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shutvik [7]
4 years ago
10

A student conducts an experiment to see how music affects plant growth. The student obtains four identical plants. Each one is p

otted in the same type of soil and receives the same
amount of sunlight and water each day. Plant A listens to classical music for three hours each day. Plant B listens to rock music for three hours each day. Plant C listens to country
music for three hours each day. Plant D does not listen to any music at all.
1. Based on the experiment in the scenario, which visual aid would be most helpful in showing the change in the plants' heights over time?
O A. A line graph
O B. A pie chart
OC. A bar graph
O D. A timeline

Chemistry
1 answer:
frez [133]4 years ago
7 0

Answer:

A. A line graph  

Explanation:

You use line graphs to track changes over time. Line graphs are better when the changes are small. They are also more useful when you want to compare changes over the same period for more than one group, for example, plants exposed to music and a control group.

B is wrong. A pie chart is best for comparing parts of a whole.

C is wrong. You can use a bar graph to track changes over time, but small changes are harder to spot.

D is wrong. You use a timeline to mark important points in time, for example, when you are deciding the times when you must complete various stages of a project.

Which of the charts below do you think is more helpful in showing the change in plant height over time?

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F 2.50 × 10−2 g of solid Fe(NO3)3 is added to 100. ML of a 1.0 × 10−4 M NaOH solution, will a precipitate form? (ksp=4 × 10−38 f
vladimir1956 [14]

Answer:

A precipitate will be formed

Explanation:

The Ksp equilibrium of Fe(OH)₃ is:

Fe(OH)₃ (s) ⇄ Fe³⁺(aq)+ 3OH⁻(aq)

And its expression is:

Ksp = 4x10⁻³⁸ = [Fe³⁺] [OH⁻]³

<em>Where the concentrations are concentrations in molarity in equilibrium,</em>

We can write Q as:

Q = [Fe³⁺] [OH⁻]³

<em>Where [] are actual concentrations in molarity of each specie.</em>

<em />

When Q>= Ksp; a precipitate is formed,

When Q< Ksp no precipitate is produced:

[OH⁻] = [NaOH] = 1.0x10⁻⁴M

[Fe²⁺] = 2.50x10⁻²g * (1mol / 179.85g) / 0.100L = 1.39x10⁻³M

<em>179.85g/mol is molar mass of Fe(NO₃)₂ and the volume of the solution is 0.100L = 100mL</em>

<em />

Q = [Fe³⁺] [OH⁻]³

Q = [ 1.39x10⁻³] [ 1.0x10⁻⁴]³

Q = 3.8x10⁻¹⁵

As Q >> Ksp; A precipitate will be formed

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3 years ago
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