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lana [24]
3 years ago
5

The polymer formed from the monomer ch2=ch–cn is select one:

Chemistry
1 answer:
podryga [215]3 years ago
5 0
Answer:
             None of the given options show polymer made up of H₂C=CH-CN (Acrylonitrile).

Explanation:
                   Acrylonitrile (H₂C=CH-CN) which is a monomer on self linkage results in a large chain polymer called as Polyacrylonitrile.
                   The structure of Polyacrylonitrile is as follow,

                                             --(H₂C-CHCN-)n--

Where n shows the number of Acrylonitrile units joined together in the formation of Polyacrylonitrile. This polymerization reaction can take place by different mechanisms including free radical mechanism, acid catalyzed addition or base catalyzed addition reaction.

The polymerization is shown below,

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3 years ago
Using the phase diagram for H2O, which of the following correctly describes water at 0°C and 1 atm?
SCORPION-xisa [38]

At the melting point. Draw a line up from 0 degrees and a line to the right from 1 atm. They meet at the line between solid and liquid... the melting point

6 0
3 years ago
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The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con
Aloiza [94]
Use the formula for second order reaction:

\frac{1}{C} = \frac{1}{C_0} + kt

C = concentration at time t 
C0 =  initial conc.
k = rate constant
t = time

1st equation :   \frac{1}{0.56} = \frac{1}{C_0} + 115k

2nd Equation: \frac{1}{0.033} = \frac{1}{C_0} + 735k

Find \frac{1}{C_0} from 1st equation and put it in 2nd equation:

\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k

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k = 0.046
3 0
3 years ago
The rate constant k for a certain reaction is measured at two different temperatures:
bogdanovich [222]

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are K_1  and K_2  at temperature T_1  and T_2

By using Arrhenius equation at two different two different temperature,

Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K  ;K_1=4.8*10^8;T_2=273+280=553K  ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;  

By putting value of R=2 cal/mole.K

E_a=5510.265cal/mole;

By rounding off upto 2 significant figure;

E_a=5.5Kcal/mole;

8 0
3 years ago
How many moles are in 15 grams of Li?
svetoff [14.1K]

Answer:

there are 2.2 moles of Li in 15 grams of Li

8 0
2 years ago
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