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3 years ago

The electrons in an insulator are tightly bound to their atoms.

Chemistry

2 answers:

Murrr4er [49]3 years ago

4 0

If its a true or false statement . the answer is true

Ierofanga [76]3 years ago

4 0

What is it asking?Is it supposed to be true or false?

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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

2 years ago

Practice It!

Paladinen [302]

Mitosis hope that helps i tried:)

4 0

2 years ago

Read 2 more answers

State two procedures which should be used to properly handle a light microscope

babunello [35]

hold at base and arm

5 0

2 years ago

Do all colors of light have the same energy?

icang [17]

No, darker colors like purple and blue have higher wave frequency which transmit more energy.

6 0

3 years ago

Write a balanced chemical equation for the standard formation reaction of solid sodium hydrogen carbonate nahco3 .

NNADVOKAT [17]

There are a number of methods of which we can form sodium bicarbonate. This compound is commonly known as baking soda. It can be prepared from the reaction of sodium hydroxide and carbonic acid. Carbonic acid in water dissociates into hydrogen ions and the bicarbonate ion while sodium hydroxide would ionize into sodium ions and hydroxide ions. With this, these ions would react and form the sodium bicarbonate salt and water. The chemical reaction would be expressed as follows:

NaOH + H2CO3 = H2O + NaHCO3

Sodium bicarbonate is used in cooking, as a toothpaste and as a cleaning substance. Also, it is used in medical applications like for the preparation of the dialysate solution.

5 0

2 years ago