In a solution of KBr and water; KBr is the solute and water is the solvent;
Therefore; to achieve 3% by mass; it means we are going to have 3% of the mass being the solute and the other 97 % being the solvent.
Thus; KBr (solute) = 3/100 × 300 (total mass) = 9 g
Hence; the appropriate masses will be; 9.00 g of KBr and 291 g of water.
The mass of a substance per unit volume is the substance's density.
D = m/v.
Answer:
As the temperature of the water increases, the time needed for the dye to spread decreases. This is because the kinetic energy between the liquid particles increases, therefore helping the dye to dissolve and spread throughout the water.
Explanation:
Answer:
Using the formula cards again, add the coefficient of 2 in front of the formula and have them recalculate the number of each element and the total number of atoms in each element.
Explanation:
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.