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saveliy_v [14]
3 years ago
7

PLEASE HELPPP

Chemistry
1 answer:
rusak2 [61]3 years ago
5 0

Answer; molecule

Explanation:

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What is the expected freezing point of a 0.50 m solution of li2so4 in water? Kf for water is 1.86 ∘c/m.
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Answer: -2.79°C

Explanation:

Supposing complete ionization:

Li2SO4 → 2 Li{+} + SO4{2-}  [three ions]

(0.50 m Li2SO4) x 3 = 1.50 m ions  

(1.50 m) x (1.86 °C/m) = 2.79°C change

0°C - 2.79°C = -2.79°C

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Can someone solve this problem 5
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Answer:

2

Step-by-step explanation:

A. Moles before mixing

<em>Beaker I: </em>

Moles of H⁺ = 0.100 L × 0.03 mol/1 L

                   = 3 × 10⁻³ mol

<em>Beaker II: </em>

Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.

        H⁺][OH⁻] = 1 × 10⁻¹⁴   Divide each side by [H⁺]

             [OH⁻] = (1 × 10⁻¹⁴)/[H⁺]

             [OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)

             [OH⁻] = 0.01 mol·L⁻¹

Moles of OH⁻ = 0.100 L × 0.01 mol/1 L

                      = 1 × 10⁻³ mol

B. Moles after mixing

                 H⁺    +    OH⁻   ⟶ H₂O

I/mol:      3 × 10⁻³   1 × 10⁻³

C/mol:   -1 × 10⁻³  -1 × 10⁻³

E/mol:    2 × 10⁻³          0

You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.

You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.


C. pH

 [H⁺] = (2 × 10⁻³ mol)/(0.200 L)

        = 1 × 10⁻² mol·L⁻¹

 pH = -log[H⁺ ]

       = -log(1 × 10⁻²)

       = 2

6 0
3 years ago
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