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Zigmanuir [339]
3 years ago
9

A bottle contains 6.23 g of copper turnings. The mass of a length of magnesium ribbon is 6.23 g. Are there more copper atoms or

more magnesium atoms present?
Chemistry
1 answer:
Zepler [3.9K]3 years ago
7 0
Moles = mass (g) / molar mass (g/mol)

Molar mass of copper = 63.5 g/mol
Molar mass of magnesium = 24.3 g/mol

Moles of copper = 6.23 g / 63.5 g/mol = 0.098 mol

Moles of magnesium = 6.23 g / 24.3 g/mol = 0.256 mol

according to the Avogadro constant,
                           1 mol = 6.022 x 10²³

Hence,
 atoms of copper        = 0.098 x 6.022 x 10²³ = 5.902 x 10²²
 atoms of magnesium = 0.256 x 6.022 x 10²³ = 15.416 10²²

Hence magnesium ribbon has more atoms.


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2.122 g of a solid mixture containing only potassium carbonate (FW = 138.2058 g/mol) and potassium bicarbonate (FW = 100.1154 g/
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Answer:

The weight percent of potassium carbonate is 50,8 wt% and of potassium bicarbonate 49,2 wt%

Explanation:

The reactions of potassium carbonate (K₂CO₃) and potassium bicarbonate (KHCO₃) with HCl produce:

K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O

KHCO₃ + HCl → KCl + CO₂ + H₂O

That means that you need 2 moles of HCl to titrate potassium carbonate and 1 mol to titrate potassium bicarbonate.

The moles of HCl to titrate the mixture are:

0,03416L×\frac{0,762mol}{1L} = <em>0,02603 mol of HCl</em>

If X is mass of K₂CO₃ and Y is mass of KHCO₃ in the mixture, the moles of HCl to titrate the mixture are equals to:

0,02603 mol = 2X×\frac{138,2058 g}{1mol} + Y×\frac{100,1154 g}{1mol} <em>(1)</em>

As the mass of the mixture is 2,122g:

2,122g = X + Y <em>(2)</em>

Replacing (2) in (1):

0,02603 mol = 0,01447 (2,122-Y) + 9,988x10⁻³Y

0,02603 mol = 0,0307 - 0,01447Y + 9,988x10⁻³Y

-4,6778x10⁻³ = -4,4827x10⁻³Y

1,044g = Y <em>-mass of potassium bicarbonate-</em>

Thus:

X = 1,078g <em>-mass of potassium carbonate-</em>

The weight percent of potassium carbonate is:

\frac{1,078g}{2,122g}×100 =<em> 50,8 wt%</em>

The weight percent of potassium bicarbonate is:

\frac{1,044g}{2,122g}×100 = <em>49,2 wt%</em>

<em></em>

I hope it helps!

5 0
3 years ago
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