Not sure what is wrong with the second half Lab DataVerify your calculation. In your set-up, do your units cancel?Did you report your data to the correct number of significant figures?0.015Mass of magnesium (g)Moles of magnesium (mol)0.0006223.0Temperature of water (°C)296.15Temperature of water (K)21.1Vapor pressure of water (mmHg)Barometric pressure (mmHg)781.0Observationsthe metal causes the water to bubbleVolume of hydrogen gas collected (mL)15.0Volume of hydrogen gas collected (L)0.015Pressure of hydrogen gas (mmHg)742.5Pressure of hydrogen gas (atm)58.9Moles of hydrogen gas produced (mol)0.413Ideal gas constant (L-atm-mol"K)38.5PType here to search

# 2nd half: The calculation for volume is right. V = 0.015 L# Pressure of hydrogen gas, P = Pbarometric pressure - Pwater vapor = (781.0 - 21.1) mmHg = 759.9 mmHg Pressure of hydrogen gas (mmHg), P = 759.9 (Answer) # Conversion factor: 1 atm = 760 mmHg Pressure of hydrogen gas(atm) = 759.9 mmHg * 1 atm760 mmHg= 0.99987 or 0.9999 atm Pressure of hydrogen gas(atm) = 0.99987 or 0.9999 (Answer) # Calculation for moles of hydrogen gas: Balanced equation: Mg(s) + 2HCl(aq) ----> MgCl2(aq) + H2(g) mole ratio between Mg(s) and H2(g) is 1/1 Hence moles of hydrogen gas produced = moles of magnesium = 0.00062 mol Hence moles of hydrogen gas produced(mol) = 0.00062 (Answer) # Calculation for ideal gas constant: P = 0.9999 atm V = 0.015 L T = 296.15 K n = 0.00062 mol Using ideal gas equation: PV = nRT⇒R = PVnT⇒R = 0.99987 atm * 0.015 L0.00062 mol * 296.15 K ⇒R = 0.0817 or 0.082 L.tm.mol-1.K-1 Ideal gas constant (L.atm.mol-1.K-1) = 0.082 (Answer)