Name the two gases that make up most of the air.

Chemistry

2 answers:

ddd [48]2 years ago

5 0

Answer:

Air is mostly gas

The air in Earth's atmosphere is made up of approximately 78 percent nitrogen and 21 percent oxygen. Air also has small amounts of lots of other gases, too, such as carbon dioxide, neon, and hydrogen.

Explanation:

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Karolina [17]2 years ago

5 0

The 2 gases that make up most of the air is nitrogen and oxygen.
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Anna007 [38]

I think its A. it cant be C or D b/c we are measuring the distance between Earth and Saturn, not the speed.

7 0

3 years ago

At 700 K, the reaction 2SO2(g) + O2(g) <====> 2SO3(g) has the equilibrium constant Kc = 4.3 x 106. At a certain instant, f

nadya68 [22]

Answer:

The system is not in equilibrium and will evolve left to right to reach equilibrium.

Explanation:

The reaction quotient Qc is defined for a generic reaction:

aA + bB → cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.

If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

$Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }$