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laila [671]
3 years ago
6

The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this

system? (Enter subscripts after the letters: for example, H2O = Hs2O. Also, don't forget to use proper chemical shorthand for chemical symbols. For instance, chlorine = Cl but not cl or cL.)
Chemistry
1 answer:
maks197457 [2]3 years ago
7 0
Answer is: Keq expression for this system is Keq = <span>[O</span>₂<span> ]</span> · [H₂<span>]</span>² ÷  [H₂O<span>]</span>².<span>
Chemical reaction: 2H</span>₂O(g) ⇄ O₂(g) + 2H₂(g).
The equilibrium constant<span> (Keq) is a ratio of the concentration of the products (in this reaction oxygen and hydrogen) to the concentration of the reactants (in this reaction water).</span>
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Read 2 more answers
23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp
stepan [7]

Answer:-  13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

P_1V_1=P_2V_2

From given information:-

P_1 = 58.7 Kpa

V_1 = 23.5 L

P_2 = 101.325 Kpa

V_2 = ?

Let's plug in the values and solve it for final volume.

58.7Kpa*23.5L=101.325Kpa*V_2

On rearranging the equation for V_2

V_2=\frac{58.7Kpa*23.5L}{101.325Kpa}

V_2 = 13.6 L

So, the volume of hydrogen gas at STP for the given information is 13.6 L.

3 0
2 years ago
A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in
In-s [12.5K]

Answer:

3.6667

Explanation:

<u>For helium gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{5.6}\times {3.0}={P_2}\times {9.0} atm

{P_2}=\frac {{5.6}\times {3.0}}{9.0} atm

{P_1}=1.8667\ atm

<u>The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm</u>

<u>For Neon gas:</u>

Using Boyle's law  

{P_1}\times {V_1}={P_2}\times {V_2}

Given ,  

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

{P_1}\times {V_1}={P_2}\times {V_2}

{3.6}\times {4.5}={P_2}\times {9.0} atm

{P_2}=\frac {{3.6}\times {4.5}}{9.0} atm

{P_1}=1.8\ atm

<u>The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm</u>

<u>Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.</u>

6 0
3 years ago
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