consider the following intermediate chemical equations.\n2h₂(g)+o₂(g)→2h₂o(l)\nh₂(g)+f₂(g)→2hf(g)\nin the…

consider the following intermediate chemical equations.\n2h₂(g)+o₂(g)→2h₂o(l)\nh₂(g)+f₂(g)→2hf(g)\nin the final chemical equation, hf and o₂ are the products that are formed through the reaction between h₂o and f₂. before you can add these intermediate chemical equations, you need to alter them by multiplying the\nsecond equation by 2 and reversing the first equation.\nfirst equation by 2 and reversing it.\nfirst equation by (1/2) and reversing the second equation.\nsecond equation by 2 and reversing it.

consider the following intermediate chemical equations.\n2h₂(g)+o₂(g)→2h₂o(l)\nh₂(g)+f₂(g)→2hf(g)\nin the final chemical equation, hf and o₂ are the products that are formed through the reaction between h₂o and f₂. before you can add these intermediate chemical equations, you need to alter them by multiplying the\nsecond equation by 2 and reversing the first equation.\nfirst equation by 2 and reversing it.\nfirst equation by (1/2) and reversing the second equation.\nsecond equation by 2 and reversing it.

Answer

Explanation:

Step1: Analyze the target reaction

The target reaction has $H_2O$ as a reactant and $HF$ and $O_2$ as products. The first given equation is $2H_2(g)+O_2(g)\rightarrow2H_2O(l)$ and the second is $H_2(g)+F_2(g)\rightarrow2HF(g)$. We need $H_2O$ on the reactant - side, so we should reverse the first equation to get $2H_2O(l)\rightarrow2H_2(g)+O_2(g)$. We also need more $HF$ in the final equation. To get more $HF$, we multiply the second equation by 2 to get $2H_2(g) + 2F_2(g)\rightarrow4HF(g)$.

Answer:

A. second equation by 2 and reversing the first equation.