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3 years ago

The ____ contains the highest concentration of ozone.

2 answers:

3 0

<span>C. The stratosphere. About 90% of the ozone is in the stratosphere which begins about 8 miles above the Earth's surface.</span>

zheka24 [161]3 years ago

3 0

Answer: Option (c) is the correct answer.

Explanation:

A layer above the troposphere and below the mesosphere is known as stratosphere.

This layer of stratosphere contains about 90% concentration of ozone and there is no occurrence of weather in this layer.

The occurrence of ozone in stratosphere makes it warmer.

Therefore, we can conclude that the stratosphere contains the highest concentration of ozone.

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Match the action to the effect on the equilibrium position for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g). Match Term Definition Remo

pshichka [43]

Answer:

The answers to the questions are given below.

Explanation:

According to Le Chatelier's principle, if an external constrain such as change in concentration, temperature or pressure is imposed on a chemical system in equilibrium, the equilibrium will shift in order to neutralize the effect.

A. Effective of removing ammonia, NH3.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Removing NH3 from the reaction simply means we are left with more reactants and no product. Therefore, the reactant will react to produce the product. Hence, the equilibrium position will shift to the right.

2. Effect of removing H2

N2(g) + 3H2(g) ⇌ 2NH3(g)

Remoing H2 simply means we have more products and less reactant. Therefore, the product will be convert to reactant. Hence, the equilibrium position will shift to the left.

C. Effect of adding a catalyst.

N2(g) + 3H2(g) ⇌ 2NH3(g)

Catalyst does not affect the equilibrium position. It only creates an alternative path to arrive at the product within a short time. Hence, it has no effect.

7 0

3 years ago

If 41.38 g of sodium combines with

Arlecino [84]

I think it is 3 because Sus wiw word if rn

3 0

3 years ago

What is the atomic radius

dimaraw [331]

Answer:

The atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

5 0

2 years ago

List the number of each type of atom on the left side of the equation 2Na3PO4(aq)+2CoCl2(aq)→2Co3(PO4)2(s)+6NaCl(aq) Enter your

Trava [24]

Answer:

Na(sodium atom) = 6 atoms , P(phosphorus atom) = 2 atoms, Oxygen(Oxygen atom) = 8 atoms , Co(cobalt atom) = 2 atoms, Cl(chlorine atom) = 4 atoms

Explanation:

2Na3PO4(aq) + 2CoCl2(aq) → 2Co3(PO4)2(s) + 6NaCl(aq)

The question ask us to list the number of the kind of atom found on the left side(reactant) of the chemical equation. The question also reiterated that the order of the numbers should be the same as the order of the elements on the left side of the equation.

The type of atoms found on the left side of the equation are as follows;

Na(sodium atom) = 6 atoms

P(phosphorus atom) = 2 atoms

Oxygen(Oxygen atom) = 8 atoms

Co(cobalt atom) = 2 atoms

Cl(chlorine atom) = 4 atoms

The equation on the left hand side have sodium atom, phosphorus atom, oxygen atom, cobalt atom and chlorine atom.

Note: The given equation is not balanced but it can be balanced as follows;

2Na3PO4(aq) + 3CoCl2(aq) → Co3(PO4)2(s) + 6NaCl(aq)

The balanced equation have the following number of atom on the left side:

Na(sodium atom) = 6 atoms

P(phosphorus atom) = 2 atoms

Oxygen(Oxygen atom) = 8 atoms

Co(cobalt atom) = 3 atoms

Cl(chlorine atom) = 6 atoms

5 0

3 years ago

14. An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km

emmainna [20.7K]

Answer:

27 min

Explanation:

The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:

$v = \frac{vmax[S]}{Km + [S]}$

Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.

So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min

If Km is a thousand times smaller then [S], then

v = vmax[S]/[S]

v = vmax

vmax = 1.33 μmol/min

For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:

vmax = 1.33/3 = 0.443 μmol/min

Km will still be much smaller then [S], so

v = vmax

v = 0.443 μmol/min

For 12 μmol formed:

0.443 = 12/t

t = 12/0.443

t = 27 min

7 0

3 years ago