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Brums [2.3K]
2 years ago
10

What is the relationship between layering of fluids and density?

Chemistry
1 answer:
Olin [163]2 years ago
7 0

Density of a liquid determines how it will layer (heaviest to lightest). If the liquid is least dense it will float to the bottom. Layers will remain separated because each liquid is actually floating on top of the more dense liquid beneath it.

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Avoiding an accident when driving can depend on reaction time. That time, measured in seconds from the moment the driver see dan
crimeas [40]

Answer:

The answer is below

Explanation:

A normal model is represented as (μ, σ). Therefore for (1.5, 0.18), the mean (μ) = 1.5 and the standard deviation (σ) = 0.18

The z score shows by how many standard deviations the raw score is above or below the mean. It is given as:

z=\frac{x-\mu}{\sigma}

a) For x < 1.35 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.35-1.5}{0.18}=-0.83

From the normal distribution table, the percent of drivers have a reaction time less than 1.35 seconds = P(x < 1.35) = P(z < -0.83) = 0.2033 = 20.33%

b) For x > 1.9 s

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.9-1.5}{0.18}=2.22

From the normal distribution table, the percent of drivers have a reaction time greater than 1.9 seconds = P(x > 1.9) = P(z > 2.22) = 1 - P(z<2.22) = 1 - 0.9868 = 0.0132 = 1.32%

c) For x = 1.45

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.45-1.5}{0.18}=-0.28

For x = 1.75

z=\frac{x-\mu}{\sigma}\\\\z=\frac{1.75-1.5}{0.18}=1.39

From the normal distribution table, P(1.45 < x < 1.75) = P(-0.28 < z < 1.39) = P(z < 1.39) - P(z< - 0.28) = 0.9177 - 0.3897 = 0.528 = 52.8%

d) A percentage of 10% corresponds to a z score of -1.28

z=\frac{x-\mu}{\sigma}\\\\-1.28=\frac{x-1.5}{0.18}\\\\x-1.5=-0.2034\\\\x=1.27

e) P(z < z1) - P(z< -z1) = 60%

P(z < z1) - P(z< -z1) = 0.6

P(z < -z1) = 1 - P(z < z1)

P(z<z1) - (1 - P(z < z1)) = 0.6

2P(z<z1) - 1= 0.6

2P(z<z1) = 1.6

P(z<z1) = 0.8

From the z table, z1 = 0.85

0.85=\frac{x-1.5}{0.18}and-0.85=\frac{x -1.5}{0.18}  \\\\x=1.65 \ and\ x=1.35

The reaction time between 1.35 and 1.65 seconds

8 0
3 years ago
What is cloud ? How does it affect on weather ?​
Lapatulllka [165]

Clouds are xondense form of water vapour over dust particles present in the upper atmosphere.

It effects the weather as when the clouds get heavy they get precipitate in the form pf rain, snow, hail. because of these factors temperature goes down and make us feel cold.

HOPE THIS WILL HELP U

7 0
3 years ago
Read 2 more answers
The term “electricity” in referring to static electricity, means what
sladkih [1.3K]
Is an imbalance of electric charges within or on the surface of a material. The charge remains until it is able to move away by means of an electric current or electrical discharge. Static electricity is named in contrast with current electricity, which flows through wires or other conductors and transmits energy.[1]
5 0
3 years ago
A group of students created a model showing a type of matter which of the following discribes what is shows by the model?
Lady_Fox [76]

Answer:

liquid

Explanation:

6 0
3 years ago
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
3 years ago
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