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3 years ago

In a reaction between 1-pentene and cl2, what are the products?

1 answer:

7 0

In the reaction of an alkene and a halogen, each halogen occupies the space of a hydrogen. This is an addition type of reaction. in this case, we are given with 1-pentene that is C5H10 reacting with Cl2. THe product is C5H8Cl2 or pentenyl chloride

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Consider the Gibbs energies at 25 ∘C.

zysi [14]

Answer:

A)ΔGrxn∘=55.7kJ/mol

B)Ksp=1.75×10^−10

C)ΔGrxn∘=70.0kJ/mol

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=ΔGf,products∘−ΔGf,reactants∘

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.

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Cl(aq)−∘]−ΔGf,

AgCl(s)⇌Ag(aq)++Cl(aq)−

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

ΔGrxn∘=55.7kJ/mol

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7kJ/mol=−(8.3145×10−3J/molK)(298.15K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

ΔGrxn∘=[ΔGf,Ag(aq)+∘+ΔGf,Br(aq)−∘]−ΔGf,

ΔGrxn∘=[77.1kJ/mol+(−104.0kJ/mol)]−(−96.90kj/mol

ΔGrxn∘=70.0kJ/mol

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

ΔGrxn∘=−RTlnKsp

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

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

Ksp=5.45×10^−13

8 0

3 years ago

Question 3

hammer [34]

Answer:

in the excited state

Explanation:

Because in excited state an atom has more energy

5 0

3 years ago

Read 2 more answers

Who discovered copper and in what year?

Vesnalui [34]

8,700 B.C.
No clue who discovered it.

4 0

3 years ago

Read 2 more answers

What is the number of molecules present in 1.12 dm^3 of nitrogen gas at STP

Anuta_ua [19.1K]

Answer:

$\huge\boxed{\sf No.\ of\ molecules = 3 * 10\²\² \ molecules}$

Explanation:

<u>Given Data:</u>

Volume = v = 1.12 dm³ = 1.12 L

Density of nitrogen at STP = D = 1.25 g / L

Molar mass = M = 14 * 2 = 28 g / mol

Avogadro's Number = $\tt{N_{A}}$ = 6.023 * 10²³ mol⁻¹

<u>Required:</u>

No. of molecules = ?

<u>Formula:</u>

$\tt{No. \ of \ molecules = \frac{Density * Volume}{Molar\ Mass} * N_{A}}$

<u>Solution:</u>

No. of molecules = (1.25*1.12) / 28 * (6.023 * 10²³)

No. of molecules = ( 1.4 / 28 ) * 6.023 * 10²³

No. of molecules = 0.05 * 6.023 * 10²³

No. of molecules = 0.3 * 10²³

No. of molecules = 3 * 10²² molecules

$\rule[225]{225}{2}$

Hope this helped!

4 0

3 years ago

Why did scientists need to establish an international language for elements

Ulleksa [173]

So it could be used in every country(different languages) yet still understood

5 0

3 years ago