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Ann [662]
3 years ago
10

Doreen Schmidt is a chemist. She needs to prepare 28 ounces of a 12​% hydrochloric acid solution. Find the amount of 14​% soluti

on and the amount of 7​% solution she should mix to get this solution.
Chemistry
1 answer:
Tpy6a [65]3 years ago
8 0

Answer:

20 ounces of 14% solution and 8 ounces of 7% solution.

Explanation:

Let the amount of 14% solution required be A and the amount of 7% solution required be B.

Mixture of A and B must give 28 ounces:

A + B = 28 ................. Eqn 1

Also, mixture of 14% A and 7% B must give 12%, 28 ounces solution:

0.14A + 0.07B = 0.12(28) = 3.36...................Eqn 2

From eqn 1, B = 28 - A, eqn 2 becomes:

0.14A + 0.07(28 - A) = 3.36

0.07A = 3.36 - 1.96

A = 20 ounces

Substitute the value of A into eqn 1 to get B:

20 + B = 28

B = 8 ounces.

Hence, 20 ounces of 14% hydrochloric acid and 8 ounces of 7% hydrochoric acid should be mixed together in order to get 28 ounces of 12% solution.

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The natural distribution of the isotopes of a hypothetical element is 60.795% at a mass of 281.99481 u, 22.122% at a mass of 283
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<u>Answer:</u> The average atomic mass of the element is 283.291 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

  • <u>For isotope 1:</u>

Mass of isotope 1 = 281.99481 amu

Percentage abundance of isotope 1 = 60.795 %

Fractional abundance of isotope 1 = 0.60795

  • <u>For isotope 2:</u>

Mass of isotope 2 = 283.99570 amu

Percentage abundance of isotope 2 = 22.122 %

Fractional abundance of isotope 2 = 0.22122

  • <u>For isotope 3:</u>

Mass of isotope 3 = 286.99423 amu

Percentage abundance of isotope 3 = [100 - (60.795 + 22.122)] = 17.083 %

Fractional abundance of isotope 1 = 0.17083

Putting values in equation 1, we get:

\text{Average atomic mass of element}=[(281.99481\times 0.60795)+(283.99570\times 0.22122)+(286.99423\times 0.17083)]\\\\\text{Average atomic mass of element}=283.291amu

Hence, the average atomic mass of the element is 283.291 amu

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4 years ago
Name one Importance of solar power
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Because the sun is a sustainable source of energy that can be used to power homes and businesses globally. 

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A student heats up a substance from 100 C to 145 C, which requires thermal energy of 20000 J. If the mass of the chemical is 100
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Answer:

THE SPECIFIC HEAT CAPACITY IS 4.44 J/G C

Explanation:

Heat energy = 20000 J

Mass = 100 g

Change in temperature = 145 C - 100 C = 45 C

Specific heat capacity = unknown

In calculating the specific heat capacity, we use the formula:

Heat = mass * specific heat capacity * change in temperature

Re-arranging the formula by making the specific heat capacity the subject of the equation, we have:

Specific heat capacity = Heat / mass * change in temperature

Specific heat capacity = 20000 J / 100 g * 45 C

Specific heat = 20000 / 4500

Specific heat = 4.44 J/g C

The specific heat capacity of the substance is therefore 4.44 J/g C

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3 years ago
The natural gas in a storage reservoir, under a pressure of 1.00 atmosphere, has a volume of 2.74 × 109 L at 20.0°C. The tempera
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Answer : The final volume of gas will be, 2.36\times 10^9L

Explanation :

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

V\propto T

or,

\frac{V_1}{V_2}=\frac{T_1}{T_2}

where,

V_1 = initial volume of gas = 2.74\times 10^9L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 20.0^oC=273+20.0=293K

T_2 = final temperature of gas = -20.0^oC=273+(-20.0)=253K

Now put all the given values in the above formula, we get the final volume of the gas.

\frac{2.74\times 10^9L}{V_2}=\frac{293K}{253K}

V_2=2.36\times 10^9L

Therefore, the final volume of gas will be, 2.36\times 10^9L

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