C2H4 is oxidized and O2 is reduced in both reactions.
<h3>What is oxidation/reduction?</h3>
Oxidation is defined in several ways. Some of the definitions are:
- The addition of oxygen or removal of hydrogen
- Increase in the oxidation number of atoms
- Addition of electronegative or the removal of electropositive elements
Reduction, on the other hand, is defined as:
- Removal of oxygen or addition of hydrogen
- Decrease in the oxidation number of atoms
- Addition of electropositive elements or the removal of electronegative elements.
In the two reactions, oxygen is being added to C2H4. Thus, C2H4 is being oxidized.
The oxidizing agent is O2. In oxidation reactions, the oxidizing agents usually get reduced. Thus, O2 is reduced in both reactions.
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35°c is equal to 95°f
To do this multiply 35 and 1.8
35 x 1.8=63
Now add 32
Resulting in the answer 95
(The equation for to solve for c and f is c1.8+32=f
<h3>
Answer:</h3>
3.67 mol Al
<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>
2.21 × 10²⁴ atoms Al
<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>
3.66988 mol Al ≈ 3.67 mol Al
The concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
What is meant by concentration?
Concentration is the total amount of solute present in the given volume of solution. this is expressed in terms of molarity, molality, mole fraction, normality etc. The term concentration mostly refers to the solvents and solutes present in the solution.
Concentration of hydroxide ions can be calculated by,
M (OH^-) = V (NaOH) x M (NaOH) / V (total) = 50ml x 0.250M / 50ml + 120ml = 0.0735M = 7.35 x 10^-2 M.
where M (OH^-) = concentration of hydroxide ions, V(NaOH) = volume of NaOH, M(NaOH) = concentration of NaOH.
Therefore, the concentration of the hydroxide ions after 50 ml of 0.250M NaOH is added to 120ml of 0.200M Na2SO4 is 7.35 x 10^-2 M.
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A. Petrified fossil should be the answer
