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3 years ago

Which organic reaction produces rubber and plastics? A) polymerization B) esterification C) saponification D) fermentation

Chemistry

2 answers:

Karolina [17]3 years ago

6 0

The answer would be, A) Polymerization.

Anika [276]3 years ago

6 0

Answer: Option (A) is the correct answer.

Explanation:

When monomers in a chemical reaction combine together then it results in the formation of long chains. This process is known as polymerization.

For example, rubber and plastic are formed by polymerization.

Esterification is defined as a process in which esters are formed when an acid is heated with alcohol.

Saponification is defined as a process in which aqueous alkali is heated with a fat or oil resulting in the formation of a soap.

Fermentation is a process in which bacteria like microorganisms, yeast etc help in the breakdown of a substance into simpler substances.

Thus, we can conclude that polymerization is an organic reaction which produces rubber and plastics.

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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from

GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $200mL+200L=400mL$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 400 g

$T_{final}$ = final temperature of water = $27.78^oC=273+25.10=300.78K$

$T_{initial}$ = initial temperature of metal = $25.10^oC=273+27.78=298.1K$

Now put all the given values in the above formula, we get:

$q=400g\times 4.18J/g^oC\times (300.78-298.1)K$

$q=4480.96J$

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

$\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0

3 years ago

How many moles of air are in a 2 liter bottle at a pressure of 1 atmosphere and a temperature of 288°k (2 liter bottle before ad

creativ13 [48]

Answer:

0.000835225 mol

Explanation:

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