When dealing with making diluted solutions from concentrated solutions, we can use the following formula
c1v1 = c2v2
where c1 and v1 are the concentration and volume of the concentrated solution respectively.
c2 and v2 are the concentration and volume of the diluted solution respectively
substituting these values in the above formula,
20 mL x 0.200 M = C x 250.0 mL
C = 0.0160 M
Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g
Answer:
2 51 × 10^-5mol/L
Explanation:
The concentration of hydrogen ions can be calculated using the formula below :
pH = -log [H+]
pH = 4.6
[H+] = ?
[H+] = Antilog (-4.6)
[H+] = 2 51 × 10^-5mol/L
2.13 miles
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