The number of moles of the magnesium (mg) is 0.00067 mol.
The number of moles of hydrogen gas is 0.0008 mol.
The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.
<h3>
Number of moles of the magnesium (mg)</h3>
The number of moles of the magnesium (mg) is calculated as follows;
number of moles = reacting mass / molar mass
molar mass of magnesium (mg) = 24 g/mol
number of moles = 0.016 g / 24 g/mol = 0.00067 mol.
<h3>Number of moles of hydrogen gas</h3>
PV = nRT
n = PV/RT
Apply Boyle's law to determine the change in volume.
P1V1 = P2V2
V2 = (P1V1)/P2
V2 = (101.39 x 146)/(116.54)
V2 = 127.02 mL
Now determine the number of moles using the following value of ideal constant.
R = 8.314 LkPa/mol.K
n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)
n = 0.0008
<h3>Volume of 1 mole of hydrogen gas at STP</h3>
V = nRT/P
V = (1 x 8.314 x 273) / (101.325)
V = 22.4 L
Learn more about number of moles here: brainly.com/question/13314627
#SPJ1
Answer:
ΔH = -470.4kJ
Explanation:
It is possible to sum 2 or more reactions to obtain the ΔH of the reaction you want to study (Hess's law). Using the reactions:
1. CaC2(s) + 2H2O(l) → C2H2(g) + Ca(OH)2(s)ΔH = −414kJ
2. 6C2H2(g) + 3CO2(g) + 4H2O(g) → 5CH2CHCO2H(g)ΔH = 132kJ
6 times the reaction 1.
6CaC2(s) + 12H2O(l) → 6C2H2(g) + 6Ca(OH)2(s)ΔH = −414kJ*6 = -2484kJ
This reaction + 2:
6CaC2(s) + 3CO2(g) + 16H2O(l) → + 6Ca(OH)2(s) + 5CH2CHCO2H(g) ΔH = -2484kJ + 132kJ = -2352kJ
As we want to calculate the net change enthalpy in the formation of just 1 mole of acrylic acid we need to divide this last reaction in 5:
6/5CaC2(s) + 3/5CO2(g) + 16/5H2O(l) → + 6/5Ca(OH)2(s) + CH2CHCO2H(g) ΔH = -2352kJ / 5
<h3>ΔH = -470.4kJ</h3>
Answer:
wla dyan ang sagut ehh baka mali yan
