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balu736 [363]
3 years ago
15

A student dissolved a 40-gram block of a salt in 100 grams of warm water at 45°C. The solution was allowed to cool down to 24°C.

The student noticed that some of the salt came out of the solution and settled to the bottom of the beaker. It is later determined that 12 grams of the salt came out of the solution. How many grams of the salt were dissolved in the solution at 24°c 100 grams of warm water , PLEASE HELP ME FAST I ONLY HAVE A COUPLE MIN LEFT TO
FINISH I WILL GIVE YU BRAINLIEST!
Chemistry
2 answers:
skad [1K]3 years ago
5 0

Answer: 28 grams

Explanation:

sweet [91]3 years ago
5 0

Answer:

28

Explanation:

The grams of salt dissolved in the solution was 28, because 40 minus 12 would be 28. I know I'm suppose to subtract 40 by 12 considering the fact, it is 12 grams of salt that came out. The rest of the numbers are irrelevant. (apologies this is a week late.)

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2 years ago
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Calculate the pH of a 0.0150 M HNO3 solution. Assume HNO3
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Answer: 1.824

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4 0
3 years ago
If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?
Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

  • <u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol

Volume of solution = 50 + 60 = 110 mL = 0.11 L    (Conversion factor:  1 L = 1000 mL)

  • To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of TRIS acid = 8.3

[\text{TRIS acid}]=\frac{0.005}{0.11}

[\text{TRIS base}]=\frac{0.012}{0.11}

pH = ?

Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

Hence, the pH of resulting solution is 8.7

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Explanation:

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