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Free_Kalibri [48]
3 years ago
11

What percentage of gases are in the earth?

Chemistry
1 answer:
ioda3 years ago
5 0

Answer:

By volume, dry air contains 78.09% nitrogen, 20.95% oxygen, 0.93% argon, 0.04% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% at sea level, and 0.4% over the entire atmosphere.

Explanation:

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The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra
Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

             Br2(g)    ⇌  2Br(g)

Initially  0.0345M  0.0416M

Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016

Q_C =0.05016 >>> K_c(0.00584)

Thus, the given reaction will proceed in the  backward direction

 The I.C.E table is as follows:

                       Br2(g)    ⇌  2Br(g)

I               0.0345                 0.0416

C                 +x                        -2x

E             (0.0345+x)            (0.0416 -2x)

K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908  - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] =  (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] =  (0.0416 -2(0.0126)) = 0.0164 M

4 0
3 years ago
What is ΔE in kJ for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same t
kenny6666 [7]

Answer: 20.7 kJ

Explanation:

According to first law of thermodynamics:

\Delta E=q+w

\Delta E=Change in internal energy

q = heat absorbed or released

w = work done or by the system

w = work done on the system=-P\Delta V  {Work is done on the system is positive as the final volume is lesser than initial volume}

w = 4.51 kcal = 4.51\times 4.184kJ=18.9kJ    (1kcal = 4.184kJ)

q = +1.79 kJ   {Heat absorbed by the system is positive}

\Delta E=+1.79+(18.9)=20.7kJ

Thus \Delta E for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ

4 0
3 years ago
10. Aqua means "water" or "a light blue color." Marine refers to the sea. Use this information
Zielflug [23.3K]

Answer:

\huge\blue{ \mid{ \underline{ \overline{ \tt  ♡ AQUAMARINE ♡ }} \mid}}

=> The colour of this stone is usually a pale greenish blue, owing to the presence of iron impurities. Stones that are treated with heat look more blue than green. On the Mohs scale of hardness, aquamarine ranges between 7.5 and 8 making it a relatively hard gemstone.

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3 0
2 years ago
Read 2 more answers
Dalton hypothesized that atoms are indivisible and that all atoms of an element are identical. It is now known that _____.
34kurt
Dalton hypothesized that atoms are indivisible and that all atoms of an element are identical. It is now known that <span>atoms are divisible. The answer is letter C</span>
8 0
3 years ago
Read 2 more answers
what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C
svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

\frac{V}{T}=k

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

\frac{V1}{T1}=\frac{V2}{T2}

In this case:

  • V1= 0.730 L
  • T1= 28 °C= 301 °K (0°C= 273°K)
  • V2= ?
  • T2= 0°C= 273 °K

Replacing:

\frac{0.730 L}{301K}=\frac{V2}{273K}

Solving:

V2=273K*\frac{0.730L}{301K}

V2=0.662 L

<u><em>The volume of the air is 0.662 L</em></u>

6 0
3 years ago
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