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Viktor [21]
3 years ago
11

Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.

Chemistry
1 answer:
Cerrena [4.2K]3 years ago
5 0

Answer:

Answer :

The total pressure inside the cylinder is, 6.7 atm

The mole fraction of N_2 in the mixture is, 0.52

Solution :

First we have to calculate the total pressure inside the cylinder.

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

P_T=p_{N_2}+p_{O_2}+p_{Ar}+p_{He}

Now put all the given values is expression, we get the total pressure inside the cylinder.

P_T=3.5+2.8+0.25+0.15=6.7atm

Now we have to calculate the mole fraction of N_2 in the mixture.

Formula used :

pN_2=XN_2 x P_T

where,

P_T = total pressure = 6.7 atm

pN_2 = partial pressure of nitrogen gas = 3.5 atm

XN_2 = mole fraction of nitrogen gas = ?

Now put all the given values in the above formula, we get

3.5atm=XN_2 x 6.7atm

XN_2=0.52

Therefore, the total pressure inside the cylinder is, 6.7 atm and the mole fraction of N_2 in the mixture is, 0.52

Hope it helps answer the question:)

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C

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In a voltaic (galvanic) cell, oxidation occurs at the _____ and is where _____ in the salt bridge moves toward.
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In a voltaic (galvanic) cell, oxidation occurs at the <u>anode</u> and is where <u>anions</u> in the salt bridge moves toward.

<h3>What is Galvanic Cell ?</h3>

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Thus from the above conclusion we can say that In a voltaic (galvanic) cell, oxidation occurs at the <u>anode</u> and is where <u>anions</u> in the salt bridge moves toward.

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4 0
2 years ago
Suppose of barium acetate is dissolved in of a aqueous solution of ammonium sulfate. Calculate the final molarity of barium cati
ollegr [7]

Explanation:

Let us assume that the given data is as follows.

   mass of barium acetate = 2.19 g

   volume = 150 ml = 0.150 L    (as 1 L = 1000 ml)

   concentration of the aqueous solution = 0.10 M

Therefore, the reaction equation will be as follows.

        Ba(C_{2}H_{3}O_{2})_{2} \rightarrow Ba^{2+} + 2C_{2}H_{3}O^{-}_{2}

Hence, moles of C_{2}H_{3}O^{-}_{2} = 2 \times Ba(C_{2}H_{3}O_{2})_{2}  .......... (1)

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

Hence, moles of Ba(C_{2}H_{3}O_{2})_{2} will be calculated as follows.                          

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

                          =  \frac{2.19 g}{255.415 g/mol}   (molar mass of Ba(C_{2}H_{3}O_{2})_{2} is 255.415 g/mol)            

                       = 8.57 \times 10^{-3}

    Moles of C_{2}H_{3}O^{-}_{2} = 2 \times 8.57 \times 10^{-3}

                          = 0.01715 mol

Hence, final molarity will be as follows.

              Molarity = \frac{\text{no. of moles}}{volume}

                             = \frac{0.01715 mol}{0.150 L}

                             = 0.114 M

Thus, we can conclude that final molarity of barium cation in the solution is 0.114 M.

5 0
3 years ago
The 85.2-g sample of the compound x4o10 contains 48.0 g of oxygen atoms. what is the molar mass of element x?
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so, 48 = 16 x 10 x [85.2 / (4y+160) ]

Solve the equation to get y.

y = 31
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Hope it helps.........:)
3 0
3 years ago
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