Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Answer:
pH = 7.8
Explanation:
The Henderson-Hasselbalch equation may be used to solve the problem:
pH = pKa + log([A⁻] / [HA])
The solution of concentration 0.001 M is a formal concentration, which means that it is the sum of the concentrations of the different forms of the acid. In order to find the concentration of the deprotonated form, the following equation is used:
[HA] + [A⁻] = 0.001 M
[A⁻] = 0.001 M - 0.0002 M = 0.0008 M
The values can then be substituted into the Henderson-Hasselbalch equation:
pH = 7.2 + log(0.0008M/0.0002M) = 7.8
<h3>
Answer:</h3>
18.9 g F₂
<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>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.00 × 10²³ molecules F₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of F₂ (Diatomic) - 38.00 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
18.9306 g F₂ ≈ 18.9 g F₂
Answer:
Condenses at 27.25K.
Freezes at 24.65K.
Explanation:
In order to solve this above question, there is is need to make use of the following equation. The main idea here is to convert degree celsius to Kelvin. Hence,
0°C + 273.15 = 273.15K---------------------(1).
Therefore, we will make use of the above equation (1) and slot in the values for at degree celsius at which it condenses and at degree celsius at which it freezes.
So, we have at temperature at which it condenses:
-245.9°C + 273.15 = 27.25K.
Also, we have at temperature at which it freezes.
-248.5°C + 273.15 = 24.65K.
