All categories

2 years ago

Compare the composition of sucrose purified from the sugar cane with the composition of sucrose purified from sugar beets.

2 answers:

8 0

The problem ask to compare the composition of sucrose purified from the sugar cane with the composition of sucrose purified from beets and the answer is that the two sucrose molecule will be chemically and structurally the same. I hope you are satisfied with my answer and feel free to ask for more

Vladimir79 [104]2 years ago

3 0

Answer: they are the same.

Explanation:

1) Sucrose is a compound with chemical formula C₁₂H₂₂O₁₁

2) That means that all molcules of sucrose will have the same kind of atoms in the same proportion, whic is to say same composition:

12 atoms of C: 22 atoms of H: 11 atoms of O, per each molecule of sucrose.

3) For this question you can rely in the definition of compound: a pure substance formed by the combination of two or more elements always in the same ratio (same composition).

4) That also implies, that all the molecules of sucrose have the same properties.

You might be interested in

An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g

jekas [21]

Explanation:

A mixture is defined as the substance that contains two or more different number of substances that are physically mixed together.

For example, a mixture of air which contains oxygen, nitrogen and other gases.

A mixture in which solute particles are unevenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A solution is defined as the substance in which two or more substances are mixed together.

A compound is defined as the substance that contains two or more different elements that chemically combined together in a fixed ratio by mass.

A element is defined as the substance that contains only one type of atoms.

For example, a piece of sodium element will contain only atoms of sodium.

Whereas a pure substance is defined as the substance which contains only one type of molecule or one type of atom.

For example, $O_{2}$ , $N_{2}$ etc are pure substances.

Thus, we can conclude that the terms which could be used to describe the given sample of air is as follows.

pure chemical substance.

heterogenous mixture.

mixture.

4 0

2 years ago

During a titration the following data were collected. A 10. mL portion of an unknown monoprotic acid solution was titrated with

Liula [17]

Answer:

8.0 moles

Explanation:

Since the acid is monoprotic, 1 mole of the acid will be required to stochiometrically react with 1 mole of NaOH.

Using the formula: $\frac{concentration of acid X volume of acid}{concentration of base X volume of base} = \frac{mole of acid}{mole of base}$

Concentration of acid = ?

Volume of acid = 10 mL

Concentration of base = 1.0 M

Volume of base = 40 mL

mole of acid = 1

mole of base = 1

Substitute into the equation:

$\frac{concentration of acid X 10}{1.0 X 40} = \frac{1}{1}$

Concentration of acid = 40/10 = 4.0 M

To determine the number of moles of acid present in 2.0 liters of the unknown solution:

Number of moles = Molarity x volume

molarity = 4.0 M

Volume = 2.0 Liters

Hence,

Number of moles = 4.0 x 2.0 = 8 moles

8 0

2 years ago

The double bond between carbon and oxygen is similar to an alkene C-C, except that C o is: a) shorter and weaker. b) shorter and

mina [271]

Answer:

the double bond between c and o is shorter and weaker

Explanation:

this is because the bond between c and o involves unequal sharing of electrons whole c and c involves hybridization sp2 of orbitals and also catenation phenomenon in which carbon could form long chain with it's other carbon

5 0

2 years ago

The concentration of ozone in a sample of air that has a partial pressure of O3 of 0.33 torr and a total pressure of air of 695

Goryan [66]

Answer:

0.047 %

Explanation:

Step 1: Given data

Partial pressure of ozone (pO₃): 0.33 torr

Total pressure of air (P): 695 torr

Step 2: Calculate the %v/v of ozone in the air

Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.

%v/v = 0.33 torr/695 torr × 100%

%v/v = 0.047 %

8 0

2 years ago

A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns c

makkiz [27]

0.216g of aluminium compound X react with an excess of water water to produce gas. this gas burn completely in O2 to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure

0.108 / n = 24 / 1
n = 0.0045 mole ( CO2 >>0.0045 mole
0.216 - 0.0045 = 0.2115
so Al = 0.2115 / 27 => 0.0078 mole
C = 0.0045 * 1000 => 4.5 and Al = 0.0078 * 1000 = 7.8

7 0

2 years ago