Answer :
(1) pH = 1.27
(2) pH = 13.35
(3) The given solution is not a buffer.
Explanation :
<u>(1) 53.1 mM HCl</u>
Concentration of HCl = 
As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.
So, Concentration of hydrogen ion= 
pH : It is defined as the negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


<u>(2) 0.223 M KOH</u>
Concentration of KOH = 0.223 M
As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.
So, Concentration of hydroxide ion= 0.223 M
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)


Now we have to calculate the pH.

<u>(3) 53.1 mM HCl + 0.223 M KOH</u>
Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.
It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.
As we know that the pH of strong acid and strong base solution is always 7.
So, the given solution is not a buffer.
D, the rate increases as concentrations increase.
Typically, reaction rates decrease with time because reactant concentrations decrease as reactions are converted to products. Reaction rates generally increase when reactant concentrations are increased.
Answer:

Explanation:
Hello,
In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

Best regards.
Answer: Object B will heat up more.
Explanation:
The formula for specific heat is as follows.
Q = 
Where,
Q = heat provided
m = mass
C = specific heat
= change in temperature
Now, both the objects have same mass and equal amount of heat is applied.
According to the formula, the equation will be as follows.
= 
= 
Cancel m from both sides, as mass is same. Therefore,
= 
Cancel out the initial temperature and put the values of specific heat, then the equation will be as follows.
= 
Therefore, from the above equation it can be concluded that the object with low specific heat will heat up more as its specific heat will be inversely proportional to its final temperature.
Hence, object B will heat up more.
Answer:
I am so confused about this i am so sorry but i cant there is to much going on
Explanation: