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3 years ago

Polymerization starts at the ________ end of the template.

Chemistry

1 answer:

balandron [24]3 years ago

3 0

Answer:

B. 3'

Explanation:

Polymerization is the process of forming large molecules by joining together many small molecules. The small molecules are known as the monomers, and the macromolecules are called polymers.

There are two(2) types of polymerization.

a. addition polymerization

b. condensation polymerization

Polymerization is characterized by eliminating small molecules such as water, carbon(IV)oxide, methanol e.t.c.

In complex compounds, polymerization starts at the 3' end of the template.

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Write the balanced NET IONIC equation for the reaction that occurs when perchloric acid and calcium hypochlorite are combined. I

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Answer:

$H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}$

Explanation:

Hello there!

In this case, since perchloric acid is HClO4 and is a strong acid and calcium hypochlorite is Ca(ClO)2, the undergoing molecular chemical reaction turns out:

$2HClO_4{(aq)}+Ca(ClO)_2{(aq)}\rightarrow 2HClO{(aq)}+Ca(ClO_4)_2{(aq)}$

Thus, since the resulting hypochlorous acid is weak, it does not fully ionize, so it remains unionized, however, we can write the ions for the other species:

$2H^++2(ClO_4)^-+Ca^{2+}+2(ClO)^-\rightarrow 2HClO{(aq)}+Ca^{2+}+2(ClO_4)^-$

Now, we can cancel out the spectator ions, calcium and perchlorate, to obtain:

$2H^+(aq)+2(ClO)^-(aq)\rightarrow 2HClO{(aq)}\\\\H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}$

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3 years ago

How many grams of pure naoh must be used to prepare 10.0 l of a solution that has a ph of 13? __________g?

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40 g NaOH. You must use 40 g NaOH to prepare 10.0 L of a solution that has a pH of 13.

Step 1. Calculate the pOH of the solution

pOH = 14.00 – pH = 14.00 -13 = 1

Step 2. Calculate the concentration of NaOH

[NaOH] = [OH^(-)] = 10^(-pOH) mol/L = 10^(-1) mol/L = 0.1 mol/L

Step 3. Calculate the moles of NaOH

Moles of NaOH = 10.0 L solution × (0.1 mol NaOH/1 L solution) = 1 mol NaOH

Step 4. Calculate the mass of NaOH

Mass of NaOH = 1 mol NaOH × (40.00 g NaOH/1 mol NaOH) = 40 g NaOH

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