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Assoli18 [71]
3 years ago
12

Calculate the mass of 120cc of nitrogen gas at ntp.How many number of molecules are present in it.Easy step of mole concept​

Chemistry
1 answer:
bekas [8.4K]3 years ago
6 0

Explanation:

The formula of nitrogen gas is given as;

          N₂

  Given volume of the gas = 120cm³

Now,

 We need to find the number of molecules and the mass of this compound

Solution:

At STP;

     1 mole of gas occupies a volume of 1dm³ ;

          1000cm³  = 1dm³

          120cm³  gives 0.12dm³;

Now;

       

              1dm ³ of gas contains 1 mole of substance at STP

             0.12dm³ will contain 0.12mole of nitrogen gas at STP

Mass of N₂ gas = number of moles x molar mass of N₂

    Molar mass of N₂ = 2(14)  = 28g/mol

    Mass of N₂  = 0.12 x 28  = 3.36g

1 mole of a substance contains 6.02 x 10²³ molecules

0.12 mole of N₂ will contain 0.12 x 6.02 x 10²³  = 7.22 x 10²²molecules

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Explanation:

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A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. a 100.0 mL volume of the so
Oksana_A [137]

Answer : The molar mass of unknown substance is, 39.7 g/mol

Explanation : Given,

Mass of unknown substance = 9.56 g

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Formula used :

\text{Molarity}=\frac{\text{Mass of unknown substance}\times 1000}{\text{Molar mass of unknown substance}\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

2.41M=\frac{9.56g\times 1000}{\text{Molar mass of unknown substance}\times 100.0mL}

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Therefore, the molar mass of unknown substance is, 39.7 g/mol

5 0
3 years ago
A city continuously disposes of effluent from a wastewater treatment plant into a river. The minimum flow in the river is 130 m3
Vera_Pavlovna [14]

Answer:

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Explanation:

C = Allowable concentration = 1.1 mg/L

Q_1 = Flow rate of river = 130\ \text{m}^/\text{s}

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C_1 = Background concentration = 0.69 mg/L

C_2 = Maximum concentration that of the pollutant

The concentration of the mixture will be

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The maximum concentration that of the pollutant (in mg/L) that can be safely discharged from the wastewater treatment plant is 2.54\ \text{mg/L}.

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