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gizmo_the_mogwai [7]
3 years ago
12

A student is trying to calculate the density of a ball. She already knows the mass, but she needs to determine the volume as wel

l. Which of the following formulas can be used to calculate the volume of the ball?
Group of answer choices

V equals four-thirds times pi times r cubed

V equals one-third times pi times r squared times h

V = s3

V = πr2h
Chemistry
1 answer:
murzikaleks [220]3 years ago
7 0

Answer:

V equals four-thirds times pi times r cubed

Explanation:

Volume = a³ , where a is length of each side. Volume = l × w × h , where l is length, w is width and h is height. Volume = 4/3 πr³ , where r is the radius. Volume = πr²h , where r is the radius and h is the height.

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Smog usually forms from gound-level ozone and what other human-made pollutant?
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This happens when pollutants emitted by cars, power plants, industrial boilers, refineries, chemical plants, and other sources chemically react in the presence of sunlight. Ozone at ground level is a harmful air pollutant, because of its effects on people and the environment, and it is the main ingredient in “smog."

4 0
3 years ago
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Provide two written definitions of an oxidation-reduction reaction and for each definition tell which reactant is reduced and wh
vesna_86 [32]

Answer:

Explanation:

Oxidation:

1) Oxidation involve the removal of electrons.

2) Oxidation occur when oxidation state of atom of an element is increased.

Reduction:

1) Reduction involve the gain of electron.

2) Reduction is occur when number of electrons increased in an atom and oxidation state decreased.

Consider the following reactions.

4KI + 2CuCl₂  →   2CuI  + I₂  + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced.

CO + H₂O   →  CO₂ + H₂

the oxidation state of carbon is +2 on reactant  side and on product side it becomes  +4 so carbon get oxidized.

Na₂CO₃ + H₃PO₄  →  Na₂HPO₄ + CO₂ + H₂O

The oxidation state of carbon on reactant side is +4. while on product side is  also +4 so it neither oxidized nor reduced.

H₂S + 2NaOH → Na₂S + 2H₂O

The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.

The  most comprehensive definition is 2nd definition which involves the increase or decrease in oxidation state.

4 0
4 years ago
Chemical equation of sodium carbonate on hydrochloric acid<br>​
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Answer: Na2CO3 + 2 HCl → 2 NaCl + H2CO3

I hope this helped!

<!> Brainliest is appreciated! <!>

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5 0
3 years ago
36 g of water vapor takes up how many liters at room temperature and pressure of 293K and 100 K PA
crimeas [40]

100kpa is correct .100 kpa is correct

6 0
4 years ago
Read 2 more answers
A 1.30M solution of BaCl2 has a density of 1.230 g/ml. a) What is the mole fraction of BaCl2 in this solution?
Elodia [21]

<u>Answer:</u> The mole fraction of barium chloride in the solution is 0.024

<u>Explanation:</u>

We are given:

Molarity of barium chloride solution = 1.30 M

This means that 1.30 moles of barium chloride is present in 1 L or 1000 mL of solution.

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.230 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.230g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.230g/mL\times 1000mL)=1230g

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

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

Moles of barium chloride = 1.30 moles

Molar mass of barium chloride = 208 g/mol

Putting values in equation 1, we get:

1.30mol=\frac{\text{Mass of barium chloride}}{208g/mol}\\\\\text{Mass of barium chloride}=(1.30mol\times 208g/mol)=270.4g

Mass of water = Mass of solution - Mass of barium chloride

Mass of water = 1230 - 270.4 = 959.6 g

<u>Calculating the moles of water:</u>

Given mass of water = 959.6 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{959.6g}{18g/mol}\\\\\text{Moles of water}=53.31mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

  • <u>For barium chloride:</u>

\chi_{\text{(barium chloride)}}=\frac{n_{\text{(barium chloride)}}}{n_{\text{(water)}}+n_{\text{(barium chloride)}}}

\chi_{\text{(barium chloride)}}=\frac{1.30}{1.30+53.31}\\\\\chi_{\text{(barium chloride)}}=0.024

Hence, the mole fraction of barium chloride in the solution is 0.024

6 0
4 years ago
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