The volume in liters occupied by 22.6 g of I₂ gas at STP is 1.99 L (answer A)
<u><em>calculation</em></u>
Step: find the moles of I₂
moles= mass÷ molar mass
from periodic table the molar mass of I₂ is 253.8 g/mol
moles = 22.6 g÷253.8 g/mol =0.089 moles
Step 2:find the volume of I₂ at STP
At STP 1 moles =22.4 L
0.089 moles= ? L
<em>by cross multiplication</em>
={ (0.089 moles x 22.4 L) /1 mole} = 1.99 L
Answer: The half-reactions represents reduction are as follows.
Explanation:
A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.
For example, the oxidation state of Cr in
is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

Similarly, oxidation state of Mn in
is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.

Thus, we can conclude that half-reactions represents reduction are as follows.
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
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