<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>
Answer:
The majority of the air we breathe is made up of nitrogen and oxygen, though you'll also find argon, carbon dioxide and other gases in trace amounts.
Explanation:
<h3>
Answer:</h3>
3.67 mol Al
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
2.21 × 10²⁴ atoms Al
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:

- Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
3.66988 mol Al ≈ 3.67 mol Al
<u>Answer:</u>
<em>4.5 L water we have in litres (L).</em>
<em><u></u></em>
<u>Explanation:</u>

where
= Final T - Initial T
Q is the heat energy in calories
c is the specific heat capacity (for water 1.0 cal/(g℃))
m is the mass of water
Plugging in the values

So,
Volume of water = mass/density

=4.5 L (Answer)
Answer:
Option A. It has stayed the same.
Explanation:
To answer the question given above, we assumed:
Initial volume (V₁) = V
Initial temperature (T₁) = T
Initial pressure (P₁) = P
From the question given above, the following data were:
Final volume (V₂) = 2V
Final temperature (T₂) = 2T
Final pressure (P₂) =?
The final pressure of the gas can be obtained as follow:
P₁V₁/T₁ = P₂V₂/T₂
PV/T = P₂ × 2V / 2T
Cross multiply
P₂ × 2V × T = PV × 2T
Divide both side by 2V × T
P₂ = PV × 2T / 2V × T
P₂ = P
Thus, the final pressure is the same as the initial pressure.
Option A gives the correct answer to the question.