Chemistry, 14.02.2020 03:48 mem8163

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg .

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Answer from: sin91

The given question is incomplete. The complete question is as follows.

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and then found to have a mass of 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)

Explanation:

When $Ba^{2+}$ and $Na_{2}SO_{4}$ are added then white precipitate forms. And, reaction equation for this is as follows.

$Ba^{2+} + SO^{2-}_{4} \rightarrow BaSO_{4}$

It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of $BaSO_{4}$ is 233.43.

Now, we will calculate the number of moles as follows.

No. of moles = mass × M.W

= $\frac{0.212}{233.43}$

= 0.00091 mol of $BaSO_{4}$

Hence, it means that 0.00091 mol of $Ba^{2+}$ . Now, we will calculate the mass as follows.

Mass = moles × MW

= $0.00091 \times 137.327$

= 0.124 grams or 124 mg of barium

Thus, we can conclude that mass of barium into the original solution is 124 mg.

Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

Answer from: Quest

what does the table look like

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A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the so...

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