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3 years ago

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as

Chemistry

1 answer:

zubka84 [21]3 years ago

6 0

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g 22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

= 175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

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Answer:

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3 years ago

A solution of hydrochloric acid of unknown concentration was titrated with 0.10 M NaOH. If a 100.-mL sample of the HCl solution

madam [21]

Answer: The initial pH of the HCl solution is 3

Explanation:

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where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

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$n_1=1\\M_1=?M\\V_1=100mL\\n_2=1\\M_2=0.10M\\V_2=1mL$

Putting values in above equation, we get:

$1\times M_1\times 100=1\times 0.10\times 1\\\\M_1=\frac{1\times 0.10\times 1}{1\times 100}=10M$

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$pH=-\log[H^+]$

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3 years ago

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Explanation:

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2 years ago