Answer:
4.74 × 10³ mg
Explanation:
Given data
- Health risk limit for chloroform in groundwater: 60.0 g/L
- Volume of the sample of groundwater: 79.0 mL = 79.0 × 10⁻³ L
The maximum mass of chloroform that there could be in the sample of groundwater to meet the standards are:
79.0 × 10⁻³ L × 60.0 g/L = 4.74 g
1 gram is equal to 10³ milligrams. Then,
4.74 g × (10³ mg/1 g) = 4.74 × 10³ mg
Those Hydrogen atoms which are present at alpha position to carbonyl group are mildly acidic in nature. When such acidic proton containing carbonyl compounds are treated with strong base, they yield enolates. The negative charge created on alpha carbon resonates and shifts to carbonyl oxygen resulting in formation of carbon double bond carbon.
In <span>tert−butyl methyl ketone there are two carbons at each alpha position. Among these two carbons only methyl carbon contains hydrogen atoms while the second one is bonded to further three carbons making it Quaternary carbon. The base abstracts proton from methyl group and enolate is formed.
</span>tert−butyl methyl ketone in this case acts as acid, Hydroxyl ion acts as base, while the enolate generated is the conjugate base of <span>tert−butyl methyl ketone and Water produced is the conjugate acid of hydroxide ion.</span>
N - 15 undergoes beta decay and has a half-life of less than 1 minute.
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