Answer:
m = 0.3249 g
Explanation:
First, I'm assuming you have a reaction of mercury(II) oxyde descomposition. If this is the case, then the equation to use is the following:
HgO ---------> Hg + O2
Balancing the equation:
2HgO ----------> 2Hg + O2
This means that 2 moles of HgO reacts to produce 1 mole of O2, so, we first calculate the moles of O2, then, the moles of HgO and finally the mass:
We have the volume of O2, the pressure and temperature, so let's use the ideal gas equation:
PV = nRT
Solving for n:
n = PV/RT
R: 0.082 L atm / K mole
T = 70 + 273 = 343 K
V = 83 / 1000 = 0.083 L
Calculating n:
n = 1 * 0.083 / 0.082 * 343
n = 0.003 moles
as stated before, 2 moles of HgO reacts with 1 mole of oxygen so:
2 moles HgO = 1 moles O2
moles HgO = moles O2 / 2
moles HgO = 0.003 / 2 = 0.0015 moles
Finally, to calculate the mass:
m = n * MM
the molar mass of HgO is 216.59 g/mol, so replacing:
m = 0.0015 * 216.59
m = 0.3249 g
Answer:
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Answer:
Explanation:the formula for osmotic pressure is given in the first line, all the terms are explained before solution, from the question,
Molar mass=Mm=386.6
Mass= m= 14.6
Volume= 263/1000 = 0.263dm^3
Temperature=T= 298K.
Gas constant .R= 0.082
Then substitute into the equation πV= nRT
Making π subject of formula
π= nRT/V
Then further substitution
The balanced chemical equation is P4+3O2--> 2P2O3