Any change that alters a substance without changing it into another substance.
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Answer : The mass of of water present in the jar is, 298.79 g
Solution : Given,
Mass of barium nitrate = 27 g
The solubility of barium nitrate at is 9.02 gram per 100 ml of water.
As, 9.02 gram of barium nitrate present in 100 ml of water
So, 27 gram of barium nitrate present in of water
The volume of water is 299.33 ml.
As we know that the density of water at is 0.9982 g/ml
Now we have to calculate the mass of water.
Therefore, the mass of of water present in the jar is, 298.79 g
Hi!
To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g
To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:
Have a nice day!
To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g
To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:
Have a nice day!