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3 years ago

Question 16 of 30

Chemistry

1 answer:

andrezito [222]3 years ago

4 0

Answer: B. CH₂OH

Explanation:

First off, we need a liquid solvent. Eliminate C and D.

Now, we have to consider polarity.

Carbon tetrachloride is nonpolar, so sodium chloride will be insoluble.
CH₂OH is polar, so sodium chloride will be soluble.

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Alex dragged a log across the yard in 30 the log weighted 400n she did 900 j of work how much power did I she have

In-s [12.5K]

The power used by Alex to drag the log across the yard is determined as 2,656 W.

The mass of the log is calculated as follows;

W = mg

m = W/g

m = (400)/9.8

m = 40.82 kg

<h3>Velocity of the log</h3>

K.E = ¹/₂mv²

v² = 2K.E/m

v² = (2 x 900)/(40.82)

v² = 44.096

v = 6.64 m/s

<h3>Power used by Alex</h3>

P = Fv

P = 400 x 6.64

P = 2,656 W

Learn more about power here: brainly.com/question/13881533

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2 years ago

A sample of gas at 15 atmospheres and 445 K is cooled to 250 K. What pressure will the gas be under after it is cooled? Round to

faltersainse [42]

Answer:

Explanation: 8.43

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3 years ago

The ability of a substance to dissolve in another is called

Luden [163]

Answer: Solubility

Explanation:

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4 years ago

Read 2 more answers

A mixture of CO2 and Kr weighs 35.0 g and exerts a pressure of 0.708 atm in its container. Since Kr is expensive, you wish to re

ollegr [7]

Explanation:

Since, it is given that carbon dioxide is completely removed by absorption with NaOH. And, pressure inside the container is 0.250 atm.

For Kr = 0.250 atm and pressure $CO_{2}$ will be calculated as follows.

$CO_{2}$ = (0.708 - 0.250) atm

= 0.458 atm

Now, we will calculate the mole fraction as follows.

$CO_{2} = \frac{0.458}{0.708}$

= 0.646

Kr = $\frac{0.250}{0.708}$

= 0.353

Now, we will convert into gram fraction as follows.

$CO_{2} = 0.646 \times 44$

= 28.424

Kr = $0.353 \times 83.78$

= 29.57

Therefore, total mass is calculated as follows.

Total mass = (28.424 + 29.57)

= 57.994

Hence, the percentage of $CO_{2}$ and Kr are calculated as follows.

$CO_{2} = \frac{28.424}{57.99} \times 100$

= 49%

Kr = $\frac{29.57}{57.99} \times 100$

= 51%

Hence, amount of $CO_{2}$ and Kr present i mixture is as follows.

$CO_{2}$ in mixture = $35 \times 0.49$

= 17.15 g

Kr = $35 \times 0.51$

= 17.85 g

Thus, we can conclude that 17.15 g of $CO_{2}$ is originally present and 17.85 g of Kr is recovered.

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