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Rudik [331]
3 years ago
6

For the formation of hydrogen iodide, H2(g) + 12(g) - 2HI(g) the value of the

Chemistry
1 answer:
Sauron [17]3 years ago
7 0

Answer:

t 700 K, the equilibrium constant for the reaction;

H  

2(g)

​  

+I  

2(g)

​  

⇌2HI  

(g)

​  

 is 54.8.

If 0.5 mol litre  

−1

 of HI  

(g)

​  

 is present at equilibrium at 700 K, what are the concentrations of H  

2(g)

​  

 and I  

2(g)

​  

 assuming that we initially started with HI  

(g)

​  

 and allowed it to reach equilibrium at 700 K.

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Explain why food chains do not tend to exceed four links.
Softa [21]
Energy is "lost" at each trophic level when you go up the chain. <span> Typically there are fewer organisms at higher trophic levels.

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3 0
3 years ago
A covalent bond is best described as ______(A) a bond between two polyatomic ions.(B) a bond between a metal and a nonmetal.(C)
nekit [7.7K]

Answer:

Option C. the sharing of electrons between atoms

Explanation:

Covalent bond is a type of bond in which the reacting element share their valence electrons in order to attain the noble gas configuration.

4 0
4 years ago
In fractional distillation, liquid can be seen running from the bottom of the distillation column back into the distilling flask
shepuryov [24]

Answer:

substances with a higher boiling point are returning back to the flask which allows another substances with the specific context temperature (lower boiling point) to boil over and be purified.

Explanation:

The reason it happens because the lower boiling point substance vaporizes and crosses over while the other substance is waiting for its boiling point to reach

7 0
3 years ago
How many moles of aluminum oxide will be produced from 0.50 mol of oxygen? 4 al + 3 o2 2 al2o3?
LuckyWell [14K]
From the equation;
 4 Al + 3 O2 = 2 Al2O3
The mole ratio of Oxygen is to Aluminium hydroxide is  3:2.
Therefore; moles of Al2O3 is 
 (0.5/3 )× 2 = 0.333 moles
Therefore; The moles of aluminium oxide will be 0.333 moles
5 0
3 years ago
Read 2 more answers
At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---&gt; +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
4 years ago
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