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GarryVolchara [31]
3 years ago
14

An unknown aqueous metal analysis yielded a detector response of 0.255. When 1.00 mL of a solution containing 100.0 ppm of the m

etal was mixed with 99.0 mL of the unknown, the detector signal increased to 0.502. Calculate the concentration of the metal in the unknown solution. Report your answer in ppm
Chemistry
1 answer:
AVprozaik [17]3 years ago
3 0

Answer:

1.022ppm is the unknown concentration of the metal

Explanation:

Based on Lambert-Beer law, the increasing in signal of a detector is directly proportional to its concentration.

The unknown concentration (X) produces a signal of 0.255

99mL * X + 1mL * 100ppm / 100mL produces a signal of 0.502

0.99X + 1ppm produce 0.502, thus, X is:

0.255 * (0.99X + 1 / 0.502) =

X = 0.503X + 0.508

0.497X = 0.508

X =

1.022ppm is the unknown concentration of the metal

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A certain compound has the following percent composition: 57.1% C, 4.8% H, and 38.l% O.
crimeas [40]

Answer:

C₆H₆O₃

Explanation:

Calculation sequence:

% => grams => moles => reduce => empirical Ratio

Molecular multiple = Molecular Mass / Empirical Mass

  C: => 57.1% => 57.1 g => 57.1/12 = 4.7583

  H: =>  4.8% =>  4.8 g =>   4.8/1  = 4.8000

  O: => <u>38.1% => 38.1 g </u>=> 38.1/16 = 2.3813

TTL => 100%       100 g

Reduced Mole values =>

C : H : O => 4.7583/2.3813 : 4.8000/2.3813 : 2.3813/2.3813 => 2 : 2 :  1

∴ empirical formula => C₂H₂O

empirical formula weight => 2C + 2H + 1O = [2(12) + 2(1) + 1(16)] amu = 42 amu

molecular formula weight (given in problem) = 126 g/mole

The molecular formula is a whole number multiple of the empirical formula.

molecular multiple = 126 amu / 42 amu = 3

∴ molecular formula => (C₂H₂O)₃ => C₆H₆O₃

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The slant of the earth is called what
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I think the slant would be called the tilt
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How is energy from grass passed on between animals
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An animal will eat the grass and the energy then gets passed through the food chain

Explanation:

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What is the chemical name of the compound
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Aluminum has a specific heat of 0.902 j/gC. How much heat is lost when a piece of aluminum with a mass of 23.984 g cools from a
Marizza181 [45]

Answer:

Q = - 8501.99 j

Explanation:

Given data:

Specific heat of Al = 0.902 j/g.°C

Heat lost = ?

Mass of sample = 23.984 g

Initial temperature = 415°C

Final temperature = 22°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 22°C - 415°C

ΔT = -393°C

Q = m.c. ΔT

Q = 23.984 g× 0.902 j/g.°C × -393°C

Q = - 8501.99 j

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