Answer:
1.76
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>
<em />
Step 1: Calculate the molarity of HI(aq)
M = mass of solute / molar mass of solute × liters of solution
M = 0.660 g / 127.91 g/mol × 0.300 L
M = 0.0172 M
Step 2: Write the acid dissociation reaction
HI(aq) ⇄ H⁺(aq) + I⁻(aq)
HI is a strong acid, so [H⁺] = 0.0172 M
Step 3: Calculate the pH
pH = -log [H⁺]
pH = -log 0.0172
pH = 1.76
The answer is 25 grams for this question
I want to say addition. But I have a tendency to be wrong
Explanation:
(a) As the given chemical reaction equation is as follows.

So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.
Hence, equation for rate law of reaction will be as follows.
Rate =
(b) Since, the rate equation is as follows.
Rate =
Let us assume that (
)
Putting the given values into the above equation as follows.

K =
=
Hence, the value of rate constant for the given reaction is
.
(c) Now, we will calculate the rate as follows.
Rate =
=
= 
Therefore, rate when
M and
M is
.
I believe the answer is B. PO4-3