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pishuonlain [190]
3 years ago
9

You have 0.2231 grams of an unknown monovalent salt (MCl). You treat it with an excess of AgNO3 and precipitate all the chloride

ion. 0.5471 grams of AgCl precipitates. What is the formula weight and identity of M
Chemistry
1 answer:
ludmilkaskok [199]3 years ago
4 0

Answer:

Formula weight of MCl = 58.55 g/mol

M = Sodium

Explanation:

Calculation of the moles of AgCl as:-

Mass = 0.5471 g

Molar mass of AgCl = 143.32 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.5471\ g}{143.32\ g/mol}

Moles= 0.00381\ mol

1 mole of AgCl is produced from 1 mole of MCl

So, moles of MCl = 0.00381 mol

Mass = 0.2231 grams

Molar mass = Mass/Moles = 0.2231 grams/0.00381 mol= 58.55 g/mol

Molar mass of Cl = 35.5 g/mol

Molar mass of M = Molar mass of MCl - Molar mass of Cl = 58.55 g/mol - 35.5 g/mol = 23.05 g/mol

This mass corresponds to sodium.

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<u>Answer:</u>

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III Postulate - By emitting or absorbing energy, an electron may shift from one stationary energy orbit to another.

<u>Three main Limitations -</u>

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Explanation:

<u>About Bohr's postulates-</u>

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Explanation:

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7 0
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Calculate the ph of a solution formed by mixing 200.0 ml of 0.30 m hclo with 300.0 ml of 0.20 m kclo. the ka for hclo is 2.9 × 1
velikii [3]
C(HClO) = 0,3 M.
<span>V(HClO) = 200 mL = 0,2 L.
n(HClO) = </span>c(HClO) · V(HClO).
n(HClO) = 0,06 mol.<span>
c(KClO</span>) = 0,2 M.
<span>V(KClO) = 0,3 L.
n(KClO) = 0,06 mol.
V(buffer solution) = 0,2 L + 0,3 L = 0,5 L.
ck</span>(HClO) = 0,06 mol ÷ 0,5 L = 0,12 M.
cs(KClO) = 0,06 mol ÷ 0,5 L = 0,12 M.<span>
Ka(HClO</span>) = 2,9·10⁻⁸.<span>
This is buffer solution, so use Henderson–Hasselbalch equation: 
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