Answer:
Mass released = 8.6 g
Explanation:
Provided information:
Starting moles of nitrogen= 0.950 mol
Starting volume = 25.5 L
Final nitrogen mass released = ?
Final volume = 17.3 L
Calculation:
Equation:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Starting mass of nitrogen:
Mass = moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Ending mass of nitrogen:
Mass = moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - ending mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
Response:The ethanol percentage is 0.1093%
Explanation:
As given:
t = time = 10 s
I = current = 320 mA
F = Faraday's constant = 96485.3365 C mol⁻¹
n = number of electrons transferred = 4
Molecular weight of ethanol is 46 g/mol
Question: What is the percent (by volume) of ethanol in the driver's breath, %E =?
First, calculate the ethanol mass:
The moles of ethanol:
Applying the ideal gas law formula:
Here:
T = 26°C = 299 K
P = 1 atm
Substituting in the values:
The percentage of ethanol:
%
Solution:
The molecular formula is PbSO₄, indicating lead sulfate
Option c.
Explanation:
The percentage makeup shows that in 100 g of this compound, there are:
68.3 g of Pb, 10.6 g of S, and (100 - 68.3 - 10.6) = 21.1 g of O
To find the moles of each element, we divide by their molar masses:
68.3 g Pb / 207.2 g/mol = 0.329 moles Pb
10.6 g S / 32.06 g/mol = 0.331 moles S
21.1 g O / 16 g/mol = 1.32 moles O
Next, we find the mole ratio by dividing each by the smallest number of moles:
0.329 / 0.329 = 1 Pb
0.331 / 0.329 = 1 S
1.32 / 0.329 = 4 O
Thus, the molecular formula is PbSO₄, representing lead sulfate.
