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Tpy6a [65]
3 years ago
9

Answer the above pici bet no one can solve this correctly with steps​

Chemistry
2 answers:
Sever21 [200]3 years ago
4 0

Answer:

solution given:

log k=4-\frac{200}{t}

so,

On comparison

∆S°/R = 4

∆ S° = 4R

algol [13]3 years ago
3 0

mousam brother is correct

hope it helps you have a good day

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3/7 of a sum is Rs.1800. Find the sum​
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The same would be considered and relevant

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7 0
3 years ago
You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol
defon

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀\frac{[A^{-}]}{[HA]}

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀\frac{[A^{-}]}{[HA]}

1,38 = \frac{[A^{-}]}{[HA]} <em>(1)</em>

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] <em>(2)</em>

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid  ≡ 0,15 moles of acetic acid ×\frac{60,05 g}{1mol} = <em>9,0 g of acetic acid</em>

<em></em>

I hope it helps!

5 0
3 years ago
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