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3 years ago

14

Based on the information provided here, crude oil is:

1 answer:

Anna007 [38]3 years ago

8 0

THE EXPLINATION ANSWER: I think crude oil is a mixture because it is made up of compounds that can be separated using distillation, a physical process.

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Need the names for these asap please

jeyben [28]

Answer:

3_ethylpentane

4_ethylheptane

1_pentene

3_hexene

5 0

3 years ago

Determine the chemical formulas for the two compounds below. (Carbon, Hydrogen, and Oxygen).

tester [92]

Remember that any intersection of lines is a C, and that the number of hydrogens attached are the necessary to complet the 4 bonds.

1) CH3 - CH (OH) - CH (CH3) -CH3

2) CH3 - O - CH(CH3)-CH2 - CH3

I have used the parenthesis to indicate that the radical inside is in other branch, bonded by a single line -

4 0

3 years ago

Write the oxidation number for the following elements: for brainliest answer

ollegr [7]

Explanation:

Here's an oxidation chart to help

..................

6 0

3 years ago

(2) 8.1.3 Write down the structural formula of alcohol used in this reaction.

vitfil [10]

Answer:

-OH

Explanation:

Alcohols generally have the structural formula OH

for example, ethanols structural formula is C2H5OH

7 0

3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago