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Hunter-Best [27]
3 years ago
10

Consider a sample of helium and a sample of neon, both at 30.0°C and 1.5 atm. Both samples have a volume of 5.0 liters. Which st

atement concerning these samples is not true? Consider a sample of helium and a sample of neon, both at 30.0°C and 1.5 atm. Both samples have a volume of 5.0 liters. Which statement concerning these samples is not true?
A) Each sample weighs the same amount.
B) The density of the neon is greater than the density of the helium.
C) Each sample contains the same number of atoms of gas.
D) Each sample contains the same number of moles of gas.
E) none of the above
Chemistry
1 answer:
Triss [41]3 years ago
7 0

Answer:

A.

Explanation:

Using the ideal gas equation, we can calculate the number of moles present. I.e

PV = nRT

Since all the parameters are equal for both gases, we can simply deduce that both has the same number of moles of gases.

The relationship between the mass of each sample and the number of moles can be seen in the relation below :

mass in grammes = molar mass in g/mol × number of moles.

Now , we have established that both have the same number of moles. For them to have the same mass, they must have the same molar masses which is not possible.

Hence option A is wrong

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How many molecules OF2 would have a mass of .132 g
BlackZzzverrR [31]
<h3>Answer:</h3>

1.47 × 10²¹ molecules OF₂

<h3>General Formulas and Concepts: </h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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<h3>Explanation: </h3>

<u>Step 1: Define</u>

0.132 g OF₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of O - 16.00 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of OF₂ - 16.00 + 2(19.00) = 54.00 g/mol

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 0.132 \ g \ OF_2(\frac{1 \ mol \ OF_2}{54.00 \ g \ OF_2})(\frac{6.022 \cdot 10^{23} \ molecules \ OF_2}{1 \ mol \ OF_2})
  2. Divide/Multiply:                                                                                               \displaystyle 1.47204 \cdot 10^{21} \ molecules \ OF_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.47204 × 10²¹ molecules OF₂ ≈ 1.47 × 10²¹ molecules OF₂

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