Answer:
78.2 g/mol
Step-by-step explanation:
We can use the <em>Ideal Gas Law</em> to solve this problem:
pV = nRT
Since n = m/M, the equation becomes
pV = (m/M)RT Multiply each side by M
pVM = mRT Divide each side by pV
M = (mRT)/(pV)
Data:
ρ = 2.50 g/L
R = 0.082 16 L·atm·K⁻¹mol⁻¹
T =98 °C
p = 740 mmHg
Calculation:
(a)<em> Convert temperature to kelvins
</em>
T = (98 + 273.15) = 371.15 K
(b) <em>Convert pressure to atmospheres
</em>
p = 740 × 1/760 =0.9737 atm
(c) <em>Calculate the molar mass
</em>
Assume V = 1 L.
Then m = 2.50 g
M = (2.50 × 0.082 06 × 371.15)/(0.9737 × 1)
= 76.14/0.9737
= 78.2 g/mol
<span>Silver oxalate dissociation equation is following:
</span><span>
Ag</span>₂C₂O₄(s) ⇄ 2Ag⁺(aq) + C₂O₄²⁻(aq)
According to reaction follows next stoichiometric ratio:
[Ag⁺] : [C₂O₄²⁻] = 2 : 1
[C₂O₄²⁻] = [Ag⁺] / 2
[C₂O₄²⁻] = (1.7×10⁻⁴)/2 = 8.5×10⁻⁵ M
So, solubility product constants for silver oxalate is:
Ksp = [Ag⁺]² x [C₂O₄²⁻]
Ksp = [1.7×10⁻⁴]² x [8.5×10⁻⁵]
Ksp = 2.46×10⁻¹²
12 moles of water H₂O are produced from the combustion of pentane.
Explanation:
We have the following combustion of pentane (C₅H₁₂):
C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O
Knowing the chemical reaction we devise the following reasoning:
if 1 moles of pentane C₅H₁₂ produces 6 moles of water H₂O
then 2 moles of pentane C₅H₁₂ produces X moles of water H₂O
X = (2 × 6) / 1 = 12 moles of water H₂O
Learn more about:
combustion of organic compounds
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Answer:
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