Data:
m (<span>Sample Mass) = ?
n (</span><span>Number of moles) = 0.714 mol
MM (Molar Mass) of </span>Mercury (I) Chloride (

)
Hg = 2*200.59 = 401.18 amu
Cl = 2*35.453 = 70.906 amu
----------------------------------------
Molar Mass

= 401.18 + 70.906 = 472.086 ≈ 472.09<span> amu or 472.09 g/mol
</span>
Formula:

Solving:



Answer:
By approximation would be letter
D) <span>
337.2 g</span>
Answer:
Chlorophyll is one. I can't quite remember other
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<span />
Answer:
0.0000098 should be the answer
Explanation:
Answer:
c 43.38 g
Explanation:
The reaction between MnO2 and HCl can be represented by the following balanced equation:
MnO2 + HCl ---> Cl2 + MnCl2 + H2O
From the balanced equation, the theoretically required molar ratio of MnO2 to HCl is 1:1, therefore the yields would have been expected to be equal.
For the fact that HCl gives a higher yield (65.02g) than MnO2 (60.25g) according to the problem statement, HCl should be in excess, while the limiting reagent should be MnO2 .
Thus, the theoretical yield of Cl2 will be 60.25 g.
By definition, the percentage yield is given by
% Yield = (Actual Yield) / (Theoretical Yield),
This can be simplified to
Actual Yield = % Yield * Theoretical Yield
Plugging in the given values we have
Actual Yield = 72% * 60.25 = 43.38 g