Answer:
0.52 g of KNO₃ are contained in 19.7 mL of diluted solution.
Explanation:
We can work on this problem in Molarity cause it is more easy.
Molarity (mol/L) → moles of solute in 1L of solution.
100 mL of solution = 0.1 L
We determine moles of solute: 44.7 g . 1mol /101.1 g = 0.442 mol of KNO₃
Our main solution is 0.442 mol /0.1L = 4.42 M
We dilute: 4.42 M . (11.9mL / 200mL) = 0.263 M
That's concentration for the diluted solution.
M can be also read as mmol/mmL, so let's find out the mmoles
0.263 M . 19.7mL = 5.18 mmol
We convert the mmol to mg → 5.18 mmol . 101.1 mg / mmol = 523.7 mg
Let's convert mg to g → 523.7 mg . 1 g / 1000 mg = 0.52 g
Zinc (Zn) always has a +2 charge. It is one of the exceptions.
The standard state of the elements Nitrogen and Oxygen are N2 and O2, knowing that they are diatomic elements. With that piece of information, the unbalanced equation for the formulation of NO2(g) should be as follows -
N2 + O2 ---> NO2
And if you include their states -
N2 ( g ) + O2 ( g ) ---> NO2 ( g )
To balance this chemical equation consider the number of reactants and products on other side of the equation. If you were to include a coefficient of one - half with respect to N2 on the reactant side, it would balance the reactants and products -
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.
