This should be what you need to have it balanced
Answer:
6 mins
Explanation:
The time taken for Ar to effuse can be obtained as follow:
Time for Br₂ (t₁) = 12 mins
Molar mass of Br₂ (M₁) = 2 × 80 = 160 g/mol
Molar mass of Ar (M₂) = 40 g/mol
Time for Ar (t₂) =?
t₂/t₁= √(M₂/M₁)
t₂ / 12 = √(40/160)
Cross multiply
t₂ = 12 × √(40/160)
t₂ = 12 × 0.5
t₂ = 6 mins
Therefore, it will take 6 mins for the same amount of Ar to effused out.
Answer:
pH = 12.8
Explanation:
HF + NaOH → F⁻ + Na⁺ + H₂O
<em>1 mole of HF reacts with 1 mole of NaOH</em>
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Initial moles of HF and NaOH are:
HF = 0.018L × (0.308mol / L) = 5.544x10⁻³mol HF
NaOH = 0.023L × (0.361mol / L) = 8.303x10⁻³mol NaOH
That means moles of NaOH remains after reaction are:
8.303x10⁻³mol - 5.544x10⁻³mol = <em>2.759x10⁻³moles NaOH</em>
Total volume is 18.0mL + 23.0mL = 41.0mL = 0.0410L
Molar concentration of NaOH is
2.759x10⁻³moles NaOH / 0.0410L = 0.0673M = [OH⁻]
pOH = - log [OH⁻] = 1.17
As pH = 14 - pOH
<em>pH = 12.8</em>
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Water boils at 100 degrees Celsius
Answer:
0.54 mole of H2O.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2CH3OH + 3O2 —> 2CO2 + 4H2O
From the balanced equation above,
2 moles of CH3OH reacted to produce 4 moles of water.
Finally, we shall determine the number of mole of water (H2O) produced by the reaction of 0.27 moles of CH3OH. This can be obtained as follow:
From the balanced equation above,
2 moles of CH3OH reacted to produce 4 moles of water.
Therefore, 0.27 moles of CH3OH will react to produce = (0.27 × 4)/2 = 0.54 mole of H2O.
Thus, 0.54 mole of H2O is produced from the reaction.
