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NeX [460]
3 years ago
11

Describe in detail what you know about the enthalpy, entropy, and free energy changes when a sample of gas condenses to a liquid

. How does temperature affect these changes?.
Chemistry
2 answers:
ycow [4]3 years ago
8 0
The keyword here is gas condenses to a liquid, which mean we're talking about condensation process

The enthalpy energy in condensation process is negative because it releases energy
The entropy in general will also decreases .

Temperature affect this change because it will create free energy if added with this result

hope this helps
Vera_Pavlovna [14]3 years ago
6 0

When a gas condenses to liquid the process is exothermic with the release energy and the enthalpy change is negative. As liquid has more ordered structure than gas, the entropy, which is the measure of disorder decreases or ∆ S has a negative value.  

Free energy is given by the Gibb's free energy equation,  

∆G = ∆H –T∆S

For a process to be spontaneous ∆ G must be negative.

∆H is negative and ∆S is negative, so the magnitude of |T∆S| should be less than the magnitude of |∆H| in that way ∆G is always negative.

|T∆S| should be less than the magnitude of |∆H|, so to favor that T should be low. The process is spontaneous at a lower temperature than at a higher temperature.  

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What mass of H2 is needed to react with 8.75 g of O2 according to the following equation: O2(g) + H2(g) → H2O(g)?
Alika [10]

Answer:- B: 1.10g H_2 is the right answer.

Solution:- The balanced equation is:

O_2(g)+2H_2(g)\rightarrow 2H_2O(g)

We have been given with 8.75 grams of oxygen and asked to calculate the grams of hydrogen needed to react with given grams of oxygen according to the balanced equation.

From balanced equation, 1 mole of oxygen reacts with 2 moles of hydrogen.

So, let's convert grams of oxygen to moles and multiply it by the mole ratio to calculate the moles of hydrogen that are easily converted to grams on multiplying by it's molar mass.

The complete set up looks as:

8.75g O_2(\frac{1mole}{32g})(\frac{2mole H_2}{1mole O_2})(\frac{2.02g}{1mole})

= 1.10g H_2

Hence, the right option is B: 1.10g H_2 .



7 0
3 years ago
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The President's Challenge Program is a fitness evaluation designed for __________.
lara [203]
Hi there!

I believe the answer to this is B. Elementary and High school students.

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7 0
3 years ago
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A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and
lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

F_{solution} = mass of solvent + mass of oil (i.e A+C)

<u>Feed Phase:</u>

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

 = 25kg

Mass ratio of oil to solution Y_{F} =\frac{Mass C}{Mass (A+C)}

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane  in feed = 25 ×  0.9 =22.5kg + 2.5 =25kg

therefore  Y_{F} = \frac{2.5}{25}

= 0.1

mass ratio of solid to solution Y_{A}  =  \frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}

=3

<u>Solvent Phase:</u>

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

<u>Underflow:</u>

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = \frac{mass B}{mass(A+C)}

solution in underflow E₁ = Mass (A+C)

<u>Overflow:</u>

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75=  E₁ × N₁

75 =  E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0
3 years ago
One of these notations represents the definition of one mole. That is
just olya [345]

Answer:

6.02*10^23

Explanation:

This is the number for one mole. Just like one dozen = 12, one mole = 6.02*10^23.

Fun fact, if you had a mole of pennies you could spend 1 million dollars every second of your life and not have even spent 1% of it by the time you die at 100 years old.

3 0
4 years ago
How many moles of mercury(Hg) are in 1.30 *10.7 atoms or mercury
Svetach [21]
1 mole Hg --------------- 6.02x10²³ atoms
?? moles Hg ------------ 1.30x10⁷ atoms

(1.30x10⁷) x 1 / 6.02x10²³ => 2.159x10⁻¹⁷ moles 
5 0
3 years ago
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