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3 years ago

How many methylene groups are present in 2,4-dimethylhexane?

Chemistry

2 answers:

stepan [7]3 years ago

8 0

There are zero number of methylene groups in the substance 2,4-dimethylhexane. Methylene groups are the part of a molecule that contains a carbon atom that has a double bond or =CH. Looking at the structure of 2,4-dimethylhexane, you cannot see any carbon atom that has a double bond.

dusya [7]3 years ago

4 0

Answer:

Zero methylene groups

Explanation:

Hi, on one side you have the 2.4-dimethylhexane molecule as shown in the figure.

On the other side, you have the methylene group with the following structure:

=CH2

As can be seen there isn't any methylene group in the 2.4-dimethylhexane

Important: don't confuse the methylene group with the methyl group. Methyl group is -CH3 and there are two of them in this molecule.

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How many grams of H2O can be made from the combustion of 3.75 liters of C7H14 and an excess of O2 at STP?

Kitty [74]

Answer:

21.10g of H2O

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2C7H14 + 21O2 —> 14CO2 + 14H2O

From the balanced equation above, 2L of C7H14 produced 14L of H2O.

Therefore, 3.75L of C7H14 will produce = (3.75 x 14)/2 = 26.25L of H2O.

Next, we shall determine the number of mole of H2O that will occupy 26.25L at stp. This is illustrated below:

1 mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2O will occupy

26.25L i.e

Xmol of H2O = 26.25/22.4

Xmol of H2O = 1.172 mole

Therefore, 1.172 mole of H2O is produced from the reaction.

Next, we shall convert 1.172 mole of H2O to grams. This is illustrated below:

Number of mole H2O = 1.172 mole

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O =..?

Mass = mole x molar mass

Mass of H2O = 1.172 x 18

Mass of H2O = 21.10g

Therefore, 21.10g of H2O is produced from the reaction.

3 0

3 years ago

Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...

g100num [7]

Answer:

O₂; KCl; 33.3

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

2KCl + 3O₂ ⟶ 2KClO₃

n/mol: 100.0 100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

$\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}$

(ii) From O₂

$\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}$

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

$\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}$

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

3 0

3 years ago