Answer:
2-methoxybutane
Explanation:
This reaction is an example of Nucleophilic substitution reaction. Also, the reaction of (S)-2-bromobutane with sodium methoxide in acetone, is bimolecular nucleophilic substitution (SN2). The reaction equation is given below.
(S)-2-bromobutane + sodium methoxide (in acetone) → 2-methoxybutane
Answer:
52.45g
Explanation:
The computation of the mass of pure acetic acid in 125mL of this solution is shown below:
The percentage of mass would be equivalent to the g of solute in each 100g of water
As we know that
density = mass ÷ volume
So,
Volume = mass ÷ density
V = 100g / 1.049 (g / ml)
V = 95.328 mL
Now In every 95,328 ml of C_2H_4O_2 there are 40g of C_2H_4O_2
i.e.
each 125ml of C_2H_4O_2 there are 52.45g
SO,
x = 40g. 125ml ÷ 95.328
x = 52.45g
Answer:
118
Explanation:
Of these 118 elements, 94 occur naturally on Earth. Six of these occur in extreme trace quantities: technetium, atomic number 43; promethium, number 61; astatine, number 85; francium, number 87; neptunium, number 93; and plutonium, number 94.
Answer:
0.041 L = 41.3 mL
Explanation:
This problem we will solve by considering the stoichiometry of the reaction and the definition of molarity.
Number of moles in .800 L solution:
0.800 L x 0.0240 M = 0.800 L x .0240 mol/L = 0.0192 mol Fe³⁺
to form the precipitate Fe(OH)₃ we will need 3 times .0192
mol NaOH required = 0.057
given the concentration of 1.38 mol M NaOH we can calculate how many milliliters of NaOH will contain 0.057 mol:
1.L/1.38 mol NaOH x 0.057 mol NaOH = 0.041 L
0.041 L x 1000 mL/1L = 41.3 mL
