Answer:
Only changes in temperature will influence the equilibrium constant 
. The system will shift in response to certain external shocks. At the new equilibrium 
will still be equal to , but the final concentrations will be different.
The question is asking for sources of the shocks that will influence the value of . For most reversible reactions:
- External changes in the relative concentration of the products and reactants.
For some reversible reactions that involve gases:
- Changes in pressure due to volume changes.
Catalysts do not influence the value of . See explanation.
Explanation:
.
Similar to the rate constant, the equilibrium constant depends only on:
the standard Gibbs energy change of the reaction, and
the absolute temperature (in degrees Kelvins.)
The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium 
the two processes balance each other. The concentration of each species will stay the same.
Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.
- Changes in concentration influence the number of particles per unit space.
- Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.
For reactions that involve gases,
- Changing the volume of the container will change the concentration of gases and change the reaction rate.
However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.
Answer:
Answers are in the explanation
Explanation:
Equlibrium of HF in H₂O is:
HF + H₂O ⇄ F⁻ + H₃O⁺
Now, the KOH reacts with HF, thus:
KOH + HF → F⁻ + H₂O
<em>That means after reaction, concentration of HF decrease increasing F⁻ concentration.</em>
Now, seeing the equilibrium, as moles of HF decrease and F⁻ moles increase, the equilibrium will shift to the left decreasing H₃O⁺ concentration.
For the statements:
A. The number of moles of HF will increase. <em>FALSE</em>. HF react with KOH, thus, moles of HF decrease
B. The number of moles of F- will decrease. <em>FALSE</em>. The reaction produce F⁻ increasing its moles.
C. The equilibrium concentration of H₃O⁺ will increase. <em>FALSE. </em>The equilibrium shift to the left decreasing concentration of H₃O⁺
D. The pH will decrease. <em>FALSE</em>. As the H₃O⁺ concentration decrease, pH will increase
E. The ratio of [HF] / [F-] will remain the same. <em>FALSE</em>. Because moles of HF are decreasing whereas F- moles are increasing changing, thus, ratio.
Answer:
Its B. Dilute acid + carbonate
Explanation:
hope it helps !!
<u>Answer:</u> The correct IUPAC name of the alkane is 4-ethyl-3-methylheptane
<u>Explanation:</u>
The IUPAC nomenclature of alkanes are given as follows:
- Select the longest possible carbon chain.
- For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
- A suffix '-ane' is added at the end of the name.
- If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.
We are given:
An alkane having chemical name as 3-methyl-4-n-propylhexane. This will not be the correct name of the alkane because the longest possible carbon chain has 7 Carbon atoms, not 6 carbon atoms
The image of the given alkane is shown in the image below.
Hence, the correct IUPAC name of the alkane is 4-ethyl-3-methylheptane
The temperature, pressure and volume of a gas are all related
