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2 years ago

--IF YOU PUSH SOMETHING WITH STRONG FORCE DOES THE MOTION GO SLOWER

Chemistry

2 answers:

zaharov [31]2 years ago

8 0

Answer:

i dont think so

Explanation:

is there a image to explain this ? if so it would help ALOT

postnew [5]2 years ago

5 0

If you push some thing heavier many but other wise no

Hope this helps you

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How does octect law work?

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3 years ago

How are atoms grouped on the periodic table

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3 years ago

A sample of carbon-12 has a mass of 6.00 g. How many atoms of carbon-12 are in the sample?

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<em>mC: 12g/mol</em>

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---->>> A :)

4 0

3 years ago

Read 2 more answers

A solution was prepared by dissolving 195.0 g of KCl in 215 g of water. Calculate the mole fraction of KCl. (The formula weight

defon

Answer:

Approximately $0.180$ .

Explanation:

The mole fraction of a compound in a solution is:

$\displaystyle \frac{\text{Number of moles of compound in question}}{\text{Number of moles of all particles in the solution}}$ .

In this question, the mole fraction of $\rm KCl$ in this solution would be:

$\displaystyle X_\mathrm{KCl} = \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}$ .

This solution consist of only $\rm KCl$ and water (i.e., $\rm H_2O$ .) Hence:

$\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})}\end{aligned}$ .

From the question:

Mass of $\rm KCl$ : $m(\mathrm{KCl}) = 195.0\; \rm g$ .
Molar mass of $\rm KCl$ : $M(\mathrm{KCl}) = 74.6\; \rm g \cdot mol^{-1}$ .

Mass of $\rm H_2O$ : $m(\mathrm{H_2O}) = 215\; \rm g$ .
Molar mass of $\rm H_2O$ : $M(\mathrm{H_2O}) = 18.0\; \rm g\cdot mol^{-1}$ .

Apply the formula $\displaystyle n = \frac{m}{M}$ to find the number of moles of $\rm KCl$ and $\rm H_2O$ in this solution.

$\begin{aligned}n(\mathrm{KCl}) &= \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} \\ &= \frac{195.0\; \rm g}{74.6\; \rm g \cdot mol^{-1}} \approx 2.61\; \em \rm mol\end{aligned}$ .

$\begin{aligned}n(\mathrm{H_2O}) &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \\ &= \frac{215\; \rm g}{18.0\; \rm g \cdot mol^{-1}} \approx 11.9\; \em \rm mol\end{aligned}$ .

The molar fraction of $\rm KCl$ in this solution would be:

$\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})} \\ &\approx \frac{2.61 \; \rm mol}{2.61\; \rm mol + 11.9\; \rm mol} \approx 0.180\end{aligned}$ .

(Rounded to three significant figures.)

8 0

3 years ago

If 16.4 grams of copper (II) bromide react with 22.7 grams of sodium chloride, how many grams of sodium bromide are formed?

fomenos

The amount of sodium bromide that would be formed from the reaction will be 7.5524 grams

<h3>Stoichiometric calculation</h3>

Looking at the equation of the reaction:

$CuBr_2 + 2NaCl --- > CuCl_2 + 2NaBr$

The mole ratio of CuBr2 and NaCl is 1:2.

Mole of 16.4 grams of CuBr2 = 16.4/223.37

= 0.0734 moles

Mole of 22.7 grams of NaCl = 22.7/58.44

= 0.3884 moles

Equivalent mole of NaCl = 0.1468 moles

Thus, NaCl is in excess while CuBr2 is limiting.

Mole ratio of CuBr2 and NaBr = 1:1

Mass of 0.0734 mole NaBr = 0.0734 x 102.894

= 7.5524 grams

More on stoichiometric calculation can be found here: brainly.com/question/8062886

4 0

3 years ago