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3 years ago

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Draw the major byproduct of the reaction (not the minor species discussed in q2b). At which stage did you remove it from your sa

mple?

1 answer:

givi [52]3 years ago

3 0

You need to do something like that your self so sorry can help.

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Which balanced equation represents a fusion reaction?

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What kind of mixture is a solution

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3 years ago

Read 2 more answers

In a titration of HCl with NaOH, 100mL of the base was required to neutralize 20mL of 5.0 M HCl. What is the molarity of the NaO

nikklg [1K]

Answer:

1 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Volume of base (Vb) = 100mL

Volume of ac(Va) = 20mL

Molarity of acid (Ma) = 5M

Molarity of base (Mb) =...?

Step 2:

The balanced equation for the reaction. This is given below:

HCl + NaOH —> NaCl + H2O

From the above equation, the following were obtained:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 3:

Determination of the molarity of the base.

The molarity of the base can be obtained as follow:

MaVa/MbVb = nA/nB

5 x 20 / Mb x 100 = 1

Cross multiply to express in linear form

Mb x 100 = 5 x 20

Divide both side by 100

Mb = (5 x 20)/100

Mb = 1 M

Therefore, the molarity of the base is 1 M

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3 years ago

The decomposition of nitrogen dioxide to nitrogen monoxide and oxygen gas is a second order process as suspected in the previous

lesya [120]

Explanation:

For the given reaction $2NO_{2} \rightarrow 2NO + O_{2}$

Now, expression for half-life of a second order reaction is as follows.

$t_{1/2} = \frac{1}{[A_{0}]k}$ ....... (1)

Second half life of this reaction will be $t_{1/4}$ . So, expression for this will be as follows.

$t_{1/4}$ = $\frac{1}{k} [\frac{1}{[A]_{f}} - \frac{1}{[A_{0}]}]$ ...(2)

where $[A]_{f}$ is the final concentration that is, $\frac{[A]_{0}}{4}$ here and $[A]_{i}$ is the initial concentration.

Hence, putting these values into equation (2) formula as follows.

$t_{1/4}$ = $\frac{1}{k} [\frac{4}{[A]_{0}} - \frac{1}{[A_{0}]}]$

= $\frac{3}{[A_{0}]k}$ ...... (3)

Now, dividing equation (3) by equation (1) as follows.

$\frac{t_{1/4}}{t_{1/2}}$ = $\frac{3}{[A_{0}]k} \times [A_{0}]k$

= 3

or, $t_{1/4}$ = 3 $t_{1/2}$

Thus, we can conclude that one would expect the second half-life of this reaction to be three times the first half-life of this reaction.

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3 years ago

What is the total number of moles of atoms in Pb(C2H3O2)2?

tekilochka [14]

1 mole of atoms is in <span>Pb(C2H3O2)2?</span>

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3 years ago