(E) ionic aluminum fluoride (AlF3)
Answer:
0.73L
Explanation:
The following data were obtained from the question :
V1 = 0.65 L
P1 = 3.4 atm
T1 = 19°C = 19 + 273 = 292K
V2 =?
P2 = 3.2 atm
T2 = 36°C = 36 + 273 = 309K
The bubble's volume near the top can be obtain as follows:
P1V1 /T1 = P2V2 /T2
3.4 x 0.65/292 = 3.2 x V2 /309
Cross multiply to express in linear form as shown below:
292 x 3.2 x V2 = 3.4 x 0.65 x 309
Divide both side by 292 x 3.2
V2 = (3.4 x 0.65 x 309) /(292 x 3.2)
V2 = 0.73L
Therefore, the bubble's volume near the top is 0.73L
After the first 2 min the 20 g are only 10g, (1 half-life)
after other 2 more min 5g (2 half-lives)
after other 2 more min 2.5 g (3 half-lives)
after 2 more 1.25 g (4 haf lifes)
Answer:
each of two or more forms of the same element that contain equal numbers of protons but different number of nurtrons
in there nucleir
Explanation:
.....
The normality that would be calculated will be to high because the change in volume will be greater than the actual change in volume. if the buret tip is not filled when reading the initial volume, the actual volume should be lesser with that reading. so if you will you the higher reading the change in volume or the volume you use in titration will be higher than the actual
