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3 years ago

Consider the reaction of 1-butanol with K2Cr2O7, H2SO4, heat. Draw only the organic product derived from 1-butanol.

Chemistry

1 answer:

guapka [62]3 years ago

7 0

Answer:

Butanoic acid.

Explanation:

Hello,

In this case, when a primary alcohol such as 1-butanol (OH is bonded to a primary carbon) is oxidized in the presence of a strong oxidizing media such as potassium dichromate (K2Cr2O7) and sulfuric acid, the stepwise oxidation goes to the corresponding aldehyde with a further oxidation to the corresponding carboxylic acid:

$R-CH_2-OH\longrightarrow R-COH\longrightarrow R-COOH$

Therefore, on the attached picture you can find that the formed aldehyde is butanal and the inly organic product, due to the strong oxidizing media is finally butanoic acid.

Best regards.

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Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

3 years ago

How many moles of O2 are needed to burn 2.56 moles of CH3OH?

lukranit [14]

Answer:

$n_{O_2}=3.84molO_2$

Explanation:

Hello!

In this case, since the combustion reaction of methanol is:

$CH_3OH+\frac{3}{2} O_2\rightarrow CO_2+2H_2O$

In such a way, since there is 1:3/2 mole ratio between methanol and oxygen, we can compute the moles of oxygen that are needed to burn 2.56 moles of methanol as shown below:

$n_{O_2}=2.56molCH_3OH*\frac{\frac{3}{2}molO_2}{1molCH_3OH} \\\\n_{O_2}=3.84molO_2$

Best regards!

6 0

3 years ago

Which of the following is NOT true regarding Rutherford's Gold Foil experiment?

erastova [34]

Answer:

The area around the nucleus must be of low mass.

Explanation:

Rutherford`s experiment showed that there are some positive charges in the center of the atoms, and because they are all together, they will give a great mass to the atom.

It was quite different from Thomson`s experiment, in which it was thought that the negative charges were mixed with the positive charges, around the atom (like a Pudding Model). In Rutherford`s experiment, because the direction of beta particles, it was the prediction of the positive nucleus.

Hope this info is useful.

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