Answer:
m = 0.531 molal
Explanation:
∴ m fructose = 3.35 g
∴ V water = 35.0 mL
∴ ρ H2O = 1 g/mL
- molality = moles solute / Kg solvent
∴ Mw fructose = 180.16 g/mol
⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose
⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O
⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O
⇒ m = 0.531 molal
Answer:
Structures are given below.
Explanation:
- Treatment of 2-bromo-2-methylbutane with KOH in ethanol will give elimination of HBr through E2 mechanism.
- H atoms adjacent to Br will be eliminated.
- 2-bromo-2-methylbutane has two possible adjacent H atoms that can be eliminated giving mixture of products.
- Product of this elimination reaction is alkene. Here saytzeff fule is followed during elimination. So most substituted alkene will be major product.
- Structure of alkenes are given below.
D. released a connection to the pedal. disengage is the opposite of engage so by stopping the connection he’s doing the opposite of connecting it
Yes. Whenever a chemical change is observed (a few examples being change in color, odor, or precipitate formation) a new substance is formed.