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Setler79 [48]
3 years ago
6

\The specific heat of aluminum is 0.21 cal g°C . How much heat is released when a 10 gram piece of aluminum foil is taken out of

the oven and cools from 50 to 10 °C?
Chemistry
1 answer:
atroni [7]3 years ago
8 0

<u>Answer: </u>The amount of heat released is 84 calories.

<u>Explanation: </u>

The equation used to calculate the amount of heat released or absorbed, we use the equation:

Q= m\times c\times \Delta T

where,

Q = heat gained  or released = ? Cal

m = mass of the substance = 10g

c = specific heat of aluminium = 0.21 Cal/g ° C

Putting values in above equation, we get:

\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC  

Q=10g\times 0.21Cal/g^oC\times-40^oC

Q = -84 Calories

Hence, the amount of heat released is 84 calories.

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What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of
Allisa [31]

73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

mass of solvent CHCl3 = 530 grams or 0.53 kg

Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 =  -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

ΔT = iKf.m

m=  molality

molality = \frac{moles of solute}{mass of solvent in kilograms}

putting the value in the equation:

molality= \frac{1.31}{0.53}

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Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

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6 0
3 years ago
An aqueous antifreeze solution is 60.0% ethylene glycol (HOCH2CH2OH) by mass and has a density of 1.06 g/mL. Calculate the molal
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Answer:

[HOCH₂CH₂OH] = 24.1 m

Explanation:

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100 g = 60 g + Solvent mass

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Molality are the moles of solute contained in 1kg of solvent.

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Molality → 0.967 mol / 0.04 kg = 24.1 m

4 0
3 years ago
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