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podryga [215]
3 years ago
12

Substance A has a mass of 10 g and a volume of 4 ml. Substance B has a mass of 13 g and a volume of 5.2 ml. Which substance has

a higher density?
Chemistry
1 answer:
ale4655 [162]3 years ago
3 0

Answer:

They both have the same density.

Explanation:

Density = mass/volume

10g ÷ 4 = 2.5

13g ÷ 5.2 = 2.5

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If the solubility of KCl in 100 mL of H₂O is 34 g at 20 °C and 43 g at 50 °C, label each of the following solutions as unsaturat
sertanlavr [38]

Answer:

a) Unsaturated

b) Supersaturated

c) Unsaturated

Explanation:

A saturated  solution contains the <u>maximum amount of a solute that will dissolve in a given  solvent at a specific temperature</u>.

An unsaturated solution contains <u>less solute than it  has the capacity to dissolve. </u>

A supersaturated solution, <u>contains more  solute than is present in a saturated solution</u>. Supersaturated solutions are not very  stable. In time, some of the solute will come out of a supersaturated solution as crystals.

According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:

a) 30 g in 100 mL of H₂O at 20 °C  ⇒ unsaturated

b) 65 g in 100 mL of H₂O at 50 °C  ⇒ supersaturated

c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)

8 0
3 years ago
A solution is made by mixing of water and of acetic acid . Calculate the mole fraction of water in this solution. Be sure your a
Molodets [167]

Answer:

The answer is "0.35".

Explanation:

please find the complete question in the attached file.

Mass of CH_3C0_2 H= 77 \ g \\\\

Molar mass of CH_3C0_2 H = 60.05 \  \frac{g}{mol} \\\\

No of moles in n_{CH_3CO_2H} = 77 \ g \times  \frac{1 \ mol \ CH_3C0_2H}{60.05 \ g}  

                                          = 1.28 \ mol \ CH_3CO_2H

Mass of water (H_2O)= 42 \ g

The molar mass of water (H_2O)= 18.02 \  \frac{g}{mol}

moles of water n_{H_2O}:

= 42 \ g \times \frac{1 mol H_2O}{18.02 \ g} \\\\= 2.33 \ mol \ H_2 O

Molfraction of acetic acid:

X CH_2CO_2H = \frac{n_{CH_3CO_2H}}{n_CH_3CO_2H +n_{H_2}}\\\\

                     =\frac{1.28 \ mol}{1.28 \ mo1 + 2.33 mol}\\\\ = 0.354\\\\= 0.35

7 0
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