Answer:
As water on the surface of lakes, oceans, and rivers warms up, it travels into the sky as very tiny droplets, or vapor. When the water vapor gets colder, it turns back to liquid to help form clouds.
When the liquid gets so heavy it can’t stay in the atmosphere anymore, it falls, or “precipitates,” as rain, snow, sleet, hail, or, my favorite, graupel. Once the precipitation reaches the ground or lands in lakes, oceans, and rivers, the cycle continues.
Explanation:
<span>carbon = 42.1%
hydrogen = 6.5%
oxygen = 51.4%
First lookup the molar mass of carbon, hydrogen, and oxygen.
mass of carbon = 12.0107
mass of hydrogen = 1.00794
mass of oxygen = 15.999
Now calculate the molar mass of each element in sucrose multiplying the atomic weight of each element by the number of times the element is used.
carbon = 12 * 12.0107 = 144.1284
hydrogen = 22 * 1.00794 = 22.17468
oxygen = 11 * 15.999 = 175.989
Calculate the molar mass of sucrose by adding the mass of each element used.
144.1284 + 22.17468 + 175.989 = 342.2921
Finally, calculate the percentage by mass of each element by dividing the mass used for each element by the total mass of sucrose.
carbon = 144.1284 / 342.2921 = 0.421068 = 42.1%
hydrogen = 22.17468 / 342.2921 = 0.064783 = 6.5%
oxygen = 175.989 / 342.2921 = 0.514149 = 51.4%</span>
Chlorous Acid Ionizes as,
HClO₂ ⇆ H⁺ + ClO₂⁻
Ka = [H⁺] [ClO₂⁻] / [HClO₂]
Ka of Chlorous Acid = 1.1 × 10⁻²
The concentration of H⁺ ions at equilibrium are calculated as,
Initial 0.1 M ⇆ 0 0
At Equilibrium 0.1-X ⇆ X X
So,
Ka = [X] [X] / [0.1-X]
Putting value of Ka,
1.1 × 10⁻² = X² / 0.1-X
Solving for X,
(www.cymath.com)
X = 0.028 M
Hence at equilibrium concentration of H⁺ and ClO₂⁻ is 0.028 M.
Percentage Ionization is calculated as,
= [H⁺] / [HA] × 100
= (0.028 / 0.1) × 100
= 0.28 × 100
= 28 % (Percentage Ionization)
Molar mass of TiCl₃ = (47.9 + 35.5×3) g/mol = 154.4 g/mol
No. of moles of TiCl₃ = (380 g) / (154.4 g/mol) = 2.46 mol
1 mole of TiCl₃ contains 1 mole of Ti.
No. of moles of Ti needed = (2.46 mol) × 1 = 2.46 mol
Molar mass of Ti = 47.9 g/mol
Mass of Ti needed = (2.46 mol) × (47.9 g/mol) = 118 g
Mass of sugar=Mass of solution - mass of water
=1.07X355 - 355
= 24.85g
