`H_(2)O(l)toH_(2)O(g)` , `DeltaH=+43.7 kJ` `H_(2)O(s)toH_(2)O(l)` , `DeltaH=+6.05kJ` The value of `DeltaH_((sublimation))` of ice is

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`H_(2)O(l)toH_(2)O(g)` , `DeltaH=+43.7 kJ` `H_(2)O(s)toH_(2)O(l)` , `DeltaH=+6.05kJ` The value of `DeltaH_((sublimation))` of ice is

`H_(2)O(l)toH_(2)O(g)` , `DeltaH=+43.7 kJ`
`H_(2)O(s)toH_(2)O(l)` , `DeltaH=+6.05kJ`
The value of `DeltaH_((sublimation))` of ice is
A. `49.75 kJ mol^(-1)`
B. `37.65 kJ mol^(-1)`
C. `43.7 kJ mol^(-1)`
D. none of these

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
Add the two equations to get
`H_(2)O(s) to H_(2)O(g)`

4 views Answered 2023-02-05 15:29