Answer:
Kindly check the explanation section.
Explanation:
From the description given in the question above, that is '' H subscript f to the power of degree of the reaction" we have that the description matches what is known as the heat of formation of the reaction, ∆fH° where the 'f' is a subscript.
In order to determine the heat of formation of any of the species in the reaction, the heat of formation of the other species must be known and the value for the heat of reaction, ∆H(rxn) must also be known. Thus, heat of formation can be calculated by using the formula below;
∆H(rxn) = ∆fH°( products) - ∆fH°(reactants).
That is the heat of formation of products minus the heat of formation of the reaction g specie(s).
Say heat of formation for the species is known as N(g) = 472.435kj/mol, O(g) = 0kj/mol and NO = unknown, ∆H°(rxn) = −382.185 kj/mol.
−382.185 = x - 472.435kj/mol = 90.25 kJ/mol
1p is not possible because it goes like this 1s2 2s2 2p6
Answer:
That's not possible. When a limiting reactant is called out in a stoichiometry problem, what that means is there's enough of all the other reactants for the limiting one to be completely consumed in the formation of product/products
Answer : The excess reactant in the combustion of methane in opem atmosphere is molecule.
Solution : Given,
Mass of methane = 23 g
Molar mass of methane = 16.04 g/mole
The Net balanced chemical reaction for combustion of methane is,
First we have to calculate the moles of methane.
= = 1.434 moles
From the above chemical reaction, we conclude that
1 mole of methane react with the 2 moles of oxygen
and 1.434 moles of methane react to give moles of oxygen
The Moles of oxygen = 2.868 moles
Now we conclude that the moles of oxygen are more than the moles of methane.
Therefore, the excess reactant in the combustion of methane in open atmosphere is molecule.
Answer:
you need to be more specific
try adding a picture of the samples
Explanation: