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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.
You can learn more about solutions here: brainly.com/question/2412491
Answer:
[C] carbon solid
Explanation:
Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.
The given question is incomplete. The complete question is:
Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C6H12O6) and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. Determine the mass of glucose (C6H1206) produced
Answer: 60.0 g of glucose
Explanation:
To calculate the moles, we use the equation:
a) moles of
b) moles of
According to stoichiometry :
6 moles of
require = 6 moles of
Thus 2.0 moles of
require=
of
Thus
is the limiting reagent as it limits the formation of product.
As 6 moles of
give = 1 moles of glucose
Thus 2.0 moles of
give =
of glucose
Mass of glucose =
Thus 60.0 g of glucose will be produced from 88.0 g of carbon dioxide and 64.0 g of water