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MatroZZZ [7]
2 years ago
10

One mole of an element contains which of the following?

Chemistry
2 answers:
lapo4ka [179]2 years ago
5 0
B. 6.02 x 10^23 atoms
alukav5142 [94]2 years ago
3 0

Answer : The correct option is, (B) 6.02\times 10^{23} atoms.

Explanation :

Mole : It is defined as the amount of substance that contain the same number of units as there are atoms present in exactly 12 grams of carbon-12.

Avogadro's number : It is defined as the number of atoms molecules present in a mole of any substance.

The Avogadro's number is represented as, 6.022\times 10^{23}

The relationship between a mole and Avogadro number are :

As, 1 mole of a substance contains 6.022\times 10^{23} number of atoms, molecules or ions.

Hence, the correct option is, (B) 6.02\times 10^{23} atoms.

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IgorLugansk [536]

This is a one-step unit analysis problem.  Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.

1 mole is equal to 6.022x10²³ particles as given, so:

1.5x10^{24} particles FeO_{2} (\frac{1mol}{6.022x10^{23}particles } )=2.49 molFeO_{2}

<h3>Answer:</h3>

2.49 mol

Let me know if you have any questions.

3 0
3 years ago
A sample of gas has a volume of 2.36 L at a temperature of 53.00 °C. The gas sample is heated to a temperature of 139.00 °C (ass
Liono4ka [1.6K]

Answer:

The volume increases because the temperature increases and is 2.98L

Explanation:

Charles's law states that the volume of a gas is directely proportional to its temperature. That means if a gas is heated, its volume will increase and vice versa. The equation is:

V₁/T₁ = V₂/T₂

<em>Where V is volume and T is absolute temperature of 1, initial state, and 2, final state of the gas.</em>

In the problem, the gas is heated, from 53.00°C (53.00 + 273.15 = 326.15K) to 139.00°C (139.00 + 273.15 = 412.15K).

Replacing in the Charles's law equation:

2.36L / 326.15K= V₂/412.15K

<h3>2.98L = V₂</h3>

<em />

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3 years ago
Biodiesel is produced from which type of molecules?
Ksju [112]

Answer:

Lipids

Explanation:

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3 years ago
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