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Olegator [25]
3 years ago
7

Decreasing the volume of a gas would _______________. A. Decrease the temperature. B. Increase the amount liters. C. Decrease th

e moles. D. Increase the pressure.
Chemistry
2 answers:
Mandarinka [93]3 years ago
6 0

Answer:

<u>D. Increase the pressure.</u>

Explanation:

Okay, lets break this down.

So first, lets imagine what decreasing the volume looks like. I like to think of it as squishing a balloon or water bottle. You condense what's inside it. So, with this in mind, lets look at the question.

A. Decrease temperature:

Does this happen when you squeeze something? It gets colder? No, that's not true. So we can eliminate that one.

B. Increase the liters:

Does this happen when you squeeze something? It gets larger? It takes up more space? No, that's not true. So we can eliminate that one.

C. Decrease in moles:

Okay, so moles is just a fancy way of saying "stuff." So does the amount of "stuff" decrease when you squeeze something? No, it just gets closer together. It's denser. So we can eliminate that one.

D. Increase the pressure:

Now we're getting somewhere!! When you squeeze a balloon, the pressure DOES increase because you are condensing the gas inside it. That's why balloons will pop if you squeeze too hard, the pressure is too great and has to escape. So yes, this is your answer.

<u>Answer: D. Increase the pressure</u>

Allisa [31]3 years ago
6 0

Answer:

Increases the pressure

Explanation:

Answer to founders education

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Write the structure of the product that would be formed from the S(
prisoha [69]

Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.

The product formed on reaction with water would be a 50:50 mixture of

2S-hexane-2-ol. and 2R-hexane-2-ol.

Explanation:

2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good  nucleophile .

The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction  takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.

In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.

When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .

The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.

The SN¹ reaction is a 2 step reaction , in  the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.

In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.

The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.

Kindly refer the attachment for reaction mechanism and structure of products.

8 0
3 years ago
Go'=30.5 kJ/mol
makvit [3.9K]

Answer:

a) Keq = 4.5x10^-6

b) [oxaloacetate] = 9x10^-9 M

c) 23 oxaloacetate molecules

Explanation:

a) In the standard state we have to:

ΔGo = -R*T*ln(Keq) (eq.1)

ΔGo = 30.5 kJ/moles = 30500 J/moles

R = 8.314 J*K^-1*moles^-1

Clearing Keq:

Keq = e^(ΔGo/-R*T) = e^(30500/(-8.314*298)) = 4.5x10^-6

b) Keq = ([oxaloacetate]*[NADH])/([L-malate]*[NAD+])

4.5x10^-6 = ([oxaloacetate]/(0.20*10)

Clearing [oxaloacetate]:

[oxaloacetate] = 9x10^-9 M

c) the radius of the mitochondria is equal to:

r = 10^-5 dm

The volume of the mitochondria is:

V = (4/3)*pi*r^3 = (4/3)*pi*(10^-15)^3 = 4.18x10^-42 L

1 L of mitochondria contains 9x10^-9 M of oxaloacetate

Thus, 4.18x10^-42 L of mitochondria contains:

molecules of oxaloacetate = 4.18x10^-42 * 9x10^-9 * 6.023x10^23 = 2.27x10^-26 = 23 oxaloacetate molecules

3 0
4 years ago
In calculating the concentration of [Cu(NH3)4]2+ from [Cu(H2O)4]2+, the stepwise formation constants are as follows: K1=1.90×104
pashok25 [27]

Answer:

Kf = 1.11x10¹³

Explanation:

The value of Kf for a multistep process that involves an equilibrium at each step, is the multiplication of the constant of the equilibrium of each step.

Kf = K1xK2xK3xK4

Kf = 1.90x10⁴ x 3.90x10³ x 1.00x10³ x 1.50x10²

Kf = 1.11x10¹³

4 0
3 years ago
. Hemoglobin (Hb) and oxygen gas form a complex (HbO2) that carries oxygen throughout the human body. Unfortunately, carbon mono
sp2606 [1]

Answer:

1. Removing them to an area of fresh air. This helps to prevents further poisoning by the carbon monoxide and increase the amount of oxygen entering into the body. This will help to reduce the concentration of carbon monoxide binding oxygen

2. Administering pure oxygen goes a long way to enhance ventilation and increase the oxygen saturation to 100%. This will help to overcome the effect of the carbon monoxide and promote more hemoglobin binding

5 0
3 years ago
Consider the reaction below. Mg(s) + 2HCl(aq) mc019-1.jpg H2(g) + MgCl2(aq) What is the most likely effect of an increase in pre
bonufazy [111]
1) Reaction:

Mg(s) + 2HCl (aq) ----> H2(g) + MgCl2(aq)

2) Analysis

An increase in pressure affects directly the rate of reaction involving reactiong gases. Changing the pressure where there are only solids or liquids does not affect the rate of reaction.

This reaction is not an equilibrium, the reaction is only forward. So, the reacting components, Mg(s) and HCl(aq) are a solid and a liquid.

Therefore, the reaction is not affected by the change in pressure.

Answer: the reaction is not affected at all.


4 0
4 years ago
Read 2 more answers
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