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2 years ago

28 mL of 0.10 M HCl is added to 60.0 mL of 0.10 M Sr(OH)2. Determine the concentration of OHin the resulting solution.

Chemistry

1 answer:

serg [7]2 years ago

6 0

Answer:

Explanation:

Sr(OH)₂.+ 2HCl = SrCl₂ + 2H₂O

Moles of HCl in 28mL of .10 M HCl = .028 x .1 = .0028 moles .

Moles of Sr(OH)₂ in 60mL of .10 M Sr(OH)₂ = .060 x .1 = .0060 moles

2 moles of HCl reacts with 1 mole of Sr(OH)₂

.0028 moles of HCl reacts with .0014 mole of Sr(OH)₂

moles of Sr(OH)₂ remaining = .0060 - .0014 = .0046 moles .

Sr(OH)₂ = Sr⁺ + 2OH⁻

1 mole 2 mole

.0046 .0092

Total volume of solution = 88 mL .

88 mL of solution contains .0092 moles of OH⁻

concentration of OH⁻ = .0092 / .088

= .1045 M .

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Answer:

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Explanation:

THIS IS THE COMPLETE QUESTION BELOW

Measurements show that the enthalpy of a mixture of gaseous reactants increases by 319kJ during a certain chemical reaction, which is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -92kJ of work is done on the mixture during the reaction. Calculate the change of energy of the gas mixture during the reaction in kJ.

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RideAnS [48]

Answer:

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The picture above illustrate convergent, divergent and transform boundary.

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