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luda_lava [24]
3 years ago
9

10. A solution contains 130 grams of KNO3 dissolved in 100 grams of water When 3 more grams of KNO3 is added, none of it dissolv

es, nor do any additional crystals appear. The temperature of the solution is closest to
A.65 Celsius

B. 68 Celsius

C. 70 Celsius

D. 72 Celsius
Chemistry
2 answers:
krek1111 [17]3 years ago
6 0

Answer:

B. 68 Celsius

Explanation:

Murljashka [212]3 years ago
3 0

Answer:

Option B

Explanation:

We will check the solubility graph for potassium nitrate,  KNO 3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.

Hence, option B is correct

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How much heat is required to vaporize 25g of water at 100*c
ElenaW [278]

Answer:

Heat required = 13,325 calories or 55.75 KJ.

Explanation:

To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water

Which equals ,

Q = mL,

Where, m is the mass of water present

           L = specific latent heat of vaporization

Here , m= 25 gram

L equals to 533 calories (or 2230 Joules)

So, Q = 25×533 = 13,325 Calories

Or , Q = 55,750 Joules = 55.75 KJ

so, Heat required = 13,325 calories or 55.75 KJ.

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Calculate the pH of a buffer prepared by mixing 20.0 mL of 0.10 M acetic acid and 55.0 mL of 0.10 M sodium acetate
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Calculate the pH of a buffer prepared by mixing 30.0 mL of 0.10 M acetic acid and 40.0 mL of 0.10 M sodium acetate.

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What is the compound name to C5H6?
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Read 2 more answers
A radioactive element reduces to 5.00% of its initial mass in
Stolb23 [73]

The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months

To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Amount remaining (N) = 5%

Original amount (N₀) = 100%

<h3>Number of half-lives (n) =?</h3>

N₀ × 2ⁿ = N  

5 × 2ⁿ = 100

2ⁿ = 100/5

2ⁿ = 20

Take the log of both side

Log 2ⁿ = log 20

nlog 2 = log 20

Divide both side by log 2

n = log 20 / log 2

<h3>n = 4.32</h3>

Thus, 4.32 half-lives gas elapsed.

Finally, we shall determine the half-life of the element. This can be obtained as follow.

Number of half-lives (n) = 4.32

Time (t) = 500 years

<h3>Half-life (t½) =? </h3>

t½ = t / n

t½ = 500 / 4.32

t½ = 115.74 years

Multiply by 12 to express in months

t½ = 115.74 × 12

<h3>t½ ≈ 1389 months </h3>

Therefore, the half-life of the radioactive element in months is approximately 1389 months

Learn more: brainly.com/question/24868345

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