<h3>
Answer:</h3>
2 M
<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>Unit 0</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<u>Aqueous Solutions</u>
- Molarity = moles of solute / liters of solution
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
36.7 g CaF₂
300 mL H₂O
<u>Step 2: Identify Conversions</u>
Molar Mass of Ca - 40.08 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of CaF₂ - 40.08 + 2(19.00) = 78.08 g/mol
1000 mL = 1 L
<u>Step 3: Convert</u>
<em>Solute</em>
- Set up:

- Multiply:

<em>Solution</em>
- Set up:

- Multiply:

<u>Step 4: Find Molarity</u>
- Substitute [M]:

- Divide:

<u>Step 5: Check</u>
<em>Follow sig fig rules and round.</em> <em>We are given 1 sig fig as our lowest.</em>
1.56677 M ≈ 2 M
Explanation:
According to Buoyance equation,
m = ![[m' \times \frac{1 - \frac{d_{a}}{d_{w}}}{1 - \frac{d_{a}}{d}}]](https://tex.z-dn.net/?f=%5Bm%27%20%5Ctimes%20%5Cfrac%7B1%20-%20%5Cfrac%7Bd_%7Ba%7D%7D%7Bd_%7Bw%7D%7D%7D%7B1%20-%20%5Cfrac%7Bd_%7Ba%7D%7D%7Bd%7D%7D%5D)
where, m = true mass
m' = mass read from the balance = 17.320 g
= density of air = 0.0012 g/ml
= density of the balance = 7.5 g/ml
d = density of liquid octane = 0.7025 g/ml
Now, putting all the given values into the above formula and calculate the true mass as follows.
m =
= ![[17.320 g \times \frac{1 - \frac{0.0012 g/ml}{7.5 g/ml}}{1 - \frac{0.0012 g/ml}{0.7025}}]](https://tex.z-dn.net/?f=%5B17.320%20g%20%5Ctimes%20%5Cfrac%7B1%20-%20%5Cfrac%7B0.0012%20g%2Fml%7D%7B7.5%20g%2Fml%7D%7D%7B1%20-%20%5Cfrac%7B0.0012%20g%2Fml%7D%7B0.7025%7D%7D%5D)
=
= 17.317 g
Thus, we can conclude that the true mass of octane is 17.317 g.
Answer:
antimony-121 has the highest percent natural abundance
Explanation:
percent natural abundance;
121.76 = 120.90 x + 122.90 (1 - x)
121.76 = 120.90 x + 122.90 - 122.90x
121.76 = -2x + 122.90
121.76 - 122.90 = -2x
x= 121.76 - 122.90/ -2
x= 0.57
Where x and 1 - x refers to the relative abundance of each of the isotopes
Percent natural abundance of antimony-121 = 57 %
Percent natural abundance of antimony-123 = (1 - 0.57) = 43%
Let us remember that isotopy refers to a phenomenon in which atoms of the same element have the same atomic number but different mass numbers. This results from differences in the number of neutrons in atoms of the same element.
We can clearly see that antimony-121 has the highest percent natural abundance.