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<h3>
Answer:</h3>
0.34 mol S
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
11 g S
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of S - 32.07 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
0.343 mol S ≈ 0.34 mol S
65 grams of HCl = 65/36.5 moles of HCl = 1.78 moles
1.78 moles of HCl dissolved to make a 5 litres of solution has a concentration of 1.78/5 = 0.36 mol/dm^3 (Note: 1 litre = 1 cubic decimetre)
In a strong acid, such as HCl, [H+] = [acid], so [H+] = 0.36
To calculate pH, we have to take the negative logarithm of the concentration of protons
So, -log(0.36) = 0.45
Hope I helped!! xx
Complete Question:
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride. How many grams of F− must be added to a cylindrical water reservoir having a diameter of 2.02 × 102 m and a depth of 87.32 m?
Answer:
2.23x10⁶ g
Explanation:
The concentration of the fluoride (F⁻) must be 0.800 ppm, which is 0.800 parts per million, so the water must have 0.800 g of F⁻/ 1000000 g of the solution. The density of the water at room temperature is 997 kg/m³ = 997x10³ g/m³. So, the concentration of the fluoride will be:
0.800 g of F⁻/ 1000000 g of the solution * 997x10³ g/m³
0.7976 g/m³
The volume of the reservoir is the volume of the cylinder: area of the base * depth. The base is a circumference, which has an area:
A = πR², where R is the radius = 1.01x10² m (half of the diameter)
A = π*(1.01x10²)²
A = 32047 m²
The volume is then:
V = 32047 * 87.32
V = 2.7983x10⁶ m³
The mass of the F⁻ is the concentration multiplied by the volume:
m = 0.7976 * 2.7983x10⁶
m = 2.23x10⁶ g