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3 years ago

Explain why a random copolymer is obtained when 3,3-dimethyl-1-butene undergoes cationic polymerization.

Chemistry

1 answer:

dezoksy [38]3 years ago

8 0

Solution :

In the field of chemistry, the cationic polymerization is a kind of the chain growth polymerization where the cation initiator transfers the charge to the monomer and makes it more reactive. This kind of polymerization reaction is very sensitive to the temperature. With increase in temperature, the molecular weight as well as the reaction rate decreases rapidly.

Thus in the cationic polymerization of the 3,3-dimethyl-1-butene, the carbonation intermediate is formed and it rearranges itself. The attack o the 2nd alkene ca take place both the carbonations, so that random copolymer is formed.

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How a strontium salt emits colored light

Bond [772]

Answer:

When excited electrons fall to lower energy levels, they can release energy in the form of light. metal ions in the salts used in the flame tests.

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3 years ago

The actual density of iron is 7.874 g/mL. In a laboratory investigation, Jason finds the density of a piece of iron to be 7.921

mixas84 [53]

The percent error associated with Jason’s measurement is 0.596%.

HOW TO CALCULATE PERCENTAGE ERROR:

The percentage error of a measurement can be calculated by following the following process:

Find the difference between the true value and the measured value of a quantity.
Then, divide by the true value and then multiplied by 100

The true value of the density of iron is 7.874 g/mL
Jason observed value is 7.921 g/mL

Difference = 7.921 g/mL - 7.874 g/mL

Difference = 0.047 g/mL

Percentage error = 0.047/7.874 × 100

Percentage error = 0.596%.

Therefore, the percent error associated with Jason’s measurement is 0.596%.

Learn more: brainly.com/question/18074661?referrer=searchResults

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1 year ago

Calculate the percentage of water in the hydrate manganese (ii) nitrate tetrahydrate

IRISSAK [1]

Answer:

Divide the mass of the water lost by the mass of hydrate and multiply by 100.

Explanation:

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8 0

3 years ago

Read 2 more answers

At 25 celsius, 1.00 ,ole of O2 was found to occupy a volume of 12.5 L at a pressure of 198 kPa. What value of the gas constant i

mr Goodwill [35]

Answer:

$R=8.301\frac{L*kPa}{mol*K}$

Explanation:

Hello,

In this case, assuming oxygen as an ideal gas, it is described by:

$PV=nRT$

Whereas the gas constant R is required, therefore, it is computed as shown below:

$R=\frac{PV}{nT} =\frac{198kPa*12.5L}{1.00mol*(25+273.15)K} \\\\R=8.301\frac{L*kPa}{mol*K}$

Best regards.

4 0

3 years ago

Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper

KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0

3 years ago