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Volgvan
3 years ago
13

How many rundlets are there in 218 in3?

Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
3 0

Answer:

218 "cubic inch" = 0.05238 "rundlets"

Explanation:

From the conversion factors;

1.00 gal = 231 in3 = 3.78 L

But;

1.00 rundlet = 6.81 × 104 mL = 68.1 L (Converting to L by dividing by 1000)

231 in3 = 3.78 L

218 in3 = x

x = (218 * 3.78 ) / 231

x = 3.5673L

But 1 rudlet = 68.1 L

x rundlet = 3.5673L

x = (3.5673 * 1 ) / 68.1

x = 0.05238

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So option D is correct one.

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What change in mass number occurs when a radioactive atom emits an alpha particle?
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Two moles of gas A spontaneously convert to 3 moles of gas b in a container where the temperature and pressure are held constant
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At constant temperature and pressure, If the amount of gas increases to the given value, its volume also increases to 20.85L.

<h3>What is Avogadro's law?</h3>

Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules."

It is expressed as;

V₁/n₁ = V₂/n₂

Given the data in the question;

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  • Final amount of gas n₁ = 3moles
  • Final volume v₂ = ?

V₁/n₁ = V₂/n₂

V₁n₂ = V₂n₁

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At constant temperature and pressure, If the amount of gas increases to the given value, its volume also increases to 20.85L.

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4 0
1 year ago
What is the difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm
Zanzabum

Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

Explanation:

<u>Part I :</u>

n =\frac{3.01\times 10^{24}}{6.022\times 10^{23}}

n = 4.9983

n = 4.99 moles

(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

So, 4.99 moles of gold contain = 5\times 196.96 g

4.99 moles of gold contain = 984.8 g

Mass of {3.01\times 10^{24}} atoms of gold = 984.5 g

<u>Part II :</u>

Density of Gold = 19.32 g/cm^{3}

Volume of the cuboid = length\times breadth\times height

Volume of the gold bar =6.00\times 4.25\times 2.00

Volume of the gold bar = 51cm^{3}

Using formula,

Density = \frac{mass}{Volume}

Mass = Density\times Volume

Mass = 19.32 \times 51

Mass = 985.32 g

So, A  gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

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