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3 years ago

What is the freezing point of a .743 m aqueous solution of KCl

Chemistry

1 answer:

marysya [2.9K]3 years ago

3 0

Change in freezing point = -1.86 C/m X 0.743 m X 2 = -2.76 C

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The new boiling point is 101.02°C. Suppose that you add another 2.0 mol of sucrose to this solution. What do you predict the new

solniwko [45]

Mad Lad, thanks for the answer :)

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3 years ago

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What is the mass of CO2 lost at 20 min from the limestone sample?

harkovskaia [24]

Answer:

The mass, CO2 and CO3 from the limestone sample is discussed below in details.

Explanation:

(A) mass loss of sample of limestone after 20 min

= 0.8437g-0.5979g = 0.2458 g

From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.

Therefore

184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.

1 g of sample gives 88/184.4 g of carbon dioxide

Hence 0.2458 g sample gives

= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide

(B) mole of CO 2 lost = weight/ molar mass

= 0.117 g / 44 g/mol =0.0027 mole

(C). 1 mol of limestone contain 2 mol of carbonate ion

From the reaction we know that carbonate ion of limestone is converted into carbondioxide

Hence lost carbonate ion = 0.2458 g

(D) we know that

1 mol limestone contain 1mol CaCO 3

Hence in sample present CaCO 3

= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO3

8 0

3 years ago

2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2

Harrizon [31]

It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.

<u>Explanation:</u>

2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂

We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

$100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}$

= 337.5 g AgCl

In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.

It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.

4 0

3 years ago

An experiment reveals that 125.0 grams of an unknown metal increases in temperature from 22.0 oC to 43.6 oC upon absorbing 640 j

nydimaria [60]

Answer:

Cp = 0.237 J.g⁻¹.°C⁻¹

Explanation:

Amount of energy required by known amount of a substance to raise its temperature by one degree is called specific heat capacity.

The equation used for this problem is as follow,

Q = m Cp ΔT ----- (1)

Where;

Q = Heat = 640 J

m = mass = 125 g

Cp = Specific Heat Capacity = <u>??</u>

ΔT = Change in Temperature = 43.6 °C - 22 °C = 21.6 °C

Solving eq. 1 for Cp,

Cp = Q / m ΔT

Putting values,

Cp = 640 J / (125 g × 21.6 °C)

Cp = 0.237 J.g⁻¹.°C⁻¹

3 0

3 years ago

PLZ HELP

nika2105 [10]

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4 years ago

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