Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Should be 1.8L.
2 moles of hydrogen react with 1 mole of oxygen. If 2 moles of hydrogen is 3.6L, 1 mole of oxygen should be 1.8L.
<h3><u>Answer;</u></h3>
Directly proportional
<h3><u>Explanation;</u></h3>
- <em><u>Concentration is one of the factors that determine the rate of a reaction. Reaction rates increases with increase in the concentration of the reactants, which means they are directly proportional.</u></em>
- An increase in the concentration of reactants produces more collisions and thus increasing the rate at which the reaction is taking place. Therefore, <u>Increasing the concentration of a reactant increases the frequency of collisions between reactants and will cause an increase in the rate of reaction.</u>
1.34 L of HF
Explanation:
We have the following chemical reaction:
Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)
First we calculate the number of moles of SnF₂:
number of moles = mass / molecular weight
number of moles of SnF₂ = 5 / 157 = 0.03 moles
From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.
Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:
number of moles = volume / 22.4 (L/mole)
volume of HF = number of moles × 22.4
volume of HF = 0.06 × 22.4 = 1.34 L
Learn more about:
problems with gases at STP
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