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OlgaM077 [116]
3 years ago
12

Onsider the following reaction at equilibrium:

Chemistry
1 answer:
11111nata11111 [884]3 years ago
5 0

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

C(s)+H_2O(l)\leftrightharpoons CO(g)+H_2(g)

A. C is added to the reaction mixture.

If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.

B. H_2O is condensed and removed from the reaction mixture.

If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.

C. CO is added to the reaction mixture.

If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.

D. H_2 is removed from the reaction mixture.

If the concentration of product specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of product specie occurs. So, on removing hydrogen , the equilibrium will shift in right direction.

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Molar mass of 34.00 g/mol molecular formula
alukav5142 [94]
Multiplying the subscripts within the empirical formula by this number gives you the molecular formula H2O2. 
8 0
3 years ago
Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− ,
zysi [14]

Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

  • CH₃CH₂CO₂H = HA
  • CH₃CH₂CO₂⁻ = A⁻

pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

  • (a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

10^{-0.644} = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

  • (b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

10^{-0.004} = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

  • (c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

10^{0.426} = [A⁻]/[HA]

2.667 = [A⁻]/[HA]

6 0
3 years ago
I NEED HELP ASAP IF YOU GET IT CORRECT I WILL MARK YOU AS BRAINLIEST
masya89 [10]

Answer:

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Explanation:

If 1 is tall, and 1 is short, 50% would be tall, and the other 50% would be short.

4 0
3 years ago
Read 2 more answers
The element lanthanum has two stable isotopes, lanthanum-138 with an atomic mass of 137.9 AMU and lanthanum-139 with an atomic m
dmitriy555 [2]

Answer:

A. there is an isotope of lanthanum with an atomic mass of 138.9

Explanation:

By knowing the different atomic masses of both Lanthanum atoms, we can not tell anything about their occurence in nature. Therefore, all the last three options are incorrect. Because, the atomic mass does not tell anything about the availability or natural abundance of an element.

Now, the isotopes of an element are those elements, which have same number of electrons and protons as the original element, but different number of neutrons. Therefore, they have same atomic number but, different atomic weight or atomic masses.

Hence, by looking at an elements having same atomic number, but different atomic masses, we can identify them as isotopes.

Thus, the correct option is:

<u>A. there is an isotope of lanthanum with an atomic mass of 138.9.</u>

7 0
3 years ago
You wish to make a 0.375 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid
cluponka [151]

We need to add 2.09 mL of concentrated acid to obtain 75 mL of 0.335 M HBr solution.

You are performing a dilution of HBr going from a concentration of 12M to 0.335M, and you want to end up with a final volume of 75 ml of the dilute solution.

Consider the dilution formula: M1V1 = M2V2.

The basis behind this formula is that the number of moles of the acid before and after the dilution must remain constant.

M1 = the molarity of the stock solution,

M2 = the molarity of the diluted solution,

V2 = the final volume of the diluted solution.

In this case, we need to determine V1, which is the volume of the stock solution used to prepare the diluted sample. With this knowledge, we can plug our numbers into the equation and we obtain the following:

(12 mol/L)*V1 = (0.335mol/L)*(0.075L).

Keeping in mind that molarity is the moles of a substance in one liter of solution, we will use mol/L instead of M. By doing this, we are reminded that in order to use this equation, we must convert 75 mL into units of liters.

After rearranging the equation and solving for V1, we find that V1 = 0.00209L.

Finally, we must convert back from liters to mL by multiplying the final answer by 1000.

This way we end up with V1 = 2.09 mL.

This means that we need to add 2.09 mL of concentrated acid to obtain 75 mL of 0.335 M HBr solution.

Learn more about molarity here: brainly.com/question/23243759

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