Answer:
33.30 grams of CaCl2 will be required
Explanation:
Given,
Volume of solution, V= 250 ml
Molarity of solution, M= 1.20 mol/L
Molecular mass of CaCL2, S= 40+(35.5 X 2)= 111
We know,
Required mass, W= SVM/1000
Now,
W = (111 X 250 X 1.20)/1000
= 33300/1000
= 33.30
Therefore, 33.30 grams of Calcium Chloride will be required.
Answer:
890 has 2 significant figures
Answer:
a. 52.8
Explanation:
To find the number of moles of HCl we use the relation M₁V₁=M₂V₂
where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.
M₁=7.91 M
M₂=2.13 M
V₁=?
V₂=196.1 mL
Replacing these values in the relationship.
M₁V₁=M₂V₂
7.91 M× V₁=2.13 M×196.1 mL
V₁=(2.13 M×196.1 mL)/7.91 M
=52.8 mL
The concentration of ClO₂⁻ at equilibrium if the initial concentration of HClO₂ is 0.0654.
<h3>What is concentration?</h3>
The concentration of any substance is the quantity of that substance in per square of the space or container.
The reaction is
HClO₂ + H₂O <=> H₃O⁺ + ClO₂⁻
The pH is 0.454 M
Ka = [H₃O⁺][ClO₂⁻ ] / [HClO₂]
2. 25 × 10⁻² m = [x][x] / 0.454-x]
2 + 0.011 - 0.004994 = 0
solve the quadratic equation
x = 0.0654 = [H3O+] = [ClO2-]
pH = -log (H3O+)
pH = -log(0.0654)
pH = 1.2
equilibrium concentrations of
[HClO2] = 0.454 -x = 0.454 -0.0654 = 0.3886 M
[ClO2- ] = x = 0.0654
Thus, the equilibrium concentrations is 0.0654.
To learn more about concentration, refer to the link:
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