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vesna_86 [32]
3 years ago
15

What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water t

o make a 500.0 ml solution?
Chemistry
1 answer:
const2013 [10]3 years ago
4 0

Answer:

C = 0.08M

Explanation:

molar mass of AlCl3

Al =27

Cl = 35.5

27+3(35.5) =133.5g/mol

n= mass/Molar mass

n =CV

CV = mass/molar mass

C x 500 x 10^-³ = 5/133.5

C x 500 x 10^-³ = 0.04

C = 0.04/500 x 10^-³

C = 0.08M

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ascorbic acid is a diprotic acid (Ka= 8.0x10^-5 and Ka2= 1.6x10^-12). What is the pH of a 0.260 M solution of ascorbic acid
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The pH of a 0.260 M solution of ascorbic acid is 0.585. Details about pH can be found below.

<h3>How to calculate pH?</h3>

The pH of a solution can be calculated using the following expression:

pH = - log {H+}

According to this question, ascorbic acid is a diprotic acid and posseses a concentration of 0.260M. The pH can be calculated as follows;

pH = - log {0.260}

pH = 0.585

Therefore, the pH of a 0.260 M solution of ascorbic acid is 0.585.

Learn more about pH at: brainly.com/question/15289741

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Which arrow represents the change of state described above? <br>L<br>N<br>O<br>P​
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Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid
jasenka [17]

Answer: a) The K_a of acetic acid at 25^0C is 1.82\times 10^{-5}

b) The percent dissociation for the solution is 4.27\times 10^{-3}

Explanation:

CH_3COOH\rightarrow CH_3COO^-H^+

 cM              0             0

c-c\alpha        c\alpha          c\alpha

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.10 M and \alpha = ?

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Putting in the values we get:

K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}

K_a=1.82\times 10^{-5}

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