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frez [133]
3 years ago
12

Two parts that make up a solution are... what

Chemistry
1 answer:
hjlf3 years ago
5 0
Answer: All solutions have two parts: the solute and the solvent. The solute is the substance that dissolves, and the solvent is the substance that dissolves the solute.
Hope that helps (:
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The use of crucible​
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a ceramic or metal container in which metals or other substances may be melted or subjected to very high temperatures.

"the crucible tipped and the mold filled with liquid metal"

a situation of severe trial, or in which different elements interact, leading to the creation of something new.

"their relationship was forged in the crucible of war"

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What intermolecular force attracts two non polar molecules to each other
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Convert 281 K to degrees Celsius.<br><br> 281 K = _____<br><br> -8°C<br> 554°C<br> 8°C
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8°C

Explanation:

3 0
3 years ago
Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907
melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent.\frac{^0 C}{m}

m = molality of solute (m or \frac{mol}{Kg} )

Given : kf = 1.86 \frac{^0 C*Kg}{mol}

m = 0.907 \frac{mol}{Kg} )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 \frac{^0 C*Kg}{mol} * 0.907\frac{mol}{Kg} )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0
3 years ago
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