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3 years ago

State two reasons atom X is different than atom X+1

Chemistry

1 answer:

insens350 [35]3 years ago

6 0

Explanation:

atom X

it is neutral

and may not exist independently

atom X+1

it is ion which has a charge on it positive or negative

it exists independently

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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

Which of the following mixtures would result in a buffered solution when 1.0 L of each of the two solutions are mixed?

drek231 [11]

Answer:

c. 0.2 M HNO₃ and 0.4 M NaF .

Explanation:

A buffer is defined as the mixture of a weak acid with its conjugate base or a weak base with its conjugate acid.

A weak acid or weak base are defined as an acid or base that partially dissociates in aqueous solution. in contrast, a strong acid or base are acids or bases that is dissociated completely in water.

Thus:

a. 0,2M HNO₃ and 0.4 M NaNO₃. This is a mixture of a strong acid with its conjugate base. IS NOT a buffer.

b. 0.2 M HNO₃ and 0.4 M HF . This is a mixture of two strong acids. IS NOT a buffer.

c. 0.2 M HNO₃ and 0.4 M NaF . NaF is the conjugate base of a weak acid as HF is.

The reaction of HNO₃ with NaF is:

HNO₃ + NaF → HF + NaNO₃

That means that in solution you will have a weak acid (HF) with its conjugate base (NaF). Thus, this mixture IS a buffer.

d. 0.2 M HNO₃ and 0.4 M NaOH. This is the mixture of a strong acid with a strong base, thus, this IS NOT a buffer.

I hope it helps!

7 0

3 years ago

Aqueous potassium chloride will react with which one of the following to form a precipitation reaction?

Svetllana [295]

Answer:

lead (II) chlorate

Explanation:

precipitate means not soluble in water

potassium chlorate is not very soluble in water, it precipitates and may be collected by filtration

lead(II) chloride can be made as a white precipitate by adding a solution containing chloride ions to lead(II) nitrate solution

thoughtcocom chemguidecouk

5 0

2 years ago

A certain amount of hydrogen peroxide was dissolved in 100. ml of water and then titrated with 1.68 m kmno4. what mass of h2o2 w

Aleksandr [31]

The solution for the problem is:

First, use the concentration of the volume of the thing you know to compute for the moles of that substance. Then, use the coefficient in the balanced equation to relay moles of that to moles of anything else in the chemical equation. Lastly, translate moles into mass by means of its molar mass, or into a concentration using the volume.

Applying what I have said earlier:

0.0133 L X 1.68 mol/L = 0.0223 mol KMnO4 X (1 mol H2O2 / 2 mol KMnO4) = 0.0112 mol H2O2

Mass H2O2 = 0.0112 mol H2O2 X 34.0 g/mol = 0.380 grams H2O2

8 0

4 years ago

If your lawn is 21.0 ft wide and 20.0 ft long, and each square foot of lawn accumulates 1350 new snow flakes every minute. How m

GaryK [48]

First we can calculate the area of the rectangular lawn using the formula:
Area = Width x Length = 21 ft x 20 ft = 420 square feet

And the total number of snow flakes per minute on the entire lawn is:

(1350 snowflakes per minute per square foot) x (420 square feet) = 567,000 snowflakes per minute

In one hour (or 60 minutes) we get a total of:

(567,000 snowflakes per minute) x (60 minutes / 1 hour) = 34,020,000 snowflakes

The total mass of which would be:

34,020,000 snowflakes x 1.60 mg = 54,432,000 mg = 54.432 kg (as 1 kg = 1,000,000 mg).

So 54.432 kg of snow accumulates every hour on the lawn.

7 0

3 years ago