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shtirl
2 months ago
11

If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4

is 2.40×10−5. Express your answer to three significant figures and include the appropriate units.
Chemistry
1 answer:
Anarel [2.9K]2 months ago
5 0

Answer:

The amount of calcium sulfate that precipitates is 6.14 grams.

Explanation:

Step 1: Provided data

We are mixing 500.0 mL of 0.10 M Ca^2+ with 500.0 mL of 0.10 M SO4^2−

The Ksp for CaSO4 is 2.40×10^−5.

Step 2: Determine moles of Ca^2+

Moles of Ca^2+ = Molarity of Ca^2+ * Volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles of Ca^2+ = 0.05 moles

Step 3: Determine moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

Step 4: Compute total volume

Total volume = 500.0 mL + 500.0 mL = 1000 mL = 1L

Step 5: Compute Q

Q = [Ca2+] [SO42-]

[Ca2+] = 0.050 M and [SO42-]

Qsp = (0.050)(0.050) = 0.0025 >> Ksp

This indicates that precipitation will take place.

Step 6: Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M (molar solubility)

Step 7: Determine total dissolved CaSO4

Total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 g/mol = 0.667 g

Step 8: Calculate initial mass of CaSO4

Initial moles of CaSO4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

Step 9: Calculate precipitate mass

6.807 - 0.667 = 6.14 grams.

The mass of calcium sulfate that will emerge as a precipitate is 6.14 grams.

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Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% O by mass and has a molar mass of 206 g/mol. Express your a
castortr0y [3046]

Answer:

A) The molecular formula for ibuprofen isC_{13}H_{18}O_2

B) The molecular formula for Cadaverine is C_{5}H_{14}N_2

C) The molecular formula for Epinephrine is C_9H_{13}O_3N_1

Explanation:

Element percentage in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100

A) The composition of ibuprofen, used for headaches, consists of 75.69% carbon, 8.80% hydrogen, and 15.51% oxygen by weight.

Ibuprofen has a molar mass of 206 g/mol.

The proposed molecular formula for ibuprofen is =C_xH_yO_z

Count of carbon atoms in one ibuprofen molecule;

75.69\%=\frac{x\times 12 g/mol}{206 g/mol}\times 100

x=\frac{75.69\times 206 g/mol}{100\times 12 g/mol}=12.99\approx 13

Count of hydrogen atoms in one ibuprofen molecule;

8.80\%=\frac{y\times 1 g/mol}{206 g/mol}\times 100

y=\frac{8.80\times 206 g/mol}{100\times 1 g/mol}=18.12\approx 18

Count of oxygen atoms in one ibuprofen molecule;

15.51\%=\frac{z\times 16 g/mol}{206 g/mol}\times 100

z=\frac{15.51\times 206 g/mol}{100\times 16 g/mol}=1.99\approx 2

Molecular formula for ibuprofen:

= C_xH_yO_z= C_{13}H_{18}O_2

B) Cadaverine consists of 58.55% carbon, 13.81% hydrogen, and 27.40% nitrogen by weight

Cadaverine has a molar mass of 102.2 g/mol.

The proposed molecular formula for Cadaverine is =C_xH_yN_z

Count of carbon atoms in one Cadaverine molecule;

58.55\%=\frac{x\times 12 g/mol}{102.2 g/mol}\times 100

x=\frac{58.55\times 102.2 g/mol}{100\times 12 g/mol}=4.98\approx 5

Count of hydrogen atoms in one Cadaverine molecule;

13.81\%=\frac{y\times 1 g/mol}{102.2 g/mol}\times 100

y=\frac{13.81\times 102.2 g/mol}{100\times 1 g/mol}=14.11\approx 14

Count of nitrogen atoms in one Cadaverine molecule;

27.40\%=\frac{z\times 14 g/mol}{102.2 g/mol}\times 100

z=\frac{27.40\times 102.2 g/mol}{100\times 14 g/mol}=2.00\approx 2

Molecular formula for Cadaverine:

= C_xH_yN_z= C_{5}H_{14}N_2

C) Epinephrine includes 59.0% carbon, 7.1% hydrogen, 26.2% oxygen, and 7.7% nitrogen by weight

Epinephrine has a molar mass of 180 g/mol.

The proposed molecular formula for Epinephrine is =C_xH_yO_zN_w

Count of carbon atoms in one Epinephrine molecule;

59.0\%=\frac{x\times 12 g/mol}{180 g/mol}\times 100

x=\frac{59.0\times 180 g/mol}{100\times 12 g/mol}=8.85\approx 9

Count of hydrogen atoms in one Epinephrine molecule;

7.1\%=\frac{y\times 1 g/mol}{180 g/mol}\times 100

y=\frac{7.1\times 180 g/mol}{100\times 1 g/mol}=12.78\approx 13

Count of oxygen atoms in one Epinephrine molecule;

26.2\%=\frac{z\times 16 g/mol}{180 g/mol}\times 100

z=\frac{26.2\times 180 g/mol}{100\times 16 g/mol}=2.94\approx 3

Count of nitrogen atoms in one Epinephrine molecule;

7.7\%=\frac{w\times 14 g/mol}{180 g/mol}\times 100

w=\frac{7.7\times 180 g/mol}{100\times 14 g/mol}=0.99\approx 1

Molecular formula for Epinephrine:

= C_xH_yO_zN_w= C_9H_{13}O_3N_1

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While the original inquiry is incomplete, the comprehensive question is:

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