All categories

3 years ago

How can Newton's Third Law of Motion be used to explain why it is difficult to walk on a slippery surface, such as ice?

1 answer:

6 0

E ground pushes you forward. But that interaction is friction. Reduce friction and it doesn't matter how strong your legs are, the surface is incapable of pushing you accordingly. The coefficient of static friction is very low so it is easy to slide your foot rather than push.

You might be interested in

Give another mixture of liquids that is separated on an industrial scale by fractional

iren [92.7K]

Answer:

dna or diabeties can be separated

7 0

3 years ago

Express 7.25 in scientific notation

sergeinik [125]

Considering that scientific notation has to move the decimal over so there is only one digit in the Tens place, that is the scientific notation.

8 0

3 years ago

Read 2 more answers

What type of map should have a north arrow ,key , and map scale?

Radda [10]

<h2>Answer : Political Map</h2><h3>Explanation :</h3>

A political map is a kind of map which contains a north arrow, key and a map scale.

This kind of map usually indicates the governmental boundaries of countries, states, or any counties,

And, they give information about the location of major cities, also they usually include significant bodies of water.

8 0

3 years ago

Read 2 more answers

Interpreting Velocity vs. Time Graphs

NemiM [27]

Answer:first part is A and C, second part is C, third part is 40 m/s and the fourth part is zero

Explanation: I guessed and got them right lol

4 0

3 years ago

Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of alumin

Lerok [7]

<u>Answer:</u> The percentage yield of HF is 73.36 %

<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)

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg = 6250 g (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium fluoride}=\frac{6250g}{78.07g/mol}=80.05mol$

For the given chemical reaction:

$CaF_2+H_2SO_4\rightarrow CaSO_4+2HF$

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = $\frac{2}{1}\times 80.05=160.1mol$ of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

$160.1mol=\frac{\text{Theoretical yield of hydrofluoric acid}}{20.01g/mol}=3203.6g=3.20kg$

To calculate the percentage yield of hydrofluoric acid, we use the equation:

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

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

$\%\text{ yield of hydrofluoric acid}=\frac{2.35g}{3.20g}\times 100\\\\\% \text{yield of hydrofluoric acid}=73.36\%$

Hence, the percentage yield of HF is 73.36 %

4 0

3 years ago