Answer:
-205.7kj
Explanation:
Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of the reaction.
The expression for enthalpy for the following reaction will be,
where,
n = number of moles
Now put all the given values in the above expression, we get:
Therefore, the enthalpy of the following reaction is, -205.7kj
Answer:
Answer to A. helium, neon, argon, krypton, xenon, and radon, B. Elemental hydrogen (H, element 1), nitrogen (N, element 7), oxygen (O, element 8), fluorine (F, element 9), and chlorine (Cl, element 17) are all gases at room temperature, and are found as diatomic molecules (H2, N2, O2, F2, Cl2). C. Elements Compounds
Ar (argon) HBr (hydrogen bromide) C 3H 8 (propane)
Kr (krypton) HI (hydrogen iodide) C 4H 10 (butane)
Xe (xenon) HCN (hydrogen cyanide)* CO (carbon monoxide)
Rn (radon) H 2S (hydrogen sulfide) CO 2 (carbon dioxide)
Explanation:
Hepta is seven and tetra is four, so option c is your answer
It is not directly over a flame because it depends on the substance you might not want to heat it too much.you never know what could happen
2H2 (g) + O2 (g) -->2H2 O(g)
mole ratio of H2:O2=2:1
7.25/2=3.625
