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Alexus [3.1K]
2 years ago
8

What is the of 0.068 M of OH^ - ion?

Chemistry
1 answer:
Marina CMI [18]2 years ago
4 0

Answer:

reeeeeeeeeeeeee

Explanation:

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Hydroxide is an example of _______ and it's formula is OH-.
Ksju [112]

Answer:

polyatomic ion

Explanation:

It is polyatomic ion have a great day marry christmass

6 0
3 years ago
A negative △H means an exothermic reaction.<br><br> True<br> False
Nostrana [21]
Your answer is False I think
6 0
3 years ago
Fritz-Haber process
maks197457 [2]

Answer:

5×10⁵ L of ammonia (NH3)

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above, we can say that:

3 L of H2 reacted to produce 2 L of NH3.

Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:

From the balanced equation above,

3 L of H2 reacted to produce 2 L of NH3.

Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.

Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.

6 0
3 years ago
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
Which compound is produced by a neutralization?
olchik [2.2K]

Explanation:

HNO3(aq) is the compound produced by a neutralization

6 0
3 years ago
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