Question

lab data
moles of magnesium (mol) 0.00058
verify your calculation. in your set - up, do your units cancel?
did you report your data to the correct number of significant figures?
vapor pressure of water (mmhg) 18.6
barometric pressure (mmhg) 771.0
observations
it have bubbles
volume of hydrogen gas collected (ml) 14.00
volume of hydrogen gas collected (l)
pressure of hydrogen gas (mmhg) 752.4
pressure of hydrogen gas (atm)
moles of hydrogen gas produced (mol)
ideal gas constant (l·atm·mol⁻¹·k⁻¹)

Explanation:

Step1: Convert volume from mL to L

Since \(1\ L = 1000\ mL\), for \(V = 14.00\ mL\), we have \(V=\frac{14.00}{1000}\ L=0.01400\ L\) (units: \(mL\times\frac{L}{mL}=L\)).

Step2: Convert pressure from mmHg to atm

We know that \(1\ atm = 760\ mmHg\). For \(P = 752.4\ mmHg\), \(P=\frac{752.4}{760}\ atm\approx0.99\ atm\) (units: \(mmHg\times\frac{atm}{mmHg}=atm\)).

Step3: Assume room - temperature \(T=(25 + 273.15)K=298.15\ K\)

Using the ideal gas law \(PV=nRT\), we can solve for \(n\) (moles of gas). Rearranging for \(n\), we get \(n=\frac{PV}{RT}\).
Substitute \(P = 0.99\ atm\), \(V=0.01400\ L\), \(R = 0.0821\ L\cdot atm\cdot mol^{-1}\cdot K^{-1}\), \(T = 298.15\ K\)
\(n=\frac{0.99\times0.01400}{0.0821\times298.15}\ mol\)
First calculate the numerator: \(0.99\times0.01400 = 0.01386\)
Then calculate the denominator: \(0.0821\times298.15\approx24.48\)
\(n=\frac{0.01386}{24.48}\ mol\approx5.66\times10^{-4}\ mol\) (units: \((atm\times L)\div(L\cdot atm\cdot mol^{-1}\cdot K^{-1}\times K)=mol\))

Answer:

• Volume of hydrogen gas collected (\(L\)): \(0.01400\ L\)
• Pressure of hydrogen gas (\(atm\)): \(0.99\ atm\)
• Moles of hydrogen gas produced (\(mol\)): \(5.66\times 10^{-4}\ mol\)
• Ideal gas constant (\(L\cdot atm\cdot mol^{-1}\cdot K^{-1}\)): \(0.0821\ L\cdot atm\cdot mol^{-1}\cdot K^{-1}\)