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3 years ago

Identify whether the following example is qualitative or quantitative data.

Chemistry

1 answer:

OLga [1]3 years ago

8 0

I think that it is qualitative data

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Crude oil may contain hundreds of different types of hydrocarbons. Some examples include: Butane (C₄H₁₀)

miv72 [106K]

Answer:

The answer to your question is:

Explanation:

Butane (C₄H₁₀) Compound It's composed by C and H

Dodecane (C₁₂H₂₆) Compond It's composed by C and H

Octane (C₈H₁₈) Compound It's composed by C and H

Benzene (C₆H₆) Compound It's composed by C and H

Oxygen (O₂) Element It's only oxygen

Carbon dioxide (CO₂) Compound It's composed by C and O

Water (H₂O) Compound It's composed by H and O

Gasoline Mixture It's composed by many hydrocarbons

Kerosene Mixture It's a mixture of hydrocarbons

4 0

3 years ago

Read 2 more answers

Help I Will Mark Brainlist !! (ATTACHED IMAGE BELOW) Which of the diagrams below would be the FASTEST? _____ Why? Which of the d

VLD [36.1K]

Answer:

Fastest C. Reaction time is faster

Slowest A. Reaction time is slower and it takes longer to finish

Explanation:

4 0

2 years ago

It takes 90 N of force to lift a box of books off my smiths table. If you can only leave the box 25% of the way up, how many new

leva [86]

Answer:3.6 I think sorry if wrong

Explanation:

90 divided by 25

4 0

3 years ago

If an atom has 7 protons, 8 neutrons, and 10 electrons, then it has a net charge of . On the other hand, if an atom has 3 proton

zloy xaker [14]

Answer:

Neutrons play no role in the net charge of an atom, so you can ignore them here.

Proton = 7

Electron = 10

Net Charge = 7 - 10

= -3

Again,

Proton = 3

Electron = 2

Net Charge = 3 - 2

= 1

4 0

2 years ago

has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a

Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

$A + B \rightarrow C$

Initial : 0.3 0.4 0

Change: -x -x x

Equilbm: (0.3 - x) (0.4 - x) x

We know that, relation between standard free energy and equilibrium constant is as follows.

$\Delta G = -RT ln K$

Putting the given values into the above formula as follows.

$\Delta G = -RT ln K$

$-3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})$

x = 0.1417

Hence, at equilibrium

[A] = 0.3 - 0.1417

= 0.1583 M

[B] = 0.4 - 0.1417

= 0.2583 M

[C] = 0.1417 M

5 0

3 years ago