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erma4kov [3.2K]
3 years ago
13

Bugaboo can be oxidised to a carboxylic acid by heating with acidified potassium manganate (VII). Give the name and structural f

ormula of the carboxylic acid.
Chemistry
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer:

See below.

Explanation:

CH3CH2CH2CH2OH is butanol.

The carboxylic acid formed , butyric acid has the formula:

CH3CH2CH2COOH.

Structural formula:

       H     H    H       O                  

        |       |      |        ||                      

H  -  C  -  C  -  C   -  C  -  OH

       |       |       |    

      H     H     H

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2 years ago
Which of the following lists includes only molecules?
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We are unable to see any list? Can you provide one?
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3 years ago
A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.
Bingel [31]

Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO

And consequently the moles:

n_O=0.12molO

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1

Thus, the empirical formula, taken the nearest whole number is:

C_5H_{10}O_2

Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.

Best regards!

6 0
2 years ago
given the following list of densities, which material would float in a molten vat of lead provided that they do not themselves m
SIZIF [17.4K]

Answer:

Given the following list of densities, which materials would float in a molten vat of lead provided that they do not themselves melt?

Densities (g/mL): lead = 11.4, glass = 2.6, gold = 19.3, charcoal = 0.57, platinum = 21.4.

glass and charcoal

Explanation:

The density of molten lead is about 10.65Kg/m^3

By Archimedes principles, the buoyancy of an object in a fluid is proportional to the mass of fluid displaced

which in turn is proportional to the object's density

Generally an object well float when placed on a denser medium

glass =2.6, and charcoal =0.57 are both less dense so they will float on Lead

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3 years ago
How many protons dose p have
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Answer:

i have to see the question

Explanation:

5 0
3 years ago
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