We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).

Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).

To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
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Answer:
I think that As is larger
Answer:
4KNO3 ==> 2K2O + 2N2 + 5O2
Explanation:
It's a decomposition, but not a simple one.
KNO3 ==> K2O + N2 + O2 I don't usually do this, but I think the easiest way to proceed is to balancing the K and N together. That will require a 2 in front of KNO3
4KNO3 ==> 2K2O + 2N2 + 5O2
Now you have (3*4) = 12 oxygens. Two are on the K2O. So the other 10 must be on the O2
That should do it.
Because at atmospheric pressure the boiling point is below room temperature, just like Oxygen, Nitrogen etc