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3 years ago

Which is a saturated solution?

Chemistry

1 answer:

ser-zykov [4K]3 years ago

7 0

Answer:

C）52g KCl in 100g water at 80°C

Explanation:

A saturated solution is one that contains as much solute as it can dissolve in the presence of excess solute at that particular temperature.

A solutibility curve is a graph that shows the variability with temperature of the solubility of a solute in a given solvent. A solutibility curve can provide information of whether a solution formed frommthe solute and solvent are saturated or not at a given temperature.

From the solubility curve in the attachment below:

A) A saturated solution of NH₄Cl will contain about 52 g solute per 100 g sat 50 °C. Thus, a solution of 40 g NH₄Cl in 100 g water at 50 °C is an unsaturated solution.

B) A saturated solution of SO₂ at 10°C will contain about 70 g of solute in 100 g of water. Thus a solution of 2g SO₂ in 100g water at 10°C is an unsaturated solution.

C) A saturated solution of KCl at 80 °C will contain about 52 g of solute in 100 g of water. Thus, a solution of 52g KCl in 100g water at 80°C is a saturated solution.

D) A saturated solution of Kl at 20 °C will contain about 145 g of solute in 100 g of water. Thus, a solution of 120g KI in 100g water at 20°C is an unsaturated solution.

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______________ is a reaction that causes the growth of a polymer chain. In this reaction, a relatively small molecule is formed

WITCHER [35]

Answer:

Condensation polymerization

Explanation:

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3 years ago

How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?

GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

Volume = 5.00L

Pressure = 4.00× 10³mmHG

Temperature =23°c

The formula we would be applying is Ideal gas law

PV = nRT

Step 1

We find the number of moles of Flourine gas present.

T = 23°C

Converting to Kelvin

= °C + 273k

= 23°C + 273k

= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K

n = 1.0834112811moles

Step 2

We calculate the mass of Flourine gas

The molar mass of Flourine gas =

F2 = 19 × 2

= 38 g/mol

Mass of Flourine gas = Molar mass of Flourine gas × No of moles

Mass = 38g/mol × 1.0834112811moles

41.169628682grams

Approximately = 41.17 grams.

3 0

3 years ago

Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT

marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L = L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

Therefore;

L = 60 + α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

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