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lubasha [3.4K]
3 years ago
13

Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O

Chemistry
1 answer:
pashok25 [27]3 years ago
3 0

Answer:

HgSO₄

Explanation:

% => g => moles => ratio => reduce => empirical ratio

%Hg = 67.6% => 67.6g/201g/mol = 0.34mol

%S    = 10.8% => 10.8g/32g/mol = 0.34mol

%O   = 21.6% => 21.6g/16g/mol = 1.35mol

Hg:S:O => 0.34:0.34:1.35

Reduce to whole number ratio by dividing by the smaller mole value...

Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4

∴ Empirical Formula is HgSO₄

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Answer:

The section of the bar is 2.92 inches.

Explanation:

Mass of the steel cut ,m = 1.00 kg = 1000 g

Volume of the steel bar = V = Area × height

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Cross sectional area of the steel mass = A

A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2

V = 2.71 inches^2\times h

Density of the steel = d =7.70 g/cm^3

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\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}

h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}

h = 2.92 inches

The section of the bar is 2.92 inches.

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