Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The combustion of
will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

Now we are reversing reaction 1 and then adding both the equations, we get :
(1)

(2)

The expression for enthalpy change for the reaction will be,



Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol
Francium oxide Fr₂O
Francium, as well as sodium, is in 1A group. Its valency is equal to unit.
As we know that 760 mmHg is equal to 1 atm.
So,
If 760 mmHg is equal to = 1 atm
Then
738 mmHg will be equal to = X atm
Solving for X,
X = (738 mmHg × 1 atm) ÷ 760 mmHg
X = 0.971 atm
Result:
738 mmHg is equal to 0.971 atm.
Answer: The answer would be D.
Explanation: When a screw is tightened into an object, it is wedge in between.