A solution contains 1.694 mg CoSO4 (155.0 g/mol) per milliliter. Calculate (a) the volume of 0.08640 M EDTA needed to titrate a 25.00-mL aliquot of this solution. (b) the volume of 0.009450 M Zn2 needed to titrate the excess reagent after addition of 50.00 mL of 0.08640 M EDTA to a 25.00-mL aliquot of this solution.

Answers

Answer 1
Answer:

Moles of any substance is given by the molar mass and the mass. The volume of EDTA required to titrate is 3.16 mL and the volume of zinc required to titrate is 22.8 mL.

What is molarity?

Molarity is the property of the solution that gives the concentration of the solute present in the solution.

Given,

Mass of cobalt sulfate = 1.697 gm

The molar mass of cobalt sulfate = 155 g/mol

In the first part, the volume of the aliquot is 25 mL and the molarity is 0.08640 M.

The reaction is shown as:

Co²⁺ + H₄Y → CoH₂Y + 2H⁺

Moles of cobalt: n = 0.001694 ÷ 155 = 0.0000109 moles

In 25 ml aliquot moles of cobalt are, 0.000273 moles.

The volume of EDTA solution is calculated as:

V = moles ÷ Molarity

= 0.000273 mole ÷ 0.0864 mol/ L

= 3.16 mL

Hence, 3.16 mL of EDTA is required.

For the second part, moles of EDTA are calculated as:

n = 0.008640 × 0.050 = 4.32 ×10⁻⁴

In a 25 mL, sample moles of EDTA are 2.16 × 10⁻⁴ moles.

Excess moles of EDTA in the solution:

0.000432 - 0.000216 = 0.000216 moles

The volume of EDTA is calculated as:

V = 0.000216 ÷ 0.009450

= 0.0228 L

Hence, 22.8 mL of zinc is required.

Learn more about molarity here:

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Answer 2
Answer:

Answer:

the answer is in the screenshot

Explanation:

hope this helps


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What is the molarity of the following solutions?a. 19.5 g NaHCO3 in 460.0 ml solution
b. 26.0 g H2SO4 in 200.0 mL solution
c. 15.0 g NaCl dissolved to make 420.0 mL solution

Answers

Answer:

a) NaHCO3 = 0.504 M

b) H2SO4 = 1.325 M

c) NaCl = 0.610 M

Explanation:

Step 1: Data given

Moles = mass / molar mass

Molarity = moles / volume

a. 19.5 g NaHCO3 in 460.0 ml solution

Step 1: Data given

Mass NaHCO3 = 19.5 grams

Volume = 460.0 mL = 0.460 L

Molar mass NaHCO3 = 84.0 g/mol

Step 2: Calculate moles NaHCO3

Moles NaHCO3 = 19.5 grams / 84.0 g/mol

Moles NaHCO3 = 0.232 moles

Step 3: Calculate molarity

Molarity = 0.232 moles / 0.460 L

Molarity = 0.504 M

b. 26.0 g H2SO4 in 200.0 mL solution

Step 1: Data given

Mass H2SO4 = 26.0 grams

Volume = 200.0 mL = 0.200 L

Molar mass H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = 26.0 grams / 98.08 g/mol

Moles H2SO4 = 0.265 moles

Step 3: Calculate molarity

Molarity = 0.265 moles / 0.200 L

Molarity =1.325 M

c. 15.0 g NaCl dissolved to make 420.0 mL solution

Step 1: Data given

Mass NaCl = 15.0 grams

Volume = 420.0 mL = 0.420 L

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles NaCl

Moles NaCl = 15.0 grams / 58.44 g/mol

Moles NaCl = 0.256 moles

Step 3: Calculate molarity

Molarity = 0.256 moles / 0.420 L

Molarity =0.610 M

What is the solubility of substance?

Answers

Explanation:

Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. The resulting solution is called a saturated solution.

5.75 x 1021 molecules of oxygen is the equivalent of how many
moles?

Answers

765.75

................................

Can anyone tell me about sulfuric acid in shampoo/soap?

Answers

The term sulfate is used in chemistry to denote a salt of sulfuric acid. ... This is the sulfate type which can be found in many cleaning and hygiene products including shampoos. The main reasons why it is added to shampoos are because this sulfate produces foam and is a powerful detergent.

A student conducts an experiment to see how music affects plant growth. The student obtains four identical plants. Each one is potted in the same type of soil and receives the sameamount of sunlight and water each day. Plant A listens to classical music for three hours each day. Plant B listens to rock music for three hours each day. Plant C listens to country
music for three hours each day. Plant D does not listen to any music at all.
1. Based on the experiment in the scenario, which visual aid would be most helpful in showing the change in the plants' heights over time?
O A. A line graph
O B. A pie chart
OC. A bar graph
O D. A timeline

Answers

Answer:

A. A line graph  

Explanation:

You use line graphs to track changes over time. Line graphs are better when the changes are small. They are also more useful when you want to compare changes over the same period for more than one group, for example, plants exposed to music and a control group.

B is wrong. A pie chart is best for comparing parts of a whole.

C is wrong. You can use a bar graph to track changes over time, but small changes are harder to spot.

D is wrong. You use a timeline to mark important points in time, for example, when you are deciding the times when you must complete various stages of a project.

Which of the charts below do you think is more helpful in showing the change in plant height over time?

Consider the following reaction at 298K.I2 (s) + Pb (s) = 2 I- (aq) + Pb2+ (aq)
Which of the following statements are correct?
Choose all that apply.
ΔGo > 0
The reaction is product-favored.
K < 1
Eocell > 0
n = 2 mol electrons
B-

Answers

Answer:

Eªcell > 0; n = 2

Explanation:

The reaction:

I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)

Is product favored.

A reaction that is product favored has ΔG < 0 (Spontaneous)

K > 1 (Because concentration of products is >>>> concentration reactants).

Eªcell > 0 Because reaction is spontaneous.

And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:

Eªcell > 0; n = 2