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Rudik [331]
3 years ago
6

Why can't krypton and xenon separate during the separation of gases in the air?

Chemistry
2 answers:
Tasya [4]3 years ago
7 0

Answer:

They're in Group 0 as they have full outer shells meaning they are chemically unreactive as they already have full outer shells meaning they don't want to lose or gain any electrons. Thus, they can't separate during the separation of gases in the air.

In-s [12.5K]3 years ago
5 0

Answer:

The remaining gas is a mixture of argon, neon, krypton, and xenon. Because these elements are chemically very unreactive, chemical means cannot be used to separate them.

Explanation:

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A reaction contains two reactants, A and B. If A is doubled, there will be a greater number of effective collisions between reac
liubo4ka [24]

Answer:

True

Explanation:

The complete question is:

<u><em>"A reaction contains two reactants, A and B. If A is doubled, there will be a greater number of effective collisions between reactants. TRUE FALSE"</em></u>

Collision Theory indicates that chemical reactions take place because molecules, atoms or ions collide with each other.

Furthermore, the molecules must collide effectively, that is, not all reagent collisions lead to product formation. Effective shock means that the reagent molecules have enough kinetic energy at the time of the shock for their bonds to break and product bonds to form. In addition, the molecules of the reagents must be properly oriented for the reaction to take place.

As the concentration increases, the number of shocks increases. In other words, by increasing the concentration of the reactants, the probability of collision between their molecules increases, and therefore the number of effective collisions.So the statement is true-

6 0
3 years ago
Read 2 more answers
1. Consider the reaction: N2 (g) + O2 (g)  2NO (g) + Cl2 (g) A reaction mixture at 400 C has the following composition: [NO] =
Olegator [25]

Answer:

The reaction is not at the equilibrium, we have more products than reactants.

Explanation:

Step 1: Data given

Temperature = 400 °C

When Q=K, this means the system is at equilibrium. There will not be a shift to either the left or the right.

When Q<K,it means we have more reactants than products. The system will, as reaction, make more products, this will cause the equilibrium to move to the right.

When Q>K,it means we have more products than reactants. The system will, as reaction, make less products and more reactants, this will cause the equilibrium to move to the left.

[NO] = 0.01 M

[N2] = 0.001 M

[O2] = 0.001 M

Kc = 0.0156

Step 2: The balanced equation

N2 (g) + O2 (g) ⇆ 2NO (g) + Cl2 (g)

Step 3: Calculate Q

Q = [NO]²/[N2][O2]

Q = 0.01²/(0.01*0.001)

Q = 10

Q>>Kc

This means we have more products than reactants. The system will, as reaction, make less products and more reactants, this will cause the equilibrium to move to the left.

The reaction is not at the equilibrium, we have more products than reactants.

4 0
3 years ago
Mica peels apart in parallel sheets because the chemical bonds between sheets are weak and the chemical bonds within sheets are
Makovka662 [10]
The answer is supposed to be true
7 0
3 years ago
Make a connection between bacteria and extreme environment
Ugo [173]
<span>One example of bacteria that live in an extreme environment are the thermophilic species that live in the hot springs of Yellowstone National Park. Thermophilic means heat-loving and the waters in which these bacteria live can rise above 70 degrees Celcius after being heated by lava below the surface. Most life would die at that temperature however some bacteria have evolved to survive the extreme heat.</span>
3 0
3 years ago
Calculate the number of grams of solute needed to make each of the following solutions:
AleksAgata [21]

Explanation:

(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

1) 100 g of 0.500% (w/w) NaI

Mass of solution = 100 g

Mass of solute = x

Required w/w % of solution = 0.500%

0.500\%=\frac{x}{100 g}\times 100

x=\frac{0.500\times 100 g}{100}=0.500 g

0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.

2) 250 g of 0.500% (w/w) NaBr

Mass of solution = 250 g

Mass of solute = x

Required w/w % of solution = 0.500%

0.500\%=\frac{x}{250 g}\times 100

x=\frac{0.500\times 250 g}{100}=1.25 g

1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr

3) 500 g of 1.25% (w/w) glucose

Mass of solution = 500 g

Mass of solute = x

Required w/w % of solution = 1.25%

1.25\%=\frac{x}{500 g}\times 100

x=\frac{1.25\times 500 g}{100}=6.25 g

6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)

4) 750 g of 2.00% (w/w) sulfuric acid.

Mass of solution = 750 g

Mass of solute = x

Required w/w % of solution = 2.00%

2.00\%=\frac{x}{750 g}\times 100

x=\frac{2.00\times 750 g}{100}=15.0 g

15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.

3 0
3 years ago
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