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3 years ago

Why are iodoalkanes very reactive

Chemistry

1 answer:

makvit [3.9K]3 years ago

5 0

Answer:

Iodoalkanes are very reactive because they have the weakest bond of C-I, thus they are most easily broken, making them most reactive. B. The carbon where the halogen is bonded is only attached to one other carbon.

Explanation:

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1: At which temperature would a reaction withΔH = -102 kJ/mol, ΔS = -0.188 kJ/(mol×K) be spontaneous? 2: At which temperature wo

Naddik [55]

Answer:

1: At temperatures below 542.55 K

2: At temperatures above 660 K

Explanation:

Hello there!

In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:

$\Delta G=\Delta H-T\Delta S$

Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:

$0=\Delta H-T\Delta S\\\\T=\frac{-102kJ/mol}{-0.188kJ/mol-K} \\\\T=542.55K$

It means that at temperatures lower than 542.55 K the reaction will be spontaneous.

$0=\Delta H-T\Delta S\\\\T=\frac{132kJ/mol}{0.200kJ/mol-K} \\\\T=660K$

It means that at temperatures higher than 660 K the reaction will be spontaneous.

Best regards!

7 0

2 years ago

What does it regulate? One sentence will help!

Mandarinka [93]

<span>a regultate is to control or direct by a rule, principle, or method.</span>

7 0

3 years ago

Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lith

alexandr402 [8]

The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>

Let Li6 be isotope A
Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>

Molar mass of isotope A (Li6) = 6.02 u
Molar mass of isotope B (Li7) = 7.02 u
Atomic mass of lithium = 6.94 u
Abundance of A = A%
Abundance of B = (100 - A)%

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

Abundance of A (Li6) = 8%
Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

Learn more about isotope:

brainly.com/question/24311846

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4 0

2 years ago

HEEELLP WILL GIVE 100 PIINTS ASAP

adell [148]

Answer:

Explanation:

8 0

2 years ago

Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE

nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let " $\alpha$ is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

$HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}$

Initial C 0 0

Equilibrium c(1- $\alpha$ ) c $\alpha$ c $\alpha$

Dissociation constant = $Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}$

$=\frac{c\alpha^{2}}{(1-\alpha)}$

In this case weak acids $\alpha$ is very small so, (1- $\alpha$ ) is taken as 1.

$Ka=C\alpha^{2}$

$\alpha=\sqrt\frac{ka}{c}$

From the given the concentration = 0.050 M

Substitute the given value.

$\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}$

$[H_{3}O^{+}]=c\alpha$

$[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}$

$pH= -log[H_{3}O^{+}]$

$=-log[4.9\times10^{-6}]$

$=6-log 4.9= 5.31$

Therefore, The pH of the solution is 5.31.

7 0

3 years ago

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