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3 years ago

Why is 6-decene not possible?

Chemistry

1 answer:

faltersainse [42]3 years ago

3 0

Well because the lowest possibility of having a decen is 3-decene so 6-decene is not possible. Hope this helps!! :)

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Calculate the ΔHrxn for the following

Vesnalui [34]

Answer:

2 NO (g) → N2 (g) + O2 (g)

2 NOCl (g) → 2 NO (g) + Cl2 (g)

____________________________

2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)

ΔH = [90.3 kJ x 2 x -1] + [-38.6 kJ x -1 x 2] = -103.4 kJ

The ΔH for the reaction is -103.4 kJ

6 0

3 years ago

Which coefficient before potassium (K) balances this chemical equation?

miskamm [114]

Answer:

The coefficient before potassium (K) balances this chemical equation is 2.

Explanation:

_K +Cl₂ → 2KCl

K =1 ; Cl =2

K=1 × 2 = 2

Cl = 1 × 2 = 2

2 K +Cl₂ = 2 KCl

5 0

3 years ago

Read 2 more answers

A gas is contained in a thick-walled balloon. When the pressure changes from 45.0 bar to 96.0 bar, the volume changes from 1.20

notsponge [240]

P1 * V1 ÷ T1 = P2 * V2 ÷ T2

45 * 1.20 ÷ 314 = 96 * V2 ÷ 420

30,144 * V2 = 22,680

V2 = 22,680 ÷ 30,144

The new volume is approximately 0.75 liter.

I hope I helped

5 0

3 years ago

Read 2 more answers

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

3 years ago

Unstable isotopes undergo radioactive decay. what occurs during radioactive decay?

mel-nik [20]

Here is what radioactive decay is:
<span>Radioactive decay is the spontaneous breakdown of an atomic nucleus resulting in the release of energy and matter from the nucleus. Remember that a radioisotope has unstable nuclei that does not have enough binding energy to hold the nucleus together.</span>

5 0

3 years ago