Chemistry, 24.03.2020 18:08 damondgriswold12

Anybody good at chemistry?

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Answer from: loganjune5033

Explanation:

Given data:

Mass of lead = 25 g

Initial temperature = 40°C

Final temperature = 95°C

Cp = 0.0308 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 95°C - 40°C

ΔT = 55°C

Q = 25 g × 0.0308 j/g.°C × 55°C

Q = 42.35 j

Given data:

Mass = 3.1 g

Initial temperature = 20°C

Final temperature = 100°C

Cp = 0.385 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 100°C - 20°C

ΔT = 80°C

Q = 3.1 g × 0.385 j/g.°C × 80°C

Q = 95.48 j

Given data:

Mass of Al = ?

Initial temperature = 60°C

Final temperature = 30°C

Cp = 0.897 j/g.°C

Heat released = 120 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 30°C - 60°C

ΔT = -30°C

120 j = m × 0.897 j/g.°C × -30°C

120 j = m × -26.91 j/g

m = 120 j / -26.91 j/g

m = 4.46 g

negative sign show heat is released.

Given data:

Mass of ice = 1.5 g

Change in temperature = ?

Cp = 0.502 j/g.°C

Heat added= 30.0 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

30.0 j = 1.5 g × 0.502 j/g.°C × ΔT

30.0 j = 0.753 j/°C × ΔT

30.0 j /0.753 j/°C = ΔT

39.84 °C = ΔT

Answer from: Quest

answer

the ions in solute supperate from eachouthere

Answer from: Quest

Ithink d hope that

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