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3 years ago

What 2 processesslow down the motion of the molecules

Chemistry

1 answer:

fenix001 [56]3 years ago

6 0

I believe it would be transfering heat into a cold class or vise versa

Hope this helpesss (':

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An aqueous potassium iodate ( KIO3 ) solution is made by dissolving 562 562 grams of KIO 3 KIO3 in sufficient water so that the

maxonik [38]

Answer:

Molarity of $KIO_{3}$ solution is 0.612 mol/L

Explanation:

Number of moles of a substance = (mass of substance)/(molar mass of the substance)

Molar mass of $KIO_{3}$ = 214 g/mol

So, 562 g of $KIO_{3}$ = $\frac{562}{214}$ moles of $KIO_{3}$ = 2.63 moles of $KIO_{3}$

Molarity of a solution = (number of moles of solute in solution)/(total volume of solution in liter)

Here solute is $KIO_{3}$ and solvent is water

Total volume of solution is 4.30 L

So, molarity of $KIO_{3}$ solution = $\frac{2.63}{4.30}mol/L$ = 0.612 mol/L

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

An experiment requires that each student use an 8.5 cm length of magnesium ribbon. How many students can do the experiment if th

melisa1 [442]

570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students

7 0

3 years ago

Read 2 more answers

Assessment started: undefined.

Vilka [71]

B) the bacteria that live in the intestine of a rabbit

6 0

4 years ago

A student increases the temperature of a 200cm3 balloom from 60 degress C to 180 degrees C. What will the new volume of the ball

kkurt [141]

Answer:

The new volume of the balloon is 600cm³

Explanation:

Given parameters:

Initial Volume, which is V₁ = 200cm³

Initial temperature, T₁ = 60°C

Final temperature T₂ = 180°C

Final Volume V₂ =?

To solve this kind of problem, we apply one of the gas laws that shows the relationship between volume and temperature.

This law is the Charles law, it states that " the volume of a fixed mass of a gas is directly proportional to its absolute temperature if pressure is constant".

It is simply expressed as:

V₁/T₁ = V₂/T₂

Since our unknown is V₂, we make it the subject of the expression given above:

V₂ = V₁T₂/T₁

Now input the corresponding values and solve:

V₂ = 200 x 180 / 60

V₂ = 36000/60

V₂ = 600cm³

The new volume of the balloon is 600cm³

4 0

4 years ago