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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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Do you use sig figs in percent errors??
leonid [27]

Answer:

I think you use sig figs in percent error.

3 0
2 years ago
Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of
harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

Where:    

m: is the mass

M: is the molar mass

  • For <em>hydrogen </em>we have:

n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

  • And for <em>nitrogen</em>:

n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

6 0
2 years ago
Molecular formula for C4H10
Gnesinka [82]
I think this is what you mean:

    H H H H
H-C-C-C-C-H
    H H H H

OR 

<span>CH3CH2CH2CH3
</span>
If not, clarify and I will be happy to help.
6 0
3 years ago
You are given an unknown gaseous binary compound (that is, a compound consisting of two different elements). when 10.0 g of the
Luda [366]

Answer is: a possible identity for the unknown compound is C₃H₈.

m(binary compound) = 10.0 g.

m(H₂O) = 16.3 g; mass of water.

M(O₂) = 32 g/mol; molar mass of oxygen.

M(binary compound) = 1.38 · 32 g/mol.

M(binary compound) = 44.16 g/mol.

n(binary compound) = 10 g ÷ 44.16 g/mol.

n(binary compound) = 0.225 mol; amount of substance.

n(H₂O) = 16.3 g ÷ 18 g/mol.

n(H₂O) = 0.9 mol; amount of water.

m(H₂O) : n(binary compound) = 0.9 mol ÷ 0.225 mol.

m(H₂O) : n(binary compound) = 4 : 1.

Unknown cpmpound has 4 times more hydrogen than water, it has 8 hydrogen atoms.

Second element in compound is carbon:

M(X) = 44.16 g/mol - 8 · 1.01 g/mol.

M(X) = 36.08 g/mol ÷ 3.

M(C) = 12.01 g/mol.

8 0
2 years ago
Which of the following combinations will result in a reaction that is spontaneous at all temperatures?
Katen [24]

Answer:

\huge\boxed{Option \ 2}

Explanation:

A reaction is spontaneous at all temperatures by the following combinations:

=> A negative enthalpy change ( \triangle H < 0 )

=> A positive entropy change ( \triangle S > 0 )

See the attached file for more better understanding!

8 0
3 years ago
Read 2 more answers
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