The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
What is meant by concentration?
Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.
Concentration of hydroxide ions can be calculated by,
M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.
where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.
Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
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Answer:
% purity of limestone = 96.53%
Explanation:
Question (4).
Weight of impure CaCO₃ = 25.9 g
Molecular weight of CaCO₃ = 40 + 12 + 3(16)
= 100 g per mole
We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters
From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of
CO₂.
∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g
∴ 1 liter of CO₂ will be produced by CaCO₃ = 
∴ 5.6 liters of CO₂ will be produced by CaCO₃ = 
= 25 g
Therefore, % purity of CaCO₃ = 
= 
= 96.53 %
The ratios which are needed to determine the mass of oxygen produced from the decomposition of 10 grams of potassium chlorate are;
- 31.998 g O2 : 1 mole O2
- 3 mole O2 : 2 mole KClO3
- 112.55 g KClO31 mole KClO3
From stoichiometry;
- We can conclude that according to the reaction;
3 moles of oxygen requires 2 moles of KClO3 to be produced.
And from molar mass analysis;
- 31.998 g O2 is equivalent to 1 mole O2
- O2112.55 g KClO3 is equivalent to 1 mole KClO3
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