For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:
Kd = c </span>× α²
α = √(Kd/c) × 100%
Kd = 6.0×10⁻⁷
c(HA) = 0.1M
α = √(6.0×10⁻⁷/0.1) × 100% = 0.23%
So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>
Answer: -
0.1 ml of bleach should be added to each liter of test solution.
Explanation:-
Let the volume of bleach to be added is B ml.
Density of stock solution = 1.0 g/ml
Mass of stock solution = Volume of stock x density of stock
= B ml x 1.0 g/ml
= B g
Amount of NaOCl in this stock solution = 5% of B g
= 
x B g
= 0.05 B g
Now each test solution must be added 5 mg/l NaOCl.
Thus each liter of test solution must have 5 mg.
Thus 0.05 B g = 5 mg
= 0.005 g
B = 
= 0.1
Thus 0.1 ml of bleach should be added to each liter of test solution.
Total atoms is 9 ( 2 carbon atoms, 5 hydrogen atoms, 1 oxygen atom and 1 hydrogen atom = 9 atoms)
Element is 3 ( Carbon, Hydrogen and Oxygen)
(B), because 1.0 moles would be 6.02 x 10^23 molecules. So you have half a mole.<span>
</span>
Answer:
9.82 g of Mg(NO₃)₂
Explanation:
Let's determine the reaction:
2AgNO₃ + MgBr₂ → Mg(NO₃)₂ + 2AgBr
2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.
We determine the moles of AgNO₃
22.5 g . 1mol / 169.87g = 0.132 moles
Ratio is 2:1.
2 moles of silver nitrate can produce 1 mol of magnesium nitrate
Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles
We convert moles to mass:
0.0662 mol . 148.3 g/ mol = 9.82 g
