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 .
I believe it is "empirical formula"
For better representation, let me rewrite the electronic configuration:
<span>1s</span>²<span>2s</span>²<span>2p</span>⁶<span>3s</span>²<span>3p</span>⁶<span>4s</span>²<span>3d</span>⁴
The exponents represent the number of electrons in the designated subshell. Thus, the total number of electrons are:
# of electrons = 2+2+6+2+6+2+4 = 24
Assuming this is in neutral state, the element with an atomic number of 24 is Chromium. Thus, the answer is Cr.
The reaction for the combustion of methane can be expressed as follows.
CH4 + 2O2 --> CO2 + 2H2O
We solve first for the amount of carbon dioxide in moles by dividing the given volume by 22.4L which is the volume of 1 mole of gas at STP.
moles of CO2 = (5.6 L) / (22.4 L/1 mole)
moles of CO2 = 0.25 moles
Then, we can see that every mole of carbon dioxide will need 1 mole of methane
moles methane = (0.25 moles CO2) x (1 moles O2/1 mole CO2)
= 0.25 moles CH4
Then, multiply this by the molar mass of methane which is 16 g/mole. Thus, the answer is 4 grams methane.
