All categories

2 years ago

What is the pressure of 0.5 mol nitrogen (N2) gas in a 5.0 L container at 203 K?

Chemistry

1 answer:

Yuki888 [10]2 years ago

6 0

Answer:

The pressure of 0.5 mole of nitrogen gas= $168.78 k pa$

Given:

Moles of nitrogen gas=0.5mole

Volume of nitrogen gas=5L

Temperature of nitrogen gas=203K

To Find:

Pressure of 0.5g mole of nitrogen gas

Step by Step by Explanation:

Formula used for calculating the pressure of a gas is achieved through ideal gas equation

According to Ideal Gas;

$P V=n R T$

From this above equation, pressure can be calculated as

$P=\frac{n R T}{V}$

Where P=Pressure of the gas

N=number of moles=0.5moles

R=Gas constant $=8.3144598 L k p_{a} k^{-1} m o l^{-1}$

V=Volume of nitrogen gas in litres=5L

T=Temperature of nitrogen gas=203K

Substitute all the values in the above equation we get

$P=\frac{0.5 \times 8.3144598 \times 203}{5}$

$P=\frac{843.9177}{5}$

$P=168.78 k pa$

Result:

Thus the pressure of 0.5 mole of nitrogen gas is $168.78 k pa$

You might be interested in

4. How does the kinetic theory of gases explain the weather changes happening in the troposphere? You can also research on the I

uranmaximum [27]

According to the kmt pressure is directly proportional to the number of collision between particles

4 0

2 years ago

Read 2 more answers

Some instant cold packs contain ammonium nitrate and a separate pouch of water. When the pack is activated by squeezing to break

lubasha [3.4K]

Answer:

$T_f=-7.81^0C$

Explanation:

Hello,

a) In this case, since the heat associated with the dissolution of ammonium nitrate is positive, such reaction is endothermic as it absorbs heat.

b) Now, for computing the temperature once the dissolution is done, we apply (considering that it is a cooling process):

$T_f=T_0-\frac{\Delta H}{mCp}$

Nonetheless, we should first compute the moles of the mixture as:

$n_{mix}=135.0gH_2O*\frac{1molH_2O}{18gH_2O}+50.0gNH_4NO_3*\frac{1molNH_4NO_3}{80gNH_4NO_3}=8.125mol$

Thus, the total absorbed heat is:

$\Delta H=25.4kJ/mol*8.125mol=206.375kJ$

Now, the temperature is:

$T_f=25.0^0C-\frac{25.4kJ}{(135.0+50.0)g*4.184x10^{-3}kJ/g^0C} \\\\T_f=-7.81^0C$

Best regards.

3 0

2 years ago

Ethylene (C2H4) is the starting material for the preparation of polyethylene. Although typically made during the processing of p

Solnce55 [7]

Answer:

1.17 grams

Explanation:

Let's consider the balanced equation for the combustion of ethylene.

C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)

We can establish the following relations:

1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
The molar mass of C₂H₄ is 28.05 g/mol.

The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

$-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}$

8 0

2 years ago

Which of the following are examples of plasma?

artcher [175]

The answers are "tails of comets, the ionosphere, and a neon sign." or options A, D, and E. Tails of comments are mainly made of plasma. Ice cubes are examples of a solid. A gas fire key word gas is example of a gas. The ionosphere is also mainly made of plasma. A neon sign uses plasma in order to work. A flashlight uses a light bulb which isn't a example of plasma therefore thats a no.

Hope this helps!

4 0

2 years ago

Read 2 more answers

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

fredd [130]

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:

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

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

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⁺.

3 0

2 years ago