The best answer I could find was when you Google it, that the fuse is of poor quality. I cannot leave you a link, but you can find it for yourself. Put in poor quality fireworks and all sorts of things will pop up. No pun intended.
J. J. Thomson is the corect awncer
12 % salt is present in 125 g mixture of salt and sand.
Keep in mind that the total percentage is always 100 %
Therefore, if 12 % is the salt, remaining 88 % must be
sand.
a. The amount of mixture is 125 q. Here, 12 % of 125 is
12 * 125 / 100 = 15 g of salt is present in 125 g mixture.
b. The amount of sand can be calculated similarly, 88 %
of 125 g is 88 * 125 / 100 = 110 g of sand is present in
125 g mixture.
Answer:
For this experiment we are going to take plate 1 as the control plate, so, in it there will be just E. coli in LB/agar; in plate 2, we are going to put E. coli in LB/agar and some ampicillin. Then, we have to wait for the E. coli colonies to form. After a while, the E. coli growth can be compared on both plates and determine if ampicillin affects or not the E. coli colonies.
Explanation:
If the ampicillin affects negatively E. coli colonies, we are going to observe that in plate 1 (control plate) there are E. coli colonies growing, but in plate 2, there is no E. coli colonies or, at least, there is a fewer number of colonies on it. If ampicillin doesn't affect E.coli, plate 1 (control) and plate 2 (ampicillin experiment) are going to be similar in number of colonies.
