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DanielleElmas [232]
4 years ago
6

Substances that are likely to dissociate it water

Chemistry
1 answer:
Neporo4naja [7]4 years ago
7 0

Answer:

sodium bromide (NaBr) potassium hydroxide (KOH) magnesium chloride (MgCl2) silicon dioxide (SiO2) sodium oxide (Na2O)

Explanation:

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7 0
3 years ago
Find the volume of 3.011x1023 molecules of O2 at STP.
lord [1]

Answer:

6.022 x 1023 molecules = 1 Mole O2 = 22.4L

3.022 x 1023 molecules = 1/2 Mole O2 = 11.2L

The answer is 11.2 Litres

Explanation:

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2 years ago
Tissue that makes up the outer covering in humans
Rainbow [258]
I think it's called skin
4 0
3 years ago
Read 2 more answers
What would happen if N2 were added to N2(g) + O2(g) = 2NO(9) at
DerKrebs [107]

Balanced chemical equation is :

N_2(g)+O_2(g)-->2NO(g)

It is given that the equation is in equilibrium.

We need to find what will happen if we add more N_2 is added .

By Le Chatelier's principle :

Changing the concentration of a chemical will shift the equilibrium to the side that would counter that change in concentration.

It means production of the side where content is added will decrease and concentration on other side will increase .

So , more NO would form .

Therefore, option B. is correct.

Hence, this is the required solution.

4 0
3 years ago
What is the value of the rate constant k for this reaction?When entering compound units, indicate multiplication of units explic
vekshin1

The table with the data is in the picture attached.

Answer:

  • k=0.0033M^{-2}.s^{-1}

Explanation:

The reaction equation suggests that the law could have this form:

  • rate=k[A]^a[B]^b[C]^c

Then, the work is to find the values of the exponents that satisfy the initial rate data.

A first glance shows that for the third and fourth trials the initial rates are the same. Since for these two trials only the initial concentration of substance B changed (A and C were kept equal), you conclude that the reaction rate does not depend on B, and ist exponent (lower b) is 0.

Then, so far you can say:

  • rate=k[A]^a[C]^c

When you use trials 1 and 2, you get:

\frac{r_2}{r_1}=\frac{27M/s}{9M/s}=\frac{(0.3M)^a(0.3M)^b(0.9M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=3^c\\\\ 3=3^c\\ \\ 1=c

Now, you can use trials 1 and 3 to determine the other exponent:

\frac{r_3}{r_1}=\frac{36M/s}{9M/s}=\frac{(0.6M)^a(0.3M)^b(0.3M)^c}{(0.3M)^a(0.3M)^b(0.3M)^c}=2^a\\\\ 4=2^a\\ \\ 2^2=2^a\\ \\ 2=a

Thus, you have the rate law:

  • r=k[A]^2[C]

Now, you just use any trial to obtain k. Using trail 1:

  • k(0.3M)^2(0.3M)=9.10^{-5}M/s

Which yields:

  • k=0.0033M^{-2}.s^{-1}

4 0
3 years ago
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