Passes through the kidneys and the good stuff is kept, and the other stuff is passed through leading to the bladder. The once the bladder is full enough to send the message to the brain, urine (waste) is eliminated from the body.
3! You have to ensure balance of all the different elements.
Note down the formula below

Mass of the compound

Mass % of Hydrogen:-



Mass % of Oxygen:-



Answer:- 171 g
Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.
We know that molarity is moles of solute per liter of solution.
If molarity and volume is given then, moles of solute is molarity times volume in liters.
moles of solute = molarity* liters of solution
moles of solute = 0.5*1 = 0.5 moles
To convert the moles to grams we multiply the moles by molar mass.
Molar mass of sucrose = 12(12) + 22(1) + 11(16)
= 144 + 22 + 176
= 342 grams per mol
grams of sucrose required = moles * molar mass
grams of sucrose required = 0.5*342 = 171 g
So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.
Answer:
We need the options to answer.
Explanation:
N/A