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anyanavicka [17]
3 years ago
5

How many mL of methanol would

Chemistry
2 answers:
mr Goodwill [35]3 years ago
4 0

Answer: 36.25

Explanation:

inn [45]3 years ago
4 0

Answer:

36.25

Explanation:

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A chemist dissolves of pure hydroiodic acid in enough water to make up of solution. Calculate the pH of the solution. Be sure yo
Gre4nikov [31]

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

5 0
3 years ago
The solubility of potassium sulfate in water is 16 grams per 100
kicyunya [14]

The answer is 25 grams for this question

7 0
2 years ago
Which type of reaction does this diagram represent?
Mice21 [21]
I want to say addition. But I have a tendency to be wrong

8 0
3 years ago
Read 2 more answers
The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
LiRa [457]

Explanation:

(a)  As the given chemical reaction equation is as follows.

           OCl^{-} + I^{-} \rightarrow OI^{-1} + Cl^{-1}

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 = K \times [OCl^{-}] \times [l^{-}]

(b)  Since, the rate equation is as follows.

                    Rate = K [OCl^{-}][l^{-}]

Let us assume that ([OCl^{-}] = [l^{-}])

Putting the given values into the above equation as follows.

             1.36 \times 10^{-4} = K \times (1.5 \times 10^{-3})^2

            1.36 \times 10^{-4} = K \times (2.25 \times 10^{-6})

                   K = \frac{1.36 \times 10^{-4}}{2.25 \times 10^{-6}}

                      = 60.4 M^{-1}sec^{-1}

Hence, the value of rate constant for the given reaction is 60.4 M^{-1}sec^{-1} .

(c) Now, we will calculate the rate as follows.

                Rate = K [OCl^{-}][l^{-}]

                         = 60.4 \times (1.8 \times 10^{3}) \times (6.0 \times 10^{4})

                        = 6.52 \times 10^{5}

Therefore, rate when [OCl^{-}] = 1.8 \times 10^{3} M and [I^{-}]= 6.0 \times 10^{4} M is  6.52 \times 10^{5}.

8 0
3 years ago
Select the ion.<br><br>1)Cu<br>2)PO4-3<br>3)SO2
Lilit [14]
I believe the answer is B. PO4-3
6 0
3 years ago
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