If 71.5 moles of an ideal gas is at 5.03 atm at 6.80 °C, what is the volume of the gas?
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Answer : The masses of calcium chloride and water used were, 352.6 g and 487.0 g.
Explanation :
As we are given that 42.0 % calcium chloride by mass that means 42.0 grams of calcium chloride present in 100 grams of solution.
In the solution or mixture, 42.0 % calcium chloride and 58.0 % (100-42.0=58.0) water.
Now we have to determine the mass of calcium chloride for 839.6 grams of solution.
As, 100 grams of solution contains 42.0 grams of calcium chloride
So, 839.6 grams of solution contains grams of calcium chloride
Thus, the mass of calcium chloride used is, 352.6 grams.
Now we have to determine the mass of water.
Mass of water = Mass of solution - Mass of calcium chloride
Mass of water = 839.6 g - 352.6 g
Mass of water = 487.0 g
Thus, the mass of water used is, 487.0 grams.
The empirical formula is obtained by calculating the mole ratios of the atoms in the elements.
The number of moles =mass/ R.A.M
For hydrogen, no. of moles=8/1=8
For oxygen, no. Of moles=64/16=4
The tabular solution is attached.
The number of moles =mass/ R.A.M
For hydrogen, no. of moles=8/1=8
For oxygen, no. Of moles=64/16=4
The tabular solution is attached.