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Karo-lina-s [1.5K]
3 years ago
8

The ability of a substance to dissolve in another substance?

Chemistry
2 answers:
leva [86]3 years ago
8 0
Solubility is the answer because it is the ability for substance to dissolve in a solvent
omeli [17]3 years ago
5 0
Solubility is the ability of a substance to dissolve in another substance.
You might be interested in
Question 1 of 6
Juli2301 [7.4K]

Answer:

Option A. 2, 3, 2

Explanation:

We'll begin by balancing the equation. This can be achieved by doing the following:

Fe + Cl2 —> FeCl3

There are 2 atoms of Cl on the left side and 3 atoms on the right side. It can be balance by putting 3 in front of Cl2 and 2 in front of FeCl3 as shown below:

Fe + 3Cl2 —> 2FeCl3

There are 2 atoms of Fe on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Fe as shown below:

2Fe + 3Cl2 —> 2FeCl3

Now the equation is balanced.

The coefficients are : 2, 3, 2

8 0
3 years ago
During an experiment, ice and water were placed in a perfectly insulated thermos flask at 0 °C. Describe this system when it pha
Ivanshal [37]

Answer :By the time thermal equilibrium is attained in the system, ice is completely converted to water.

What is tehrmal equilibrium?

Thermal equilibrium is a situation in which two substances are exactly at the same temperature. That is. Heat is no longer flowing beween the bodies.

Recall that an insulated thermos flask implies that heat does not eneter or leave the system. Heat will flow from the water to the ice untill the both attains the same temperature and the ice is completely converted to water.

Have a good day

5 0
2 years ago
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at
seraphim [82]

Answer:

the solubility of CaCO3 is 0.015g/l 25 °C

is favored at equilibrium

Explanation:

The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?

solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)

CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)

in partial dissociation , we can say

2.25x 10^-8=Ca^{2+} +CO^{2-}_{3} }

let Ca^2+=CO3^-2=S

2.25x10^-8=S*S

S^2=2.25x10^-8

S=0.00015mol/L

Converting that to g/l

the relative molecular mass of CaCO3=100g/mol

0.00015*100g/mol

0.015g/l

the solubility of CaCO3 is 0.015g/l @room temperature

is favored   at equilibrium

5 0
3 years ago
I’ve gotten the answer as x=.478 but when plugging back in, .46/.478 ≠ .22. Help please
ivanzaharov [21]

Answer:

15. 2.66 moles .

16. 2.09L.

Explanation:

Molarity of a solution is simply defined as the mole of solute per unit litre of the solvent. Mathematically, it is represented as:

Molarity = mole /Volume.

With the above formula, let us answer the questions given above

15. Data obtained from the question include the following:

Volume of solution = 1.4L

Molarity = 1.9M

Mole of solute =.?

Molarity = mole /Volume

1.9 = mole / 1.4

Cross multiply

Mole = 1.9 x 1.4

Mole = 2.66 moles

Therefore, the mole of the solute present in the solution is 2.66 moles.

16. Data obtained from the question include the following:

Mole of solute = 0.46 mole

Molarity = 0.22M

Volume of solvent (water) =.?

Molarity = mole /Volume

0.22 = 0.46/Volume

Cross multiply

0.22 x Volume = 0.46

Divide both side 0.22

Volume = 0.46/0.22

Volume = 2.09L

Therefore, 2.09L of water is required.

6 0
3 years ago
I need help can someone please do so?
larisa86 [58]

Answer:

0.296 J/g°C

Explanation:

Step 1:

Data obtained from the question.

Mass (M) =35g

Heat Absorbed (Q) = 1606 J

Initial temperature (T1) = 10°C

Final temperature (T2) = 165°C

Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C

Specific heat capacity (C) =..?

Step 2:

Determination of the specific heat capacity of iron.

Q = MCΔT

C = Q/MΔT

C = 1606 / (35 x 155)

C = 0.296 J/g°C

Therefore, the specific heat capacity of iron is 0.296 J/g°C

8 0
3 years ago
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