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allsm [11]
2 years ago
12

A drinking water supply is suspected of being contaminated with lead. Samples of water aspirated directly into an air-acetylene

flame gave an absorbance of 0.68 at 283.3nm. Standard solutions containing 0.5 and 1.0 ppm were found to exhibit absorbances of 0.43 and 0.86 respectively. Assuming the Beer-Lambert law is obeyed calculate the concentration of lead within the water sample. A. 79. Oppm B. 0.79ppm C. 0.43ppm D. 0.96ppm E. 0.92ppm noeullib noilgoeps noods Whaooah
Chemistry
1 answer:
kifflom [539]2 years ago
3 0

Answer:

B. 0.79ppm

Explanation:

According to the Beer-Lambert law, there is a linear relationship between the absorbance (A) of an analyte and its concentration (c). We can represent it through the following expression.

A = k . c

where

k is a proportionality constant, and the slope of the equation.

Let's consider the ordered-pairs (0.5 ppm, 0.43) and (1.0 ppm, 0.86). The slope is:

k = ΔA / Δc = 0.86 - 0.43 / 1.0 ppm - 0.5 ppm = 0.86 ppm⁻¹

When A = 0.68

0.68 = 0.86 ppm⁻¹ c

c = 0.79 ppm

You might be interested in
HELP ASAP PLEASE!!!
sveticcg [70]
1. Pure substances cannot be separated into any other kinds of matter, while a mixture is a combination of two or more pure substances.

2. A pure substance has constant physical and chemical properties, while mixtures have varying physical and chemical properties (i.e., boiling point and melting point).

3. A pure substance is pure, while a mixture is impure.

4 0
3 years ago
What is the molar out of a solution that contains 33.5g of CaCl2 in 600.0mL of water
omeli [17]

Answer:

Here's what I got.

Explanation:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

Cl

, in calcium chloride, This will help you calculate the mass of chloride anions,

Cl

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

1

mole of calcium chloride contains

2

moles of chlorine atoms.

2

×

35.453

g mol

−

1

110.98

g mol

−

1

⋅

100

%

=

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

Cl

−

.

This implies that your sample contains

0.543

g CaCl

2

⋅

63.89 g Cl

−

100

g CaCl

2

=

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

=

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

.

ALTERNATIVE APPROACH

Alternatively, you can start by calculating the number of moles of calcium chloride present in your sample

0.543

g

⋅

1 mole CaCl

2

110.98

g

=

0.004893 moles CaCl

2

To find the molarity of this solution, calculate the number of moles of calcium chloride present in

1 L

=

10

3

mL

of solution by using the fact that you have

0.004893

moles present in

500 mL

of solution.

10

3

mL solution

⋅

0.004893 moles CaCl

2

500

mL solution

=

0.009786 moles CaCl

2

You can thus say your solution has

[

CaCl

2

]

=

0.009786 mol L

−

1

Since every mole of calcium chloride delivers

2

moles of chloride anions to the solution, you can say that you have

[

Cl

−

]

=

2

⋅

0.009786 mol L

−

1

[

Cl

−

]

=

0.01957 mol L

−

This implies that

100 mL

of this solution will contain

100

mL solution

⋅

0.01957 moles Cl

−

10

3

mL solution

=

0.001957 moles Cl

−

Finally, to convert this to grams, use the molar mass of elemental chlorine

0.001957

moles Cl

−

⋅

35.453 g

1

mole Cl

−

=

0.06938 g Cl

−

Once again, you have

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

In reference to the explanation you provided, you have

0.341 g L

−

1

=

0.0341 g/100 mL

=

0.0341% m/v

because you have

1 L

=

10

3

mL

.

However, this solution does not contain

0.341 g

of chloride anions in

1 L

. Using

[

Cl

−

]

=

0.01957 mol L

−

1

you have

n

=

c

⋅

V

so

n

=

0.01957 mol

⋅

10

−

3

mL

−

1

⋅

500

mL

n

=

0.009785 moles

This is how many moles of chloride anions you have in

500 mL

of solution. Consequently,

100 mL

of solution will contain

100

mL solution

⋅

0.009785 moles Cl

−

500

mL solution

=

0.001957 moles Cl

−

So once again, you have

0.06938 g

of chloride anions in

100 mL

of solution, the equivalent of

0.069% m/v

.

Explanation:

i think this is it

8 0
2 years ago
What did Ernest Rutherford s gold foil experiment demonstrate about atoms?
Ghella [55]
-Positively charged nucleus 
-Empty spaced
-Dense core
5 0
3 years ago
Why is elemental zinc considered a pure substance?
Veseljchak [2.6K]

Answer:  C... Because Zinc is made of only one type of atoms.

Explanation: All elements are made of one type of atom. Pure gold is made out of only gold atoms, silver out of only silver atoms, etc.

7 0
3 years ago
A 44% wt/wt solution of H2SO4 has a density of 1.343g/ml. What mass of H2SO4 is 60ml of this solution?
horsena [70]

<u>Answer:</u> The mass of sulfuric acid present in 60 mL of solution is 34.1 grams

<u>Explanation:</u>

We are given:

44 % (m/m) solution of sulfuric acid. This means that 44 grams of sulfuric acid is present in 100 grams of solution.

To calculate volume of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.343 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

1.343g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.343g/mL}=77.46mL

To calculate the mass of sulfuric acid present in 60 mL of solution, we use unitary method:

In 77.46 mL of solution, mass of sulfuric acid present is 44 g

So, in 60 mL of solution, mass of sulfuric acid present will be = \frac{44g}{77.46mL}\times 60mL=34.1g

Hence, the mass of sulfuric acid present in 60 mL of solution is 34.1 grams

3 0
3 years ago
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