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Answer:
Explanation:
It is volume-volume problems that does not require the use of molar mass.
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:
1.209g of MgO participates
Explanation:
In this problem, we have 0.030 moles of MgO that participates in a particular reaction.
And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.
To convert moles to grams we need to use molar mass of the compound:
<em>1 atom of Mg has a molar mass of 24.3g/mol</em>
<em>1 atom of O has a molar mass of 16g/mol</em>
<em />
That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol
And mass of 0.030 moles of MgO is:
0.030 moles MgO * (40.3g/mol) =
<h3>1.209g of MgO participates</h3>