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vitfil [10]
3 years ago
5

Which of the following changes would decrease the rate at which a solid solute dissolves in a liquid solvent?

Chemistry
2 answers:
Fed [463]3 years ago
4 0

Answer:

The correct answer would be B. Decreasing surface area.

Explanation:

It would be decreasing because the surface area is dissolving, so the surface area is decreasing.

andriy [413]3 years ago
3 0

Answer:

Decreasing surface area of the solid in the liquid.

Explanation:

I don't know how to explain this without using. an example. So, here it is: If you were dipping bread in water, the bread would be completely soaked if you dipped the whole thing, but if you only allowed one part of the bread to be in the water, it would take longer for the water to soak all the way into it. The amount of surface area that is allowed to touch what is making the solid dissolve determines how quickly the whole thing will be dissolved.

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1.00 M CaCl2 Density = 1.07 g/mL
Lesechka [4]

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

m = 1 mol\times 111 g/mol = 111 g

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

M=d\times V=1.07 g/mL\times 1000 mL=1,070 g

1) The value of %(m/M):

\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%

2) The value of %(m/V):

\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%

Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}

Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}

n = Equivalent mass

n = \frac{\text{molar mass of ion}}{\text{charge on an ion}}

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 CaCl_2 mole has 1 mole of calcium ion)

n=\frac{40 g/mol}{2}=20

=\frac{1 mol}{20 g/mol\times 1L}=0.050 N

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 CaCl_2 mole has 2 mole of chlorine ion)

n=\frac{35.5 g/mol}{1}=35.5

=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

\frac{1 mol}{0.959 kg}=1.043 mol/kg

Moles of calcium chloride = n_1=1mol

Mass of solvent = 959 g

Moles of water = n_2=\frac{959 g}{18 g/mol}=53.28 mol

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842

7) Mole fraction of water =

\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

m'=d\times v=1.07 g/ml\times 100 g= 107 g

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

11.1\%=\frac{m}{100 mL}\times 100

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

3 0
3 years ago
When you remove energy from a substance there will be a *
stira [4]

Answer:

b, decrease in movement of the molecules

Explanation:

removing the energy will begin making the molecules lock up and stop moving due to the loss of energy.

hope this helped

3 0
3 years ago
PLEASE HELPPPPPPPPPPPPPP
docker41 [41]

Answer:

Please take  this hepful hint :

F = m * a

6.2 = 2.3 * a

a = 2.7 m/s^2

Explanation:

4 0
3 years ago
I need this ASAP thank you
Rainbow [258]

Answer:

16,,24Mg 17,,a24.1 18a mass number of the most abundant isotope

Explanation:

atomic number of Mg is 12 ,therefore its mass number should be the value that is very close to 24.

24.1 is the value of thee most abundant isotope.

8 0
2 years ago
Where does the water in air come from?
Tanya [424]

Answer:

About 90 percent of water in the atmosphere is produced by evaporation from water bodies, while the other 10 percent comes from transpiration from plants. There is always water in the atmosphere.✌️

3 0
3 years ago
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