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DaniilM [7]
3 years ago
12

A 25.00-mL sample of 6.00 M HCl solution is diluted to a new volume of 85.00 mL. What is the concentration of the dilute solutio

n
Chemistry
1 answer:
natali 33 [55]3 years ago
6 0

The concentration of the dilute solution is 1.76 M

<h3>Data obtained from the question </h3>
  • Molarity of stock solution (M₁) = 6 M
  • Volume of stock solution (V₁) = 25 mL
  • Volume of diluted solution (V₂) = 85 mL
  • Molarity of diluted solution (M₂) =?

<h3>How to determine the molarity of the diluted solution </h3>

M₁V₁ = M₂V₂

6 × 25 = M₂ × 85

150 = M₂ × 85

Divide both side by 85

M₂ = 150 / 85

M₂ = 1.76 M

Thus, the concentration of the diluted solution is 1.76 M

Learn more about dilution:

brainly.com/question/15022582

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The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?
Alex Ar [27]

____________________________________________________

Answer:

Your answer would be a). 2.0 × 10-9

____________________________________________________

Work:

In your question the "ph" of a 0.55 m aqueous solution of hypobromous acid temperature is at 25 degrees C, and it's "ph" is 4.48.

You would use the ph (4.48) to find the ka for "hbro"

[H+]

=

10^-4.48

=

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=

[BrO-]

or: [H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-]

Then you would find ka:

(3.31 x 10^-5)^2/0.55 =2 x 10^-9

____________________________________________________

<em>-Julie</em>

6 0
3 years ago
How many grams of Ag2CO3 will precipitate when excess (NH4)2CO3 solution is added to 79.8 mL of 0.545 M AgNO3 solution
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Suppose that 25. 0 ml of 0. 10 m ch3cooh(aq) ( ka =1. 8 x 10-5) is titrated with 0. 10 m naoh(aq). (a)what is the initial ph of
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The initial ph of the 0. 10M ch3cooh  (aq) solution is  -0.64.

The salt sodium acetate is present at the equivalency point. It is a salt of a strong base and a weak acid. The expression "its pH" provides the information.

pH = 7 + 0.5p + 0.5logC

where C is the salt concentration. Monobasic acid is Acetic acid here.

25 ml of 0.10 M NaOH solution are needed for the titration of 25 ml of 0.10 M acetic acid.

Total volume = 25 + 25 = 50ml

The salt concentration = C = 0.10×  = 0.05

The pH for weak acid is given by,

pH = \frac{1}{2} (pK_{a} + logC)

pH =  \frac{1}{2} (1.8× + log0.05)

pH =  \frac{1}{2} [1.8× + (-1.30)]

pH =  \frac{1}{2} (1.8× -1.30)

pH = -0.64

Learn more about pH here;

brainly.com/question/15192160

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I believe the answer is A) Buffer
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