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Temka [501]
3 years ago
12

What is the mass of 2.15 liters of N2 gas at STP?

Chemistry
1 answer:
devlian [24]3 years ago
6 0

Answer:

2.68g

Explanation:

Given parameters:

Volume of gas at STP = 2.15L

Unkown:

Mass of Nitrogen N₂ gas at STP = ?

Solution

To find the mass of the gas at STP, we use the mole concept approach.

In using the mole approach we follow the following procedures:

1. Find the number of moles of gas at STP using the relationship below:

                  Number of moles = \frac{volume occupied}{22.4dm^{3} mol^{-1} }

                         Note: 1L = 1dm³

2. Then using the mole and mass relationship, we can find the mass of the gas using the equation below:

                Mass = number of moles x molar mass

Workings

1.              Number of moles of N₂ = \frac{2.15}{22.4}

                Number of moles of N₂ = 0.096mole

2. Given that the atomic mass of N = 14g

                       Molar mass of N₂ = 2 x 14 = 28gmol⁻¹

Mass = 0.096mol x  28gmol⁻¹ = 2.68g

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3 years ago
Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
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Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

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From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.

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Which description refers to cirrus clouds?
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Answer:

b

Explanation:

8 0
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vova2212 [387]

Answer: Velocity

Explanation:

4 0
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