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morpeh [17]
3 years ago
13

The chemical formula for the initial sample is mnso4·h2o and the chemical formula for the final sample is mnso4. use their molec

ular masses to calculate the theoretical percent mass of water in the initial sample. report your result to 3 significant figures,
e.g. 11.3%
Chemistry
1 answer:
sammy [17]3 years ago
4 0
The initial sample has a molecular formula of MnSO₄·H₂O. This molecule is a hydrate as it has a unit of water within its structure for every molecule of MnSO₄. This sample is being dehydrated to remove the water to give.

MnSO₄·H₂O → MnSO₄ + H₂O

MnSO₄·H₂O has a molecular mass of 169.02 g/mol. While MnSO₄ has a molecular mass of 151 g/mol. Water has a molecular mass of 18.02 g/mol. We now can use the ratio of the mass of water to the mass of the initial sample to determine the percentage of each component by mass.

% water by mass:

18.02/169.02 x 100% = 10.7% Water by mass.

% MnO₄ by mass:

151/169.02 x 100% = 89.3% MnSO₄ by mass.

Water makes up 10.7% of the initial mass of MnSO₄·H₂O.
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limiting reactant started with/limiting reactant needed=

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from reaction                 2 mol      1 mol

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Limiting reactant will be used completely.

So,  limiting reactant started with/limiting reactant needed=

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a cylinder rod formed from silicon is 46.0 cm long and has a mass of 3.00 kg. the density of silicon is 2.33 g/cm³. what is the
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d=\frac{m}{V}\\\\
V=\pi r^{2}H\\d=\frac{m}{\pi r^{2}H} \ \ \ |*\pi r^{2}H\\\\
d \pi r^{2}H=m \ \ \ |:d \pi H\\\\
r^{2}=\frac{m}{d \pi H}\\\\
r=\sqrt{\frac{m}{d \pi H}}\\\\
2r=d\\\\
d=2\sqrt{\frac{m}{d \pi H}}

H=43cm\\
d=2.33\frac{g}{cm^{3}}\\
m=3kg=3000g\\\\
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