CH4 (g) + 2Cl2 (g) --> CH2Cl2 (g) + 2HCl (g) <span>ΔH = -205.4</span>
Answer:
Molar Mass = 171.6 g/mol
Explanation:
Here we can use the equation PV = nRT, where n could stand for m / M, considering M = Molar Mass. Now we can rearrange this equation to isolate M, as we have to determine the " molar mass " of this vapor,
PV = ( m / M )RT,
density = mass / volume = PM / RT -
M = dRT / P
First convert celcius to kelvins and torr to " standard atmosphere " before plugging in the known values,
13 C = 286 K,
741 torr = 0.975 atm -
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M = (7.125)(0.0821) (286) / 0.975
M = 171.6 g /mol
Answer:
5.00 and 2.50 moles of aqueous sodium and sulfide ions are formed.
Explanation:
The dissociation reaction of sodium sulfide is as follows.
![Na_{2}S(aq)\rightarrow 2Na^{+}(aq)+S^{2-}(aq)](https://tex.z-dn.net/?f=Na_%7B2%7DS%28aq%29%5Crightarrow%202Na%5E%7B%2B%7D%28aq%29%2BS%5E%7B2-%7D%28aq%29)
From the reaction one mole of sodium sulfide produce 2 moles sodium ions.
Let's calculate the moles of
ions.
![Moles\,of\,Na^{+}=2.50molNa_{2}S\times \frac{2mol\,Na^{+}}{1mol\,Na_{2}S}=5.00mol\,Na^{+}](https://tex.z-dn.net/?f=Moles%5C%2Cof%5C%2CNa%5E%7B%2B%7D%3D2.50molNa_%7B2%7DS%5Ctimes%20%5Cfrac%7B2mol%5C%2CNa%5E%7B%2B%7D%7D%7B1mol%5C%2CNa_%7B2%7DS%7D%3D5.00mol%5C%2CNa%5E%7B%2B%7D)
From the reaction one mole of sodium sulfide produce one mole of sulfide ions.
Let's calculate the moles of
ions.
![Moles\,of\,S^{2-}=2.50molNa_{2}S\times \frac{1mol\,S^{2-}}{1mol\,Na_{2}S}=2.50mol\,S^{2-}](https://tex.z-dn.net/?f=Moles%5C%2Cof%5C%2CS%5E%7B2-%7D%3D2.50molNa_%7B2%7DS%5Ctimes%20%5Cfrac%7B1mol%5C%2CS%5E%7B2-%7D%7D%7B1mol%5C%2CNa_%7B2%7DS%7D%3D2.50mol%5C%2CS%5E%7B2-%7D)
Atomic mass iron ( Fe ) = 55.84 a.m.u
55.84 g ------------ 6.02x10²³ atoms
24.0 g ------------- ??
24.0 x ( 6.02x10²³) / 55.84
=> 2.58x10²³ atoms