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marta [7]
2 years ago
5

What volume will 6.745g of neon gas occupy at standard temp and pressure?

Chemistry
2 answers:
Snezhnost [94]2 years ago
6 0

Explanation:

It is given that mass of neon is 6.745 g and molar mass of neon is 20 g/mol.

Hence, calculate the number of moles as follows.

            No. of moles = \frac{mass}{\text{molar mass}}

Putting the given values into the above formula as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                                = \frac{6.745 g}{20 g/mol}

                                = 0.34 mol

Also, at standard conditions value of temperature is 298 K and pressure is 1 atm.

So, putting the values into ideal gas equation and calculate the value of volume as follows.

                          PV = nRT

          1 atm \times V = 0.34 mol \times 0.082 L atm/mol K /times 298 K

                  V = 8.31 L

Thus, we can conclude that volume of given neon gas is 8.31 L.

Thepotemich [5.8K]2 years ago
4 0
First, multiply the mass by the molar mass of neon to find out how many moles of neon there are. Then, multiply by 22.4 to find out how many liters there are.

6.745g Ne x 1 mole Ne/20 g Ne x 22.4 L/1 mole Ne = 7.5544 L
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Explanation:

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If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th
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Answer:

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be  the solubility of PbI₂.

The equation of the reaction is given as :

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 [Pb²⁺] =   s

Then [I⁻] = 2s

K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} =  4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of  }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

B)

The Concentration of Pb²⁺  in water is calculated as :

\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}

\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}

\mathbf{s} =\sqrt[3]{1.75*10^{-9}}

\mathbf{s} =\mathbf{1.21*10^{-3}  \ mol/L }

The Concentration of Pb²⁺  in 1.0 mol·L⁻¹ NaI

\mathbf{PbCl{_2}}  \leftrightharpoons    \ \ \ \ \ \ \  \mathbf{Pb^{2+}}   \ \ \ \  \ +   \ \  \ \ \ \ \ \mathbf{2 I^-}

                             \ \ \ \ \ \ \  \ \   \ \  \ \ \ \ \ \ \  \mathbf0}   \ \ \ \  \ \ \ \ \ \   \ \ \ \ \ \mathbf{1.0}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{+2x}

                            \ \ \ \ \ \ \ \ \ \ \ \ \ \    \ \ \ \ \  \mathbf{+x}   \ \ \ \  \    \ \  \ \ \ \ \ \mathbf{1.0+2x}

The equilibrium constant:

K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \  m/L

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the  common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

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The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment

Answer:

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