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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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<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M

where,

n_p = number of protons

m_p = mass of one proton

n_n = number of neutrons

m_n = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is _{15}^{31}\textrm{P}

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Putting values in above equation, we get:

\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399

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3 years ago
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The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

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The table is attached below as an image.

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Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

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a = order with respect to A

b = order with respect to B

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5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

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Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

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Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

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[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

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