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Lunna [17]
2 years ago
14

A sample of oxygen, o2, occupies a volume of 0. 65 l at 11°c under 0. 50 atm of pressure. (r = 0. 08205 latm/molk). how many mol

es of o2 are present in this sample? (pv = nrt) group of answer choices
Chemistry
1 answer:
WINSTONCH [101]2 years ago
7 0

It is calculated that 0.014 moles of oxygen (O₂) are present in the given sample by using the ideal gas law.

Calculation:

Provided that :

Temperature, T = 273+11 K = 284 K

Pressure, P = 0.5 atm

Volume, V = 0.65 L

Ideal gas constant,R = 0.08205 L atm/mol K

According to the ideal gas law,

we know the formula, PV = nRT

or, n = PV / RT = (0.5 * 0.65) / (0.08205 * 284)

or, n = 0.0139 ≈ 0.014 mole

What is the ideal gas law?

  • The universal gas equation, often known as the ideal gas law, is the formula of state for a fictitious ideal gas. Although it has several restrictions, it is a decent approximation of the behavior of numerous gases under various circumstances.
  • Benoît Paul Émile Clapeyron introduced it for the first time in 1834 as a synthesis of the experimental Boyle's law, Charles' law, Avogadro's law, and Gay-Lussac's law.

Learn more about ideal gas law here:

brainly.com/question/13821925

#SPJ4

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3 years ago
A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo
Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>
  • Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

                                   = (2.684 - 2.156) g

                                   = 0.528 g

  • Find the number of moles of the elements in the compound

The number of moles is given by

n = \frac{m}{M}

where m = given mass and

M = Molar mass

Number of moles of Zinc = \frac{2.156}{65} = 0.033 moles

Number of moles of Oxygen =\frac{0.528}{16} = 0.033 moles

  • Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

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Explanation:

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What is the percent by mass of oxygen in mg(oh)2
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54.868% because well la la la (sorry had to be 20 characters)

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