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loris [4]
3 years ago
13

Draw the mechanism of polymerization of propylene with benzoyl peroxide. Show chemical structures and arrows and remember that r

adical stability is similar to carbocation stability. Thus, 3° radicals are more stable than 2° and so on.

Chemistry
1 answer:
allochka39001 [22]3 years ago
7 0

Answer:

See explanation below

Explanation:

This mechanism of polimerization is of one type. This mechanism goes through free radical polimerization, which means that we use a initiator as the radical source, then, the polimerization begins with the breaking of the covalent bonds of that initiator. In this case the benzoyl peroxide. And this mechanism is given in three steps.

In the picture you have the mechanism in the 3 steps:

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Answer:187

Explanation:

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3 years ago
Consider the Gibbs energies at 25 ∘C.
zysi [14]

Answer:

A)Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

B)Ksp=1.75×10^−10

C)Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=​Δ​G​f​,​p​r​o​d​u​c​t​s​∘​−​Δ​G​f​,​r​e​a​c​t​a​n​t​s​∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

A)

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​C​l​(​a​q​)​−​∘​]​−​Δ​G​f​,

A​g​C​l​(​s​)​⇌​A​g​(​a​q​)​+​+​C​l​(​a​q​)​−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​B​r​(​a​q​)​−​∘​]​−​Δ​G​f​,

Δ​G​r​x​n​∘​=​[​77.1​k​J​/​m​o​l​+​(​−​104.0​k​J​/​m​o​l​)​]​−​(​−96.90kj/mol

Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K

Ksp=5.45×10^−13

8 0
3 years ago
How many grams of NO are required to produce 145 g of N2 in the following reaction?
V125BC [204]

Answer:

b. 186 g

Explanation:

Step 1: Write the balanced equation.

4 NH₃(g) + 6 NO(g) → 5 N₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 145 g of N₂

The molar mass of nitrogen is 28.01 g/mol.

145g \times \frac{1mol}{28.01 g} =5.18 mol

Step 3: Calculate the moles of NO required to produce 5.18 moles of N₂

The molar ratio of NO to N₂ is 6:5.

5.18molN_2 \times \frac{6molNO}{5molN_2} = 6.22molNO

Step 4: Calculate the mass corresponding to 6.22 moles of NO

The molar mass of NO is 30.01 g/mol.

6.22mol \times \frac{30.01g}{mol} =186 g

4 0
3 years ago
How many electrons can occupy the 4p sub level of a neutral atom of bromine?
natita [175]

Answer:

five electrons

Bromine (Z=35), which has 35 electrons, can be found in Period 4, Group VII of the periodic table. Since bromine has 7 valence electrons, the 4s orbital will be completely filled with 2 electrons, and the remaining five electrons will occupy the 4p orbital

5 0
3 years ago
Read 2 more answers
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