Explain and discuss the reaction of :CuSO4∙oH20 and CuSO4∙5H2O? What is the delta H? What is the q=.What is the mass values?

Table 1.1: title

Given:0.783 M

CuSO4  10g

M1 H2O80mL

Ti22.20 ºC

Tf32.60 ºC

ΔT10.4 ºC

ΔH= (q / # moles)(-435.24) / (0.0626)= -6952.7g

Q= -mass x (4.184 J K-1g-1) ΔT

( -10g)(4.184 J K-1g-1 )( 283 K) =

Table 1.2: title

Given:0.5 M

CuSO4 ∙ 5H20 10g

M1 H2O80mL

Ti22.85 ºC

Tf21.40 ºC

ΔT-1.45 ºC

ΔH= (q / # moles)(60.7) / (0.0400)= 1516.7

Q= -mass x (4.184 J K-1g-1) ΔT

( -10g)(4.184 J K-1g-1 )( 271.55K ) =

Table 1.3: title

Given:0.626 M

CuSO4  5g

M1 H2O50mL

Ti-22.40 ºC

Tf-34.40 ºC

ΔT12.0 ºC

ΔH= (q / # moles)(-251.0) / (0.0313)= 1516.7

Q= -mass x (4.184 J K-1g-1) ΔT

( -5g)(4.184 J K-1g-1 )( 285K ) =

Table 1.4: title

Given:0.4 M

CuSO4 ∙ 5H20 5g

M1 H2O50mL

Ti23.55 ºC

Tf22.05 ºC

ΔT-1.5 ºC