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Flura [38]
3 years ago
5

Which statements are correct relative to carbohydrates?

Chemistry
1 answer:
inna [77]3 years ago
8 0

Answer:

Hi, the statements given in the question (a and b) are combination of right and wrong answers. But i will split the statement a and b into necessary fragments.

Carbohydrates are quick source of energy relatively. That is the only correct statement in the question.

Explanation:

Carbohydrates are naturally occurring organic compounds containing carbon, hydrogen and oxygen with the general formula (Cx H2y Oy) or Cx (H2O)y. They can be classified into simple sugar or complex sugar.

Simple sugars can be further divided into Monosaccharides (e.g glucose, fructose and galactose) and Disaccharides (e.g sucrose and maltose).

Complex sugar can be referred to as Polysaccharides (e.g cellulose and starch)

The statement above that says carbohydrate contains carbon,  hydrogen and nitrogen polymers is not correct.

Carbohydrates are not mostly monosaccharides. They also contain Disaccharides and Polysaccharides.

They are not the only source of fuel.

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a compound has a percent compostion of carbon equal to 48.8383%, hydrogen equal to 8.1636%, and oxygen equal to 43.1981%. what i
Margarita [4]

Answer:

C₂H₃O

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 48.8383%

Hydrogen (H) = 8.1636%

Oxygen (O) = 43.1981%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 48.8383%

H = 8.1636%

O = 43.1981%

Divide by their molar mass

C = 48.8383 / 12 = 4.07

H = 8.1636 / 1 = 8.1636

O = 43.1981 / 16 = 2.7

Divide by the smallest

C = 4.07 / 2.7 = 2

H = 8.1636 / 2.7 = 3

O = 2.7 /2.7 = 1

Thus, the empirical formula of the compound is C₂H₃O

8 0
3 years ago
A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur
dsp73

Answer:

12.8 g of O_{2} must be withdrawn from tank

Explanation:

Let's assume O_{2} gas inside tank behaves ideally.

According to ideal gas equation- PV=nRT

where P is pressure of O_{2}, V is volume of O_{2}, n is number of moles of O_{2}, R is gas constant and T is temperature in kelvin scale.

We can also write, \frac{V}{RT}=\frac{n}{P}

Here V, T and R are constants.

So, \frac{n}{P} ratio will also be constant before and after removal of O_{2} from tank

Hence, \frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}

Here, \frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm} and P_{after}=17atm

So, n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol

So, moles of O_{2} must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of O_{2} = 32 g/mol

So, mass of O_{2} must be withdrawn = (32\times 0.40)g=12.8g

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Answer:

Explanation:

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