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Korolek [52]
3 years ago
10

Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subj

ected to the following changes?
a. the catalyst is removed
b. the temp is increased
c. the volume is decreased
d. helium is added
e. CO is added
Chemistry
1 answer:
denis23 [38]3 years ago
3 0

Answer:

a. Methanol remains the same

b. Methanol decreases

c. Methanol increases

d. Methanol remains the same

e. Methanol increases

Explanation:

Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.

a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.

b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.

c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.

d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.

e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.

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Answer:

To find the average atomic mass, you take a certain number of atoms, find the total mass of each isotope, and then divide the total mass of all the atoms by the total number of atoms. Assume that you have, say, 10 000 atoms of carbon

Explanation:

5 0
3 years ago
How many moles of 0.225 M CaOH2 are present in 0.350 L of solution?
weeeeeb [17]

Answer : The number of moles of solute Ca(OH)_2 is, 0.0788 moles.

Explanation : Given,

Molarity = 0.225 M

Volume of solution = 0.350 L

Formula used:

\text{Molarity}=\frac{\text{Moles of }Ca(OH)_2}{\text{Volume of solution (in L)}}

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

0.225M=\frac{\text{Moles of }Ca(OH)_2}{0.350L}

\text{Moles of }Ca(OH)_2=0.0788mol

Therefore, the number of moles of solute Ca(OH)_2 is, 0.0788 moles.

7 0
3 years ago
According to molecular orbital theory, the regions of the wave function with the highest probability of finding electrons are ar
Vlada [557]

According to molecular orbital theory,  regions of  wave function with highest probability of finding electrons are areas with constructive interference.

An electron is a negatively charged subatomic particle that can exist either free or bound to an atom (not bound). A bound electron is one of the three primary types of particles that make up an atom, along with protons and neutrons. Protons, neutrons, and electrons combined make up the atom's nucleus. A proton's positive charge balances an electron's negative charge. When an atom has an equal number of protons and electrons, it is said to be in a neutral state. Electrons are distinct from other particles in a number of ways. They have a much lower mass, are found outside the nucleus, and exhibit both wave- and particle-like characteristics. The electron is a basic particle.

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6 0
1 year ago
A solution that is a 4 on pH scale is a
Dominik [7]
A pH scale runs from 1 to 14 with 7 being neutral.

1-6 has base like properties
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3 0
3 years ago
Read 2 more answers
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
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