Answer:
CH₄(g) + 2O₂(g) ---> 1CO₂(g) + 2H₂O(g)
Explanation:
any combustion of a hydrocarbon equation is in form:
CₓHₐ(g) + BO₂(g) ---> YCO₂(g) + ZH₂O(g), where x,a,b,y,z are all whole number positive integers
there will be 1 CO₂ to 2 H₂O, since there is 1 C to 4 H in CH₄; it is not 1:4 since 2 H is needed in H₂O
CH₄(g) + _O₂(g) ---> 1CO₂ + 2H₂O
there is 4 total O on products side, which can make 2O₂
CH₄(g) + 2O₂(g) ---> 1CO₂(g) + 2H₂O(g)
Answer:
Option B. A
Explanation:
From the question given above, the following data were obtained:
C(s) + 2H₂ (g) —> CH₄ (g). ΔH = –74.9 kJ
From the reaction above, we can see that the enthalpy change (ΔH) is negative (i.e –74.9 KJ) which implies that the heat content of the reactants is greater than the heat content of the products. Thus, the reaction is exothermic reaction.
For an exothermic reaction, the energy profile diagram is drawn in such a way that the heat content of reactants is higher than the heat content of products because the enthalpy change
(ΔH) is always negative.
Thus, diagram A (i.e option B) gives the correct answer to the question.
The statement that best describes waves is B. Waves cannot propagate without a medium
7.5x10^14 Hz. I just worked it out on paper.
1 it a 2 it b 3 it c 4 it be 7 it a
