N=6.98*10²⁴
Nₐ=6.022*10²³ mol⁻¹
n(Mg)=N/Nₐ
m(Mg)=n(Mg)M(Mg)=M(Mg)N/Nₐ
m(Mg)=24.3g/mol*6.98*10²⁴/(6.022*10²³mol⁻¹)=281.7 g
<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %
<u>Explanation:</u>
We are given:
A chemical compound having chemical formula of 
It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms
To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:

Mass of compound = ![[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol](https://tex.z-dn.net/?f=%5B%281%5Ctimes%2014%29%2B%285%5Ctimes%201%29%2B%281%5Ctimes%2012%29%2B%283%5Ctimes%2016%29%5D%3D79g%2Fmol)
Mass of hydrogen = 
Putting values in above equation, we get:

Hence, the percent composition by mass of hydrogen in given compound is 6.33 %
Answer:
Yes the two of the answer is True
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I think the reaction that represents a balanced, double replacement chemical reaction is B which is <span>Rb2O + Cu(C2H3O2)2 → 2RbC2H3O2 + CuO </span>
Answer:
KJ8RT898TGHO7-6734354546746R476
Explanation: