Answer:
There are 0.93 g of glucose in 100 mL of the final solution
Explanation:
In the first solution, the concentration of glucose (in g/L) is:
15.5 g / 0.100 L = 155 g/L
Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.
- 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)
The concentration of the second solution is:

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:
1 L --------- 9.3 g
0.1 L--------- Xg
Xg = 9.3 g * 0.1 L / 1 L = 0.93 g
Answer:
number three is the answer moderate
Kinetic energy remains conserved in an elastic collision.
Below Mechanism shows the mechanism of formation of diazonium ion. Aniline is treated with
Nitrosonium ion which is generated
in situ by the reaction of Sodium nitrile and strong acid. The resulting <span>
benzenediazonium ion on reaction with CuBr yields
Bromobenzene and on treatment with CuCN gives
Benzonitrile. Mechanism is as folow,</span>