A silver bar with a specific heat of 0.235J/gºC has a mass of 250.g and is heated from 22.0°C to 68.5°C. How much heat will the silver bar absorb?​

2731.875 J

Explanation:

Applying

Q = cm(t₂-t₁) Equation 1

Where Q = amount of heat, c = specific heat capacity of silver, m = mass of silver, t₂ = Final temperature of silver, t₁ = Initial temperature of silver

From the question,

Given: c = 0.235 J/g°C, m = 250 g, t₁ = 22.0°C, t₂ = 68.5°C

Substitute these values into equation 1

Q = 0.235×250(68.5-22.0)

Q = 58.75(46.5)

Q = 2731.875 J

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%