You never told us how old you are so how are we supposed to answer
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
<span>D) recycling ;)
Waste Management's Aerobic-Anaerobic Bioreactor* is designed to accelerate waste degradation by combining attributes of the aerobic and anaerobic bioreactors. The objective of the sequential aerobic-anaerobic treatment is to cause the rapid biodegradation of food and other easily degradable waste in the aerobic stage in order to reduce the production of organic acids in the anaerobic stage resulting in the earlier onset of methanogenesis. In this system the uppermost lift or layer of waste is aerated, while the lift immediately below it receives liquids. Landfill gas is extracted from each lift below the lift receiving liquids. Horizontal wells that are installed in each lift during landfill construction are used convey the air, liquids, and landfill gas. The principle advantage of the hybrid approach is that it combines the operational simplicity of the anaerobic process with the treatment efficiency of the aerobic process. Added benefits include an expanded potential for destruction of volatile organic compounds in the waste mass. (*US Patent 6,283,676 B1)</span>
<h3>
Answer:</h3>
5.00 mol O₂
<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>
</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
3.01 × 10²⁴ atoms O₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
4.99834 mol O₂ ≈ 5.00 mol O₂
Answer:
FALSE
Since 0.385 < 0.526, the value for week 3 is accepted.
Explanation:
Qexp = (|Xq - Xₙ₋₁|)/w
where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data
First, the data are arranged in decreasing order, from highest to lowest:
3. 5.6
2. 5.1
8. 5.1
1. 4.9
6. 4.9
5. 4.7
7. 4.5
4. 4.3
Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3
Qexp = (|5.6 - 5.1|)/1.3 = 0.385
From tables, at 95% confidence level, for n = 8, Qcrit = 0.526
Since 0.385 < 0.526, the value for week 3 is accepted.
