I just need this checked over. I am not sure about the last part D.
Temp c KBr
20 65
40 80
60 90
80 100
100 110
At 100 C, 10.5 g KBr/10gH2O = 105 g/100 g (just multiply the top and the bottom by 10 to give you 100 g H2O)
B) Since the solubility at 100C is actually 110, no, it isn't saturated (105 is less that 110).
C) Yes, because the solubility is 80 g/100 g at 40C.
D) At 40 C, 10g of H2O could dissolve 8 g of KBr. So, 10.5 - 8 = 1.5 g of solid formed.
I am not sure if I did the last part right?
To check if the last part (part D) is correct, we can calculate the amount of KBr that can be dissolved in 10g of water at 40°C and compare it to the given amount (10.5g).
The solubility of KBr at 40°C is given as 80g/100g, which means that 80g of KBr can dissolve in 100g of water at that temperature.
To calculate the amount of KBr that can dissolve in 10g of water at 40°C, we can use the following proportion:
(80g KBr / 100g H2O) = (x g KBr / 10g H2O)
Cross-multiplying, we get:
80g KBr * 10g H2O = 100g H2O * x g KBr
800g KBr = 100g H2O * x g KBr
Dividing both sides by 100g H2O, we find:
x g KBr = (800g KBr) / (100g H2O)
x g KBr = 8g KBr
Therefore, 8g of KBr can dissolve in 10g of water at 40°C.
Now, to find the amount of solid formed, we can subtract the amount dissolved (8g) from the given amount (10.5g):
10.5g KBr - 8g KBr = 2.5g KBr
Based on this calculation, it seems that the amount of solid formed at 40°C is 2.5g, not 1.5g as stated in the original answer.