1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Ksenya-84 [330]
3 years ago
15

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar

d Fe3 solution with a concentration of 6.35 Ă— 10-4 M. You treat 10.0 mL of this reference with HNO3 and excess KSCN to form a red complex, and dilute the reference to 55.0 mL. The diluted reference is placed in a cell with a 1.00-cm light path. You then take 35.0 mL of the well water, treat with HNO3 and excess KSCN, and dilute to 100.0 mL. This diluted sample is placed in a variable pathlength cell. The absorbance of the reference and the sample solutions match when the pathlength is 4.40 cm. What is the concentration of iron in the well water? For each solution, the zero is set with a blank.
Chemistry
1 answer:
torisob [31]3 years ago
7 0

Based on Beer-Lambert's Law,

A = εcl ------(1)

where A = absorbance

ε = molar absorptivity

c = concentration

l = path length

Step 1: Calculate the concentration of the diluted Fe3+ standard

Use:

V1M1 = V2M2

M2 = V1M1/V2 = 10 ml*6.35*10⁻⁴M/55 ml = 1.154*10⁻⁴ M

Step 2 : Calculate the concentration of the sample solution

Based on equation (1) we have:

A(Fe3+) = ε(1.154*10⁻⁴)(1)

A(sample) = ε(C)(4.4)

It is given that the absorbances match under the given path length conditions, i.e.

ε(1.154*10⁻⁴)(1) = ε(C)(4.4)

C = 0.262*10⁻⁴ M

This is the concentration of Fe3+ in 100 ml of well water sample

Step 3: Calculate the concentration of Fe3+ in the original sample

Use V1M1 = V2M2

M1 = V2M2/V1 = 100 ml * 0.262*10⁻⁴ M/35 ml = 7.49*10⁻⁵M

Ans: Concentration of F3+ in the well water sample is 7.49*10⁻⁵M


You might be interested in
Please help I can't balance and Google isn't helping
DiKsa [7]
The first one is right and so is eight you have to add them together to see if they go together if they dont add what is missing
5 0
3 years ago
How many moles of atoms are in 5.00 g of 13C
Illusion [34]
  <span>Mr = 13 g / mol 
mass = 5 g 
Mol = 5/13 mol :)</span>
7 0
3 years ago
Barium phosphide Ba3P2 BaP Ba3PO4
Novosadov [1.4K]

Answer:

It A(Ba3P2)

Explanation:

I took the test

6 0
3 years ago
Please help!! i need the answer ASAP!
nirvana33 [79]

Answer:

23.34 %.

Explanation:

  • The percentage of water must be calculated as a mass percent.
  • We need to find the mass of water, and the total mass in one mole of the compound. For that we need to use the atomic masses of each element and take in consideration the number of atoms of each element in the formula unit.
  • <em>Atomic masses of the elements:</em>

Cd: 112.411 g/mol, N: 14.0067 g/mol, O: 15.999 g/mol, and H: 1.008 g/mol.

  • <em>Mass of the formula unit:</em>

Cd(NO₃)₂•4H₂O

mass of the formula unit = (At. mass of Cd) + 2(At. mass of N) + 10(At. mass of O) + 8(At. mass of H) = (112.411 g/mol) + 2(14.0067 g/mol) + 10(15.999 g/mol) + 8(1.008 g/mol) = 308.5 g/mol.

  • <em> Mass of water in the formula unit:</em>

<em>mass of water</em> = (4 × 2 × 1.008 g/mol) + (4 × 15.999 g/mol) = 72.0 g/mol.

  • <em>So, the percent of water in the compound = [mass of water / mass of the formula unit] × 100 = [(72.0 g/mol)/(308.5 g/mol)] × 100 = 23.34 %</em>
7 0
3 years ago
The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.
adelina 88 [10]

Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol

Now we have to calculate the energy required.

Q=\frac{\Delta H}{n}

where,

Q = energy required

\Delta H = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

7 0
3 years ago
Other questions:
  • Circle the letter of each expression that has four significant figures. <br> a. 1.25 104 b. 12.51
    15·1 answer
  • Which type of reaction is caused by particle accelerators?
    9·2 answers
  • A 10.0 kg crate is pushed with a horizontal force of 40 N. The crate moves at a constant velocity of 3.0 m/s. What is the value
    14·1 answer
  • Why do people use symbols and numbers instead of equations in chemistry?
    14·1 answer
  • Which of the following examples illustrates a number that is correctly rounded to three significant figures? Which of the follow
    14·1 answer
  • What is the temperature of 0.45 moles of Neon gas at 1920 mm Hg if it occupies 22,400 ml of volume?
    13·1 answer
  • Biosphere 2 was an attempt to have humans live within a
    9·1 answer
  • (07.01 MC)
    11·1 answer
  • Calculate the average atomic mass for X
    15·1 answer
  • (d) How many grams of CaO can be produced?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!