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natita [175]
3 years ago
7

What mass of HCl is needed to prepare 1.00 mole of Cl₂? 2 KMnO₄ + 16 HCl → 2MnCl₂ + 5 Cl₂ + 8 H₂O + 2 KCl

Chemistry
1 answer:
dsp733 years ago
8 0

Answer:

117 g HCl

Explanation:

use a molar ratio to cancel out the moles of Cl2

1.00 mol Cl2 x 16 mol HCl / 5 mol Cl2

multiply by the molar mass of HCl to get the mass of the HCl

3.2 mol HCl x 36.46 g HCl

then you should get around

116.672 g HCl

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SOMEONE PLS HELP
Gekata [30.6K]

Answer:

<h2>464.85 mL</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we're finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

100.7 kPa = 100,700 Pa

95.1 kPa = 95,100 Pa

We have

V_2 =  \frac{100700 \times 439}{95100}  =  \frac{44207300}{95100}  \\  = 464.8506...

We have the final answer as

<h3>464.85 mL</h3>

Hope this helps you

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^{16}_{\phantom{1}8}\text{O}.

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<h3>Explanation</h3>

The superscript of the ion says "2-". That means that the ion here carries a charge of -2.

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