<h3>
Answer:</h3>
1.07 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>Atomic Structure</u>
- Molarity = moles of solute / liters of solution
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 2.14 moles (CH₃)₂SO
[Given] 2.00 L
[Solve] Molarity
<u>Step 2: Solve</u>
- Substitute in variables [Molarity]: x M = 2.14 moles (CH₃)₂SO / 2.00 L solution
- [Molarity] Divide: x = 1.07 M
Answer:
Chlorine is limiting reactant
Explanation:
Based on the reaction:
Cl₂ + 2NaOH → NaClO + NaCl + H₂O
<em>1 mole of chlorine reacts with 2 moles of NaOH</em>
<em />
To find limiting reactant, we need to determine the moles of the reactants:
<em />
<em>Moles Cl₂ -Molar mass: 70.9g/mol-:</em>
800lb Cl₂ * (453.6g / 1lb) * (1mol / 70.90g) =
5118 moles Cl₂
<em>Moles NaOH -Molar mass: 40g/mol-:</em>
1200lb NaOH * (453.6g / 1lb) * (1mol / 40g) =
13608 moles NaOH
For a complete reaction of 13608 moles of NaOH you need:
13608 moles NaOH * (1mol Cl₂ / 2 moles NaOH) = 6804 moles of Cl₂
As the solution contains just 5118 moles of chlorine,
<h3>Chlorine is limiting reactant</h3>
Answer:
a) I, II, and III
Explanation:
For the first statement;
Solvation, is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. if the solvent is water, we call this process hydration.
This means the statement is TRUE.
For the second statement;
The negatively-charged side of the water molecules are attracted to positively-charged ions. In the case of water, the oxygen end is the negatively charged side of water. This means the statement is TRUE.
For the third statement;
The positively-charged side of the water molecules are attracted to the negatively-charged chloride ions. In the case of water, the hydrogen end is the positively charged side of water. This means the statement is TRUE.
Going through the options, we can tell that the correct option is option A.
Answer: 0.52V
Explanation:
Ecell = Ecell(standard) - [(0.0592 logQ)/n]
Q = product of the quotient
n = no of electrons transferred = 2
Ecell = 0.63 - [(0.0592*Log(1 / 2.0 * 10-4) / 2]
Ecell = 0.63 - 0.0194
Ecell = 0.5205V
