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
Well you are at a 86% so t<span>his is considered a "B" grade on an average grade scale. If you take 50 points we would need to know the Total point you could have got in that class to be able to see how much a percent was the assignment work and then take that percentage of the assignment and subtract it from the 86% to see where on the scale you would fall in after the assignment points were taken off. Hope this helps :)
</span>
The thermochemical equation is the chemical equation including the net change of enthalpy (heat).
The chemical equation for the decomposition of methanol to form methane and oxygen is:
2CH3OH --> 2CH4 + O2
The thermochemical equation is:
2CH3OH ---> 2CH4 + O2 - 252.8 kJ
Note that the heat is placed as negative at the right side because it is absorbed during the decomposition, so the environment will have 252.8 kJ less per each mole of O2 produced.
You can equivalently write:
2CH3OH + 252.8 kJ --> 2CH4 + O2
Ideal gases are hypothetical gases whose molecules occupy negligible space and have no interactions, and that consequently obeys the gas laws exactly.
Not exactly sure about the amount...
I hope this helps! :)
