Answer: 0.52 L of 15 M will be used to prepare this amount of 0.52 M base.
Explanation:
But on diluting the number of moles remain same and thus we can use molarity equation.
(to be prepared)
where,
=concentration of stock solution = 15 M
= volume of stock solution = ?
= concentration of solution to be prepared = 0.52 M
= volume of solution to be prepared = 15 L
Thus 0.52 L of 15 M will be used to prepare this amount of 0.52 M base
Answer:
CH₃CH(CH₃)CH(C₃H₇)CH₂CH(CH₃)₂:
4-isopropyl-2-methylpentane.
Explanation:
Step One: Draw the structure formula of this compound. Parentheses in the formula indicate substitute groups that are connected to the carbon atom to the left.
For example, the first (CH₃) indicates that the second carbon atom from the left is connected to:
Each carbon atom in this compound is connected to four other atoms. All bonds between carbon atoms are single bonds.
The C₃H₇ in the second pair of parentheses is the condensed form of CH₃CH₂CH₂-. See the first sketch attached. Groups in parentheses are highlighted.
Step Two: Find the carbon backbone. The backbone of a hydrocarbon is the longest chain of carbon atoms that runs through the compound. See the second sketch attached. The backbone of this compound consists of seven carbon atoms and is highlighted in green. The name for this backbone shall be heptane.
Step Three: Identify and name the substitute groups.
The two substitute groups are circled in blue in the second sketch.
Step Four: Number the atoms.
Isopropyl shall be placed before methyl. Start from the right end to minimize the index number on all substitute groups. The methyl group is on carbon number two and the isopropyl group on carbon number four. Hence the name:
4-isopropyl-2-methylheptane.
