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4 years ago

Please help!!!

Chemistry

1 answer:

Firdavs [7]4 years ago

4 0

Answer:

Option B. HgCl2

Explanation:

Data obtained from the question include:

Mass of Mercury (Hg) = 47.3g

Mass of chlorine (Cl) = 16.7g

The empirical formula for the compound can be achieved by doing the following:

Step 1:

Divide by their individual molar mass

Hg = 47.3/201 = 0.235

Cl = 16.7/35.5 = 0.470

Step 2:

Divide by the smallest to express in whole number:

Hg = 0.235/0.235 = 1

Cl = 0.470/0.235 = 2

Therefore, the empirical formula is HgCl2

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A compound is analyzed and found to contain 22.10%Al, 25.40%P, and 52.50%O. What is the empirical formula of the compound?

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We will determine the moles of each element first

moles = mass / atomic mass

Atomic weight of Al : 27 g/ mol

Atomic weight of P : 3 1g /mol

Atomic weight of O : 16 g /mol

Let the total mass of compound is 100g

The mass of each element will be

Al = 22.10 g

P = 25.40 g

O = 52.50 g

Moles of Al = mass / atomic mass = 22.10 / 27 = 0.819

Moles of P = mass / atomic mass = 25.40/ 31 = 0.819

Moles of O = mass / atomic mass = 52.50/ 16 = 3.28

Now we will divide the moles of each element with the lowest moles obtained to obtain a whole number ratio of moles of each element present

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Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as

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The question in incomplete, complete question is;

Determine the theoretical yield:

Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.

Answer:

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Explanation:

$Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)$

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According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.

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According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:

$\frac{2}{1}\times 2mol=4 mol$ sodium nitrate

Mass of 4 moles of sodium nitrate :

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