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1 year ago

Calculate the mass in grams of NaCl required to prepare 0.500 L of a 0.500 M solution.

Chemistry

1 answer:

IceJOKER [234]1 year ago

7 0

Answer:

14.61 g NaCl

Explanation:

Molarity = moles / volume of solution (litres)

0.500 = moles / 0.500 L

moles = 0.25

Molar mass of NaCl is 58.44 g NaCl /1 mol NaCl.

0.25 moles NaCl x 58.44 g NaCl /1 mol NaCl

= 14.61 g NaCl

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Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

2 years ago

PLEASEE HELP ME!!!!!

Yakvenalex [24]

Answer:

Explanation:

8.61+5.779 = 14.389 = 1.4389 × 10^1

25 - 12.5 = 1.25 x 10^1

56.35 / 13.2 = 4.2689

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