Part A What volume (in mL ) of a 0.150 M HNO3 solution will completely react with 35.7 mL of a 0.108 M Na2CO3 solution according to the following balanced chemical equation?

In the reaction in Part A, what mass (in grams) of carbon dioxide forms?

Volume HNO₃ : 51.4 ml

Mass CO₂ : 0.17 g

Reaction

Na₂CO₃ + 2 HNO₃ ⇒ 2 NaNO₃ + CO₂ + H₂O

V Na₂CO₃ = 35.7 ml

M  Na₂CO₃ = 0.108

mol Na₂CO₃ :

mol ratio Na₂CO₃  : HNO₃ = 1 : 2

mol HNO₃ :

Volume HNO₃ :

mol CO₂ = mol Na₂CO₃ = 3.8556 mlmol=0.00386 mol

mass CO₂(MW=44 g/mol) :

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.