How many milliliters of a stock solution of 6.20 M HNO3 would you have to use to prepare 0.130 L of 0.470 M HNO3?

9.85

Explanation:

M1V1 =M2V2

6.20×v1= 0.470×0.130

v1 = ( 0.470 × 0.130 ) ÷ 6.20

v1 = 0.0098 L × 1000

V1 = 9.8 ml

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.