The answer is N2 + 3H2 yields 2NH3. The oxidation-reduction reaction means that there is electrons transfer during the reaction which means that the valence changed.
Answer:
Oxide of M is
and sulfate of 
Explanation:
0.303 L of molecular hydrogen gas measured at 17°C and 741 mmHg.
Let moles of hydrogen gas be n.
Temperature of the gas ,T= 17°C =290 K
Pressure of the gas ,P= 741 mmHg= 0.9633 atm
Volume occupied by gas , V = 0.303 L
Using an ideal gas equation:


Moles of hydrogen gas produced = 0.01225 mol

Moles of metal =
So, 8.3333 mol of metal M gives 0.01225 mol of hydrogen gas.

x = 2.9 ≈ 3


Formulas for the oxide and sulfate of M will be:
Oxide of M is
and sulfate of
.
Answer:
{1s^2 2s^2 2p^6} 3s^2 3p^4
{Ne}3s^2 3p^4
Explanation:
i didnt understand the rest of that but this is the e- configuration on top and the bottom is noble gas configuration