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Aloiza [94]
3 years ago
9

How much heat was transferred from 50.0 g of water if the temperature of the water went from 30.0 ° C to 55.0 °? The specific he

at capacity of the water is 4.18 J/(g•K)
Chemistry
1 answer:
Lena [83]3 years ago
4 0

Answer:

Heat transfer = Q = 62341.6 J

Explanation:

Given data:

Heat transfer = ?

Mass of water = 50.0 g

Initial temperature = 30.0°C

Final temperature = 55.0°C

Specific heat capacity of water = 4.184 J/g.K

Solution:

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 = 55.0°C -  30.0°C

ΔT = 25°C (25+273= 298 K)

Q = 50.0 g × 4.184 J/g.K ×298 K

Q = 62341.6 J

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8. Select the lattice energy for rubidium chloride from the following data (in kJ/mol]
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Answer:

Option C

Explanation:

The chemical reactions which are involved while solving this problem is there in the file attached and each chemical reaction is represented by a certain equation number

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Equation 2 represents the sublimation reaction of rubidium

Equation 3 represents the ionization enthalpy of rubidium

Equation 4 represents the enthalpy of atomization of chlorine which means it describes the bond enthalpy of Cl2 molecule

Equation 5 represents the electron affinity of chlorine

To find the lattice energy for RbCl we have to use all the equations from 1 to 5 so that at last we get the equation 6

We have to perform operations such as

Equation 1 - equation 2 - equation 3 - equation 4 - equation 5

By performing these operations the intermediate compounds gets cancelled and at last we get equation 6

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Value corresponding to the equation 6 will be the value of lattice energy of RbCl and the value is -695·3 kJ/mol

∴ Lattice energy for rubidium chloride is approximately -695 kJ/mol

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