1000 mL/L = 2000 mL; volume of diluted sodium hydroxide.
Use equation: c₁V₁ = c₂V₂.
c₁ = c₂V₂ / V₁.
c₁ = 0.169 M · 2000 mL / 16.7 mL.
c₁(NaOH) = 20.23 M.
m(NaOH) = 20.23 mol · 40 g/ml.
m(NaOH) = 809.53 g.
The mass fraction is the ratio of one substance (in this example sodium hydroxide) with mass to the mass of the total mixture (solution).
Make proportion: m(NaOH) : m(solution) = 53.4 g : 100 g.
m(solution) = 1516 g in one liter of solution.
d(solution) = 1516 g/L = 1.52 g/mL.
The equation to be used are:
PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles
The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.
PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa
n = 0.587 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.02024 mol
(101,183.9 Pa)V = (0.02024 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.931×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.931×10⁻⁴ m³ * 1000
V = 0.493 L
Answer:
1.09 L
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the volume in liters of a 0.360 mol/L barium acetate solution that contains 100 g of barium acetate. Be sure your answer has the correct number of significant digits.</em>
<em />
The molar mass of barium acetate is 255.43 g/mol. The moles corresponding to 100 grams are:
100 g × (1 mol/255.43 g) = 0.391 mol
0.391 moles of barium acetate are contained in an unknown volume of a 0.360 mol/L barium acetate solution. The volume is:
0.391 mol × (1 L/0.360 mol) = 1.09 L
Answer:
Answer below.
Explanation:
An exothermic reaction is a chemical reaction that releases energy through light or heat. It is the opposite of an endothermic reaction. Expressed in a chemical equation: reactants → products + energy.
Answer:
Balanced chemical equation:
AgCl + 2KCN → K[Ag(CN)₂] (aq) + KCl (aq)
Balanced chemical equation:
2Al + 6NaOH →2 Na₃AlO₃ + 3H₂
Explanation:
Chemical equation:
AgCl + KCN → K[Ag(CN)₂] (aq) + KCl (aq)
Balanced chemical equation:
AgCl + 2KCN → K[Ag(CN)₂] (aq) + KCl (aq)
Net ionic equation can not be written because all are present in aqueous form and no precipitation occur.
Chemical equation:
Al + NaOH → Na₃AlO₃ + H₂
Balanced chemical equation:
2Al + 6NaOH →2 Na₃AlO₃ + 3H₂
This is oxidation reduction reaction.
Hydrogen is reduced in this reaction. while aluminium is oxidized.
