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zysi [14]
3 years ago
9

A 8.65-L container holds a mixture of two gases at 11 °C. The partial pressures of gas A and gas B, respectively, are 0.205 atm

and 0.658 atm. If 0.200 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
Chemistry
1 answer:
Vsevolod [243]3 years ago
6 0

 The  total pressure  = 1.402 atm


<u><em>calculation</em></u>

Total  pressure = partial  pressure  of gas A + partial pressure of gas B +  partial pressure  of third gas

partial  pressure  of gas A= 0.205 atm

Partial pressure of gas B =0.658 atm


partial pressure for third gas is calculated using ideal  gas equation

that is PV=nRT   where,

p(pressure)=? atm

V(volume) = 8.65 L

n(moles)= 0.200 moles

R(gas constant)=0.0821 L.atm/mol.k

T(temperature) = 11°c into kelvin =11+273 =284 k

make  p the subject of the formula by  diving both side by V

p =nRT/v


p = [(0.200 moles x 0.0821 L.atm/mol.K x 284 K)/8.65L)] =0.539 atm



Total  pressure  is therefore = 0.205 atm +0.658 atm +0.539 atm

=1.402 atm

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Answer:

The coefficients of the balanced equation are:

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Explanation:

The Law of Conservation of Matter, also called the Law of Conservation of Mass or the Lomonósov-Lavoisier Law, postulates that "mass is neither created nor destroyed, it is only transformed". This means that the reagents interact with each other and form new products with different physical and chemical properties than the reagents, but the amount of matter or mass before and after a transformation (chemical reaction) is always the same. In other words, then the mass before the chemical reaction is equal to the mass after the reaction.

Then, you must balance the following chemical equation:

v Cu + w HNO₃ → x Cu(NO₃)₂ + y NO + z H₂O

For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts

  • Left side:   1 copper, 1 hydrogen, 1 nitrogen and 3 oxygen.
  • Right side:  1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

Generally, hydrogen and oxygen balance in the end. So you balance nitrogen first, because copper is already balanced (there is the same amount on both sides of the reaction). If w = 2, then:

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But you see then that the oxygen is unbalanced and you have less quantity in the reagents. Then w must be greater. Being w = 4:

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  • Right side:  1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

Going back to the idea of ​​balancing nitrogen, being y = 2:

  • Left side:   1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
  • Right side:  1 copper, 2 hydrogen, 4 nitrogen and 9 oxygen.

Balancing the hydrogen, being z = 2:

  • Left side:   1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
  • Right side:  1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.

Since you have the same amount of each element on each side of the reaction, the reaction is balanced. Then, the balanced equation is:

Cu + 4 HNO₃ → Cu(NO₃)₂ +  2 NO + 2 H₂O

Finally:

  • <u><em>v= 1</em></u>
  • <u><em>w= 4</em></u>
  • <u><em>x= 1</em></u>
  • <u><em>y= 2</em></u>
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