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3 years ago

We double both volume and absolute temperature of a given amount of an ideal gas.

Chemistry

1 answer:

Nitella [24]3 years ago

6 0

Answer:

In a nutshell, the pressure does not change.

Explanation:

Let assume that amount of moles is conserved, then following relationship is constructed from the Equation of State for Ideal Gas:

$\frac{P_{o}\cdot V_{o}}{T_{o}} = \frac{P_{f}\cdot V_{f}}{T_{f}}$

The final pressure is now cleared in the equation:

$P_{f} = \left(\frac{V_{o}}{V_{f}} \right)\cdot \left(\frac{T_{f}}{T_{o}} \right) \cdot P_{o}$

$P_{f} = \left(\frac{1}{2} \right)\cdot (2) \cdot P_{o}$

$P_{f} = P_{o}$

In a nutshell, the pressure does not change.

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PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank

Masja [62]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration equilibrium constant is $K_c = 14.39$

Explanation:

The chemical equation for this decomposition of ammonia is

$2 NH_3$ ↔ $N_2 + 3 H_2$

The initial concentration of ammonia is mathematically represented a

$[NH_3] = \frac{n_1}{V_1} = \frac{29}{75}$

$[NH_3] = 0.387 \ M$

The initial concentration of nitrogen gas is mathematically represented a

$[N_2] = \frac{n_2}{V_2}$

$[N_2] = 0.173 \ M$

So looking at the equation

Initially (Before reaction)

$NH_3 = 0.387 \ M$

$N_2 = 0 \ M$

$H_2 = 0 \ M$

During reaction(this is gotten from the reaction equation )

$NH_3 = -2 x$ (this implies that it losses two moles of concentration )

$N_2 = + x$ (this implies that it gains 1 moles)

$H_2 = +3 x$ (this implies that it gains 3 moles)

Note : x denotes concentration

At equilibrium

$NH_3 = 0.387 -2x$

$N_2 = x$

$H_2 = 3 x$

Now since

$[NH_3] = 0.387 \ M$

$x= 0.387 \ M$

$H_2 = 3 * 0.173$

$H_2 = 0.519 \ M$

$NH_3 = 0.387 -2(0.173)$

$NH_3 = 0.041 \ M$

Now the equilibrium constant is

$K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}$

substituting values

$K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}$

$K_c = 14.39$

3 0

3 years ago

A highly concentrated solution from which dilutions are typically made for laboratory used is called a what?

Komok [63]

A stock solution is the most concentrated

4 0

3 years ago

Please help ASAP. No one could not help me with this question.

Tanya [424]

Answer:

sorry

Explanation:

I don't know the answer this is really confusing but I am really sorry you have to do this.

3 0

3 years ago

What is the difference between a crystalline material and an amorphous material? Give an example of each type.

Igoryamba

Answer :

As we know that there are two types of solids.

(1) Amorphous Solids

(2) Crystalline Solids

Amorphous Solids : It is a type of solids in which the constituent particles of the matter are arranged in the random manner.

That means there is no proper arrangement of atoms in solid lattice but the atoms or molecules are closely spaced that means they can move freely from one place to another.

The examples of amorphous solid are, plastics, glass, rubber, metallic glass, polymers, gel, fused silica, pitch tar, thin film lubricants, wax.

Crystalline Solids : It is a type of solids where the constituent particles of the matter are arranged in the specific manner.

That means there is a proper arrangement of atoms in solid lattice. They do not have space between the molecules or atoms and they can not move freely from one place to another.

The examples of crystalline solids are, quartz, calcite, sugar, mica, diamonds, snowflakes, rock, calcium fluoride, silicon dioxide, alum.

7 0

3 years ago