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Kisachek [45]
4 years ago
5

A person weighs a different amount on other surfaces in space than they do on Earth because the gravitational pull of each surfa

ce is different than that of Earth. The table and graph show the weights of objects on Earth, Mars, and the Moon. Weight on Earth and Mars Weight on Earth, e 90 pounds 120 pounds 160 pounds Weight on Mars, a 36 pounds 48 pounds 64 pounds Weight on Earth and Moon A graph has weight on earth (pounds) on the x-axis, and weight on moon (pounds) on the y-axis. A line goes through points (60, 10) and (120, 20). If a person weighs 240 pounds on Earth, what would be the difference in weight of the same person on Mars and on the Moon (based on the table and graph)? The person weighs 56 pounds more on Mars than on the Moon. The person weighs 56 pounds more on the Moon than on Mars. The person weighs 560 pounds more on Mars than on the Moon. The person weighs 560 pounds more on the Moon than on Mars.
Chemistry
2 answers:
Masja [62]4 years ago
7 0

Answer:

A. The person weighs 56 pounds more on Mars than on the moon

Explanation:

ANTONII [103]4 years ago
5 0

Answer:

A

Explanation:

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1.G → giga (1 . 10⁹)  

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Commercial hydrochloric acid (HCl) is typically labeled as being 38.0 % (weight %). The density of HCl is 1.19 g/mL. a) What is
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<u>Answer:</u>

<u>For a:</u> The molarity of commercial HCl solution is 12.39 M.

<u>For b:</u> The molality of commercial HCl solution is 16.79 m.

<u>For c:</u> The volume of commercial HCl solution needed is 2.42 L.

<u>Explanation:</u>

We are given:

Mass % of commercial HCl solution = 38 %

This means that 38 grams of HCl is present in 100 grams of solution.

To calculate the volume of solution, we use the equation:

Density=\frac{Mass}{Volume}

Density of HCl solution = 1.19 g/mL

Mass of solution = 100 g

Putting values in above equation:

1.19g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=84.034mL

  • <u>For a:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Molarity of solution = ?

Molar mass of HCl = 36.5 g/mol

Volume of solution = 84.034 mL

Mass of HCl = 38 g

Putting values in above equation, we get:

\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 84.034}\\\\\text{Molality of commercial HCl solution}=12.39M

Hence, the molarity of commercial HCl solution is 12.39 M.

  • <u>For b:</u>

To calculate the molality of solution, we use the equation:

Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (HCl) = 38 g

M_{solute} = Molar mass of solute (HCl) = 36.5 g/mol

W_{solvent} = Mass of solvent = 100 - 38 = 62 g

Putting values in above equation, we get:

\text{Molality of commercial HCl solution}=\frac{38\times 1000}{36.5\times 62}\\\\\text{Molality of commercial HCl solution}=16.79m

Hence, the molality of commercial HCl solution is 16.79 m.

  • <u>For c:</u>

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated solution

M_2\text{ and }V_2 are the molarity and volume of diluted solution

We are given:

M_1=6M\\V_1=5.00L\\M_2=12.39M\\V_2=?L

Putting values in above equation, we get:

6\times 5=12.39\times V_2\\\\V_2=2.42L

Hence, the volume of commercial HCl solution needed is 2.42 L.

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