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Sveta_85 [38]
2 years ago
12

50.0 g of nitrogen gas (N2 ) are kept under pressure in a 3.00 L container. If the pressure is kept constant, how many moles of

nitrogen gas are present in the container if gas is added until the volume has increased to 5.00 L?
Chemistry
1 answer:
kotykmax [81]2 years ago
7 0

The resultant moles of nitrogen gas that are present in the container is 2.96 moles.

<h3>How do we convert mass into moles?</h3>

Mass of any substance will be converted into moles by using the below equation as:

n = W/M, where

  • W = given mass of nitrogen gas = 50g
  • M = molar mass of nitrogen gas = 28 g/mol
  • n = 50 / 28 = 1.78 moles

From the ideal gas equation of gas we know that moles and volume is directly proportional to each other and for this question equation becomes,

V₁/n₁ = V₂/n₂, where

  • V₁ & n₁ are the initial volume and moles of gas
  • V₂ & n₂ are the final volume and moles of gas.

On putting all values, we get

n₂ = (5L)(1.78mol) / (3L) = 2.96 moles

Hence required moles of nitrogen gas are 2.96.

To know more about ideal gas equation, visit the below link:

brainly.com/question/1056445

#SPJ1

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A sample of gas occupies a volume of 67.1 mL . As it expands, it does 135.3 J of work on its surroundings at a constant pressure
Semmy [17]

Answer:

V_2=1.363x10^{-3}m^3=1363mL

Explanation:

Hello,

In this case, since the work done at constant pressure as in isobaric process is computed by:

W= P(V_2-V_1)

Thus, given the pressure, initial volume and work, the final volume is:

V_2=V_1+\frac{W}{P}

Whereas the pressure must be expressed in Pa as the work is given in J (Pa*m³):

P=783Torr*\frac{101325Pa}{760Torr} =104394Pa

And the volumes in m³:

V_1=67.1mL*\frac{1m^3}{1x10^6mL} =6.71x10^{-5}m^3

Thus, the final volume turns out:

V_2=6.71x10^{-5}m^3+\frac{135.3Pa*m^3}{104394Pa}\\\\V_2=1.363x10^{-3}m^3=1363mL

Best regards.

3 0
3 years ago
For the reaction, A(g) + B(g) =&gt; AB(g), the rate is 0.385 mol/L.s when the initial concentrations of both A and B are 2.00 mo
Tanya [424]

Answer : The rate for a reaction will be 0.14Ms^{-1}

Explanation :

The balanced equations will be:

A(g)+B(g)\rightarrow AB(g)

In this reaction, A and B are the reactants.

The rate law expression for the reaction is:

\text{Rate}=k[A]^2[B]^1

or,

\text{Rate}=k[A]^2[B]

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

\text{Rate}=k[A]^2[B]

0.385=k(2.00)^2(2.00)

k=0.0481M^{-2}s^{-1}

Now we have to calculate the initial rate for a reaction that starts with 1.48 M of reagent A and 1.32 M of reagents B.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(0.0481)\times (1.48)^2(1.32)^1

\text{Rate}=0.14Ms^{-1}

Therefore, the rate for a reaction will be 0.14Ms^{-1}

7 0
4 years ago
Which of the following is not an essential element for healthy plant growth?
GREYUIT [131]

Answer:

A oxygen

Explanation:

For plant growth nitrogen, potassium and phosphorous are the three main elements needed. oxygen is expelled from plants. Plants doesn't need oxygen

8 0
3 years ago
Is Neon a metal, nonmetal, or a metalliod
natulia [17]
Neon is a nonmetal.....

5 0
3 years ago
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Hydrogen gas and solid gold(III) sulfide are reacted. Pure solid gold metal is formed and gaseous hydrogen sulfide is released.
alexdok [17]

The reactants of the reaction will be solid gold (III) sulfide and hydrogen gas, while the products will be pure solid gold and hydrogen sulfide gas.

<h3>Chemical reactions</h3>

Reactants react together during reactions to arrive at products.

In this case, solid gold (III) sulfide and hydrogen gas react according to the following equation:

Au_2S_3(s) + 3 H_2 (g)-- > 2 Au(s) + 3H_2S(g)

Reactants = solid gold (III) sulfide, hydrogen gas

Products: pure solid gold, hydrogen sulfide gas

More on chemical reactions can be found here: brainly.com/question/1689737

8 0
2 years ago
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