Answer:- 0.800 moles of the gas were collected.
Solution:- Volume, temperature and pressure is given for the gas and asks to calculate the moles of the gas.
It is an ideal gas law based problem. Ideal gas law equation is used to solve this. The equation is:
PV=nRT
Since it asks to calculate the moles that is n, so let's rearrange this for n:

V = 19.4 L
T = 17 + 273 = 290 K
P = 746 mmHg
we need to convert the pressure from mmHg to atm and for this we divide by 760 since, 1 atm = 760 mmHg

P = 0.982 atm
R = 
Let's plug in the values in the equation to get the moles.

n = 0.800 moles
So, 0.800 moles of the gas were collected.
Yes, it mixes it and has vitamins in the tea.
Something made of pure iron<span> is softer </span>than steel<span> because the atoms can slip over one another. If other atoms like carbon are added, they are different from </span>iron<span> atoms and stop the </span>iron<span> atoms from sliding apart so easily. This makes the metal </span>stronger<span> and </span>harder<span>.</span>
Answer:
29.41% of Calcium and 47.04% of Oxygen
Explanation:
The percent composition of an atom in a molecule is defined as 100 times the ratio between the mass of the atom and the mass of the molecule.
The mass of the molecule of the problem (Ore) is 46.28g. That means the percent composition of Calcium is:
13.61g / 46.28g * 100 = 29.41% of Calcium
And percent composition of Oxygen is:
21.77g / 46.28g * 100 = 47.04% of Oxygen
<h3>
Answer:</h3>
0.819 mol Ag
<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>
4.93 × 10²³ atoms Ag
<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>
0.818665 mol Ag ≈ 0.819 mol Ag