The reaction between hydrogen (H2) and fluorine (F2) is given below,
H2 + F2 ---> 2HF
One mole of both hydrogen and fluorine yields to 2 moles of hydrogen fluoride. This can also be expressed as, 2 grams of hydrogen and 38 grams of fluorine will form 40 grams of hydrogen fluoride. From the given, only 20 grams of HF is formed with 19 g of it being fluorine. Thus, the percentage fluorine of the compound formed is 95%.
The correct answer is hydrogen<span>, and </span>oxygen<span>. </span>
Answer:
Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).
Explanation:
Hello there!
In this case, according to the given redox reaction, we rewrite it as a convenient first step:

Next, we assign the oxidation numbers as follows:

Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).
Regards!