Question #8453e

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Answer 1 · 2018-01-31T08:52:45

$= 530 m L$ $=530mL$

Write and balance the equation
$2 H g O (s) \to 2 H g (l) + O_{2} (g)$ $2HgO(s)->2Hg(l)+O_2(g)$
Find the molar mass of $H g O$ $HgO$ which relative atomic masses of the elements composing it are obtainable from the periodic table; i.e.,
$H g O = 216.6 g/mol$ $HgO=216.6" g/mol"$
Find the moles of $H g O$ $HgO$ , through molar conversion using the molar mass above as the conversion factor. Ensure that the units work out and the desired unit is attained.
$=7.8cancel(gHgO)xx(1molHgO)/(216.6cancel(gHgO))$
$=0.036molHgO$
Now, find the mole $O_2$ by referring to the balanced equation where the molar ratio of $HgO$ and $O_2$ is obtainable from; i.e.,
$=0.036cancel(molHgO)xx(1molO_2)/(2cancel(molHgO))$
$=0.018molO_2$
Then, find the volume of $O_2$ . Use the formula $PV=etaRT$ , but prior to it make sure variables are up to its standard units as shown;
$T="temperature"=150^oC+273=423K$
$P="pressure"=120.3kPa$
$R="gas constant"=(8.31446L*kPa)/(mol*K)$
$eta="number of moles"=0.018mol$
Plug in values to the formula and cancel units as required to obtain the desired unit. Rearrange it and isolate the required variable; the volume (V);
$PV=etaRT$
$V=(etaRT)/(P)$
$V=((0.018cancel(mol))((8.31446L*cancel(kPa))/cancel((mol*K)))(423cancel(K)))/(120.3cancel(kPa))$
$V=0.5264L~~0.53L~~530ml$