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
<span>The lattice structure in a metal is weaker than the lattice structure of an ionic compound. This is because the metals have free electrons which can freely move around while ionic compounds are strongly bonded. Hope this answers the question.</span>
For i: 33mL
For ii: 87-88mL
For iii:22.3mL
Answer:
109.7178g of H2O
Explanation:
First let us generate a balanced equation for the reaction. This is illustrated below:
2C3H8O + 9O2 —> 6CO2 + 8H2O
Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:
Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.
Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g
Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol
Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g
From the equation,
120.19184g of C3H8O produced 144.12224g of H20.
Therefore, 91.5g of C3H8O will produce = (91.5 x 144.12224) /120.19184 = 109.7178g of H2O
