You can boil the salt water. The water will evaporate while the salt will be left behind because it's a solid. Salt has a much higher boiling point than water (2,575° F), so that's why it won't evaporate with the water.
Answer:
The answer is 98.07848. We assume you are converting between grams H2SO4 and mole. You can view more details on each measurement unit: This compound is also known as Sulfuric Acid. The SI base unit for amount of substance is the mole. 1 grams H2SO4 is equal to 0.010195916576195 mole.
<u>Quick conversion chart of moles H2SO3 to grams</u>
1 moles H2SO3 to grams = 82.07908 grams
2 moles H2SO3 to grams = 164.15816 grams
3 moles H2SO3 to grams = 246.23724 grams
4 moles H2SO3 to grams = 328.31632 grams
5 moles H2SO3 to grams = 410.3954 grams
6 moles H2SO3 to grams = 492.47448 grams
7 moles H2SO3 to grams = 574.55356 grams
8 moles H2SO3 to grams = 656.63264 grams
9 moles H2SO3 to grams = 738.71172 grams
10 moles H2SO3 to grams = 820.7908 grams
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
