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Akimi4 [234]
3 years ago
14

Calculate the minimum concentration of Mg2+ that mustbe added to 0.10 M NaF in order to initiate a precipitate ofmagnesium fluor

ide. (For MgF2, Ksp = 6.9 x10-9.)A. 1.4 x 107M
B. 6.9 x 10-9M
C. 6.9x 10-8 MD. 1.7 x 10-7M
E. 6.9 x 10-7 M
Chemistry
1 answer:
blondinia [14]3 years ago
6 0

Answer:

E. 6.9 E-7 M

Explanation:

  • NaF  →  Na+   +     F-

      0.10M   0.10M    0.10M

  • MgF2 ↔ Mg2+  +    2F-

eq:     S             S          2S + 0.1

∴ Ksp = 6.9 E-9 = [Mg2+][F-]² = (S)(2S+0.1)²

if we compared the concentration ( 0.1 M ) with the Ksp ( 6.9 E-9 ); then we can neglect the solubility as adding:

⇒ Ksp = 6.9 E-9 = (S)(0.1)² = 0.01S

⇒ S = 6.9 E-9 / 0.01 = 6.9 E-7 M

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Natali5045456 [20]

Answer:

The molality ( m ) of a solution is the moles of solute divided by the kilograms of solvent. A solution that contains 1.0 mol of NaCl dissolved into 1.0 kg of water is a “one-molal” solution of sodium chloride. The symbol for molality is a lower-case m written in italics.

In order to calculate the molality of a solution divide the moles of solute by the volume of the solution expressed in liters.

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7 0
2 years ago
Based on the equation 3 Cu (s) + 8 HNO3 (aq) 3 Cu(NO3)2 (aq) + 2NO (g) + 4H2O (g) how many grams of Cu would be needed to react
eimsori [14]

Answer:

Mass = 381.28 g

Explanation:

Given data:

Number of moles of HNO₃ = 16 mol

Mass of Cu needed to react with 16 mol of HNO₃ = ?

Solution:

Chemical equation:

3Cu + 8HNO₃    →     3Cu(NO₃)₂ + 4H₂O + 2NO

Now we will compare the moles of Cu with HNO₃ from balance chemical equation.

                     HNO₃         :          Cu

                        8              :         3

                       16             :       3/8×16 = 6

Mass of Cu needed:

Mass = number of moles × molar mass

Mass = 6 mol × 63.546 g/mol

Mass = 381.28 g

4 0
3 years ago
What is the partial pressure of C?<br> atm C
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The partial pressure of carbon is 45 mm Hg.

Explanation:

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8 0
3 years ago
what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?
IgorLugansk [536]

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

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