Learn spelling first, then come back...
KCl
Degrees
Celcius
8g
help me on this quisdtoin
According to the CRC Handbook of Chemistry and Physics, 58th Edition, 35.7 grams of sodium chloride (NaCl) can be dissolved in 100.0 grams of pure water at 0 degrees centigrade (page B-159). At 199 degrees centigrade the amount able to be dissolved increases to 39.12 grams of NaCl.
32 g KCl
How_many_grams_of_water_are_in_the_solution_of_6.20_g_of_C2H6O2_if_freezing_point_is_-0.372_degrees_Celsius
260
100 g water dissolve 45,8 g potassium chloride at 50 o 0C.
67.0 grams of caffeine will dissolve in 100 mL of water at 100 degrees Celsius. 100mL of water = 100g of water so.. 67.0 grams of water will dissolve in 100 grams of water at 100 degrees Celsius. I pretty sure this is right, I was searching for the answer myself and it seems to work. But more accurately you will need to find the density of water at 100 ºC which is 0.9584 g/mL. So, 100g/0.9584 g/mL equals about 104 mL. Then 104 mL x 67.0g/100 mL = 69.9 g.
The solubility of poassium nitrate in water at 20 oC is 616 g/L.
Instead of waiting for the answer, I ended up solving it. lol 300grams of H20 X 110grams of KNO3/100 grams of H20 = 330 grams of KNO3 The grams of H20 both cancel out and leave you with 330 grams of KNO3 Containing the solubility of KNO3. At 60 degrees celsius the KNO3 grams were 110, which is (over) / 100grams of H20. Hope this helps with this workbook problem :)
360
20
20
400
360
90 degrees
250 is one quarter of 1000 and 90 degrees (not digrees) are there in a right angle.