Answer: 2.17 g of bromide product would be formed
Explanation:
The reaction of calcium bromide with lithium oxide will be:
To calculate the moles :
As lithium oxide is in excess, calcium bromide is the limiting reagent.
According to stoichiometry :
1 mole of produce = 2 moles of
Thus 0.0125 moles of will require= of
Mass of
Thus 2.17 g of bromide product would be formed
Answer: 0.70g
The half-life of iron-53 would be 8.51 minutes. So, in 25.53 minutes would be equal to: 25.53 min/ (8.51 minutes/ half-life)= 3 half-life.
Every half-life will reduce the original weight into half. So, the final weight would be:
final weight = original weight * 1/2 ^(time)
final weight = 5.6g * (1/2)^(3 half-life)
final weight = 5.6g * 1/8= 0.7g
Answer:
The number of collisions is independent of volume.
Answer:
A. Karl Landsteiner is the most likely for her to mention.
106.905/52.00 equals 205.58
108.905/48.00 equals 226.88