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Shtirlitz [24]
3 years ago
7

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0300 M Ag + ( aq ) . What will be the conce

ntration of Ca 2 + ( aq ) when Ag 2 SO 4 ( s ) begins to precipitate? Solubility-product constants, K sp , can be found in the chempendix.
Chemistry
1 answer:
strojnjashka [21]3 years ago
5 0

Answer : The concentration of [Ca^{2+}] ion is, 0.00371 M

Explanation :

First we have to calculate the concentration of [SO_4^{2-}] ion.

The solubility equilibrium reaction of AgSO_4 will be:

Ag_2SO_4\rightleftharpoons 2Ag^{+}+SO_4^{2-}

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^{+}]^2[SO_4^{2-}]

K_{sp}=1.20\times 10^{-5}

1.20\times 10^{-5}=(0.0300)^2\times [SO_4^{2-}]

[SO_4^{2-}]=0.0133M

Now we have to calculate the concentration of [Ca^{2+}] ion.

The solubility equilibrium reaction of AgSO_4 will be:

CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][SO_4^{2-}]

K_{sp}=4.93\times 10^{-5}

4.93\times 10^{-5}=[Ca^{2+}]\times (0.0133)  

[Ca^{2+}]=0.00371M

Thus, the concentration of [Ca^{2+}] ion is, 0.00371 M

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The actual density of iron is 7.874 g/mL. In a laboratory investigation, Jason finds the density of a piece of iron to be 7.921
mixas84 [53]

The percent error associated with Jason’s measurement is 0.596%.

HOW TO CALCULATE PERCENTAGE ERROR:

  • The percentage error of a measurement can be calculated by following the following process:
  1. Find the difference between the true value and the measured value of a quantity.
  2. Then, divide by the true value and then multiplied by 100

  • The true value of the density of iron is 7.874 g/mL
  • Jason observed value is 7.921 g/mL

Difference = 7.921 g/mL - 7.874 g/mL

Difference = 0.047 g/mL

Percentage error = 0.047/7.874 × 100

Percentage error = 0.596%.

Therefore, the percent error associated with Jason’s measurement is 0.596%.

Learn more: brainly.com/question/18074661?referrer=searchResults

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2 years ago
Electrical conductivity is an example of what property
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3 0
3 years ago
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How are the properties of sodium chloride different than sodium and chlorine
lions [1.4K]
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7 0
3 years ago
And with solution...
Tems11 [23]

Answer:

The answer to your question is: V = 6.93 L

Explanation:

Data

N₂ = 5.6 g

Volume of NH₃ = ?

                              14 g of N   ----------------  1 mol

                              5.6 g -----------------------   x

                             x = (5.6 x 1) / 14 = 0.4 mol of N

Reaction

                                N₂    +     3H₂    ⇒    2NH₃

                                1 mol of N₂   ----------------  2 moles of NH₃

                                0.4 mol of N₂ --------------   x

                               x = (0.4 x 2) / 1

                               x = 0.8 mol of NH₃

Formula

                 PV = nRT

P = 5200 torr = 6.84 atm

V = ?

n = 0.8

R = 0.082 atm L/ mol °K

T = 450°C = 723°K

Substitution

                     V = (0.8)(0.082)(723) / 6.84

                     V = 6.93 L

7 0
3 years ago
Uranium has an atomic weight of 238 amu.
shepuryov [24]

Answer:

U= 238g/mol

U2O5= 556g/mol

Explanation:

Since U= 238

O=16

U3O5= 2(238)+3(16)=556g/mol

4 0
3 years ago
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