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2 years ago

Arrange these reactions according to increasing ΔS.

Chemistry

1 answer:

Gennadij [26K]2 years ago

3 0

Explanation:

ΔS denotes change in entropy. Entropy is the degree of disorderliness of a system. An increasing ΔS would mean that the entropy of the products is greater than the entropy of the reactants.

Generally the trend of entropy with state of matter is given as;

Gas > Liquid > Solid (Increasing Entropy)

a. H 2O(g) → H 2O(l)

In this reaction, there is a decreasing ΔS, since we are moving from gas to liquid. This would be the last.

b. 2NO(g) → N 2(g) + O 2(g)

There is an increase in gaseous products. This is definitely an increasing ΔS reaction. This would be the first on the list.

C. MgCO 3(s) → MgO(s) + CO2(g)

The products formed are in solid and gaseous state. There is increasing ΔS but not up to reaction 2.

The order is given as;

b. 2NO(g) → N 2(g) + O 2(g)

C. MgCO 3(s) → MgO(s) + CO2(g)

a. H 2O(g) → H 2O(l)

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A reaction container holds 5.77 g of P4 and 5.77 g of O2.

Dvinal [7]

Answer:

a) O2 is the limiting reactant

b) 5.75 grams P4O10

c) 5.79 grams P4O6

Explanation:

Step 1: Data given

Mass of P4 = 5.77 grams

Mass of O2 = 5.77 grams

Molar mass of P4 = 123.90 g/mol

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

P4 + 3O2 → P4O6

Step 3: Calculate moles of P4

Moles P4 = mass P4 / molar mass P4

Moles P4 = 5.77 grams / 123.90 g/mol

Moles P4 = 0.0466 moles

Step 4: Calculate moles O2

Moles O2 = mass O2 / molar mass O2

Moles O2 = 5.77 grams / 32.0 g/mol

Moles O2 = 0.1803 moles

Step 5: Calculate limiting reactant

P4 is the limiting reactant in this reaction. It will completely be consumed (0.0466 moles). O2 is in excess, there will react 3*0.0466 = 0.1398 moles

There will remain 0.1803 - 0.1398 = 0.0405 moles O2

Step 6: Calculate the amount of P4O6

For 1 mol P4 we'll have 1 mol P4O6

For 0.0466 moles P4 we'll have 0.0466 moles P4O6

Step 7: The balanced equatio

P4O6 + 2O2 → P4O10

We have 0.0466 moles P4O6 and 0.0405 moles O2

Step 8: Calculate the limiting reactant

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

O2 is the limiting reactant. It will completely be consumed (0.0405 moles)

P4O6 is in excess. There will react 0.0405/2 = 0.02025 moles

There will remain 0.0466 - 0.02025 = 0.02635 moles P4O6

This is 0.02635 * 219.88 g/mol = 5.79 grams P4O6

Step 9: Calculate moles and mass of P4O10

For 1 mol P4O6 we need 2 moles O2 to produce 1 mol P4O10

For 0.0405 moles O2 we'll have 0.02025 moles P4O10

This is 0.02025 * 283.89 g/mol = 5.75 grams P4O10

3 0

2 years ago

A mass of 5.4 grams of aluminum (al) reacts with an excess of copper (ii) chloride (cucl2) in solution, as shown below. 3cucl2 2

horrorfan [7]

Answer:

Mass of copper produced is 19.07g

Explanation:

Let's bring out the chemical equation;

3CuCl2 + 2Al → 2AlCl3 + 3Cu

3 : 2 : 2 : 3

Upon confirming that the reaction is indeed balanced, we can proceed.

The questions asks to calculate mass of Cu formed when a mass of 5.4g of Al is being used.

From the equation, what is the relationship between Al and Cu?

2 mol of Al would react to form 3 mol of Cu

Expressing this in terms of mass, we have;

mass = no. of moles * molar mass

Mass of Aluminium = 2 * 26.98 = 53.98 g

Mass of Copper = 3 * 63.546 = 190.638g

This means;

53.98 g of Al would react to form 190.638g of Cu

So how much Cu would form from 5.4 g of Al?

This leads us to;

53.98 = 190.638

5.4 = x

Upon cross multiplication, we are left with;

x = (190.638 * 5.4) / 53.98

x = 19.07 g

Mass of copper produced is 19.07g

8 0

2 years ago

"The compound K2O2 also exists. A chemist can determine the mass of K in a sample of known mass that consists of either pure K2O

o-na [289]

Answer:

Yes, the chemist can determine which compound is in the sample.

Explanation:

In 1 mole of K₂O, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O is 94.2 g. The mass ratio of K to K₂O is 78.2 g / 94.2 g = 0.830.

In 1 mole of K₂O₂, the mass of K is 2 × 39.1 g = 78.2 g and the mass of K₂O₂ is 110.2 g. The mass ratio of K to K₂O₂ is 78.2 g / 110.2 g = 0.710.

If the chemist knows the mass of K and the mass of the sample, he or she must calculate the mass ratio of K to the sample.

If the ratio is 0.830, the compound is pure K₂O.
If the ratio is 0.710, the compound is pure K₂O₂.
If the ratio is not 0.830 or 0.710, the sample is a mixture.

6 0

2 years ago

Which has not been suggested as a reasonably practical way to store large amounts of hydrogen in relatively small spaces for its

Yakvenalex [24]

Answer: A. Liquefy hydrogen under pressure and store it much as we do with liquefied natural gas today.

Explanation:

Current Hydrogen storage methods fall into one of two technologies;

<em>physical storage</em> where compressed hydrogen gas is stored under pressure or as a liquid; and
<em>chemical storage</em>, where the hydrogen is bonded with another material to form a hydride and released through a chemical reaction.

Physical storage solutions are commonly used technologies but are problematic when looking at using hydrogen to fuel vehicles. Compressed hydrogen gas needs to be stored under high pressure and requires large and heavy tanks. Also, liquid hydrogen boils at -253°C (-423°F) so it needs to be stored cryogenically with heavy insulation and actually contains less hydrogen compared with the same volume of gasoline.

Chemical storage methods allow hydrogen to be stored at much lower pressures and offer high storage performance due to the strong binding of hydrogen and the high storage densities. They also occupy relatively smaller spaces than either compressed hydrogen gas or liquified hydrogen. A large number of chemical storage systems are under investigation, which involve hydrolysis reactions, hydrogenation/dehydrogenation reactions, ammonia borane and other boron hydrides, ammonia, and alane etc.

Other practical storage methods being researched that focuses on storing hydrogen as a lightweight, compact energy carrier for mobile applications include;

Metal hydrides e.g. LiH

Nanostructured metal hydrides

Non-metal hydrides

Carbohydrates

Synthesized hydrocarbons

Aluminum

Liquid organic hydrogen carriers (LOHC)

Encapsulation , e.t.c.

5 0

2 years ago

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago