Let's note that 1 pint = 473.1765 mL, so 11 pints should be 5204.9415 mL.
We make a proportion out of the word problem
(85 mg glucose/ 100 mL) times (1 g/ 1000 mg) = 4.4242 grams of glucose
Answer:
c maybe sorry if its wrong
Explanation:
Answer:
111.15 g are required to prepare 500 ml of a 3 M solution
Explanation:
In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.
Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of
(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.
The formula of butane is C4H10 but I don't how many atoms it contains though
