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3 years ago

If a man has a mass of 83 kilograms on Earth, what will the force of gravity on his body be on the moon?

Chemistry

1 answer:

VladimirAG [237]3 years ago

3 0

The answer is (A.) 135.6 N

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Here is the formula you will need to answer the question.

Formula:
Force(N) = Mass(kg) x Acceleration(m/s²)
or... F(N) = m(kg) x a(m/s²)

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We need to be certain of the information presented. The problem says the man has a "Mass" of 83 kilograms on Earth. If a man has a Mass of 83 kilograms on Earth, he has a mass of 83 kilograms everywhere. The mass for this problem (even on the moon) is 83.

Mass(kg): 83

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To find the acceleration of object on other planets, moons, ect, you'll need to search the internet or ask a teacher. There are ways to calculate this, but the problem didn't supply the necessary information to do so.

Acceleration: 1.625 m/s²

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Now just plug the information into the formula.

F(N) = m(kg) x a(m/s²)

N = 83 x 1.625
N= 135.875

When considering that there could be a slighty more accurate number for acceleration, it can be concluded that the best answer is (A.)

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The molarity of a solution prepared by dissolving 4.11 g of NaI in enough water to prepare 312 mL of solution is

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Answer:

The correct answer is 8.79 × 10⁻² M.

Explanation:

Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,

No. of moles of NaI = Weight of NaI/ Molecular mass

= 4.11 / 149.89

= 0.027420

The vol. of the solution given is 312 ml or 0.312 L

The molarity can be determined by using the formula,

Molarity = No. of moles/ Volume of the solution in L

= 0.027420/0.312

= 0.0879 M or 8.79 × 10⁻² M

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3 years ago

What is the SI unit for energy? How is it abbreviated?

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Answer:

Joule - J

Explanation:

As energy is defined via work, the SI unit of energy is the same as the unit of work – the joule (J).

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3 years ago

0.000068 g Express your answer as an integer.

ivann1987 [24]

6.8x10^5? Scientific notation

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4 years ago

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A 1.26 m aqueous solution of an ionic compound with the formula mx2 has a boiling point of 101.63 ∘c. part a calculate the van't

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Answer is: Van't Hoff factor (i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT =i · Kb · b.
Kb - molal boiling point elevation constant is 0.512°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 1.26 m.
ΔT = 101.63°C - 100°C = 1.63°C.

i = 1.63°C ÷ (0.512°C/m · 1.26 m).

i = 1.051.

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3 years ago

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Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc

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Answer:

The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%
The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = $\frac{4 }{3} \pi r^2$

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = $\frac{a\sqrt{2} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3$ = $\frac{\pi *a^3\sqrt{2}}{24}$

Number of atoms/unit cell = 4

Total volume of the atoms = $4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6}$ = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = $\frac{a\sqrt{3} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3$ = $\frac{\pi *a^3\sqrt{3}}{16}$

Number of atoms/unit cell = 2

Total volume of the atoms = $2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8}$ = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = $\frac{a\sqrt{3} }{8}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3$ = $\frac{\pi *a^3\sqrt{3}}{128}$

Number of atoms/unit cell = 8

Total volume of the atoms = $8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16}$ = 0.3401

= 0.3401 X 100% = 34.01%

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3 years ago