To know if something is an acid or base, you must test its pH level
Answer:
1.21 mol KClO₃
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Mole Ratio
<u>Stoichiometry</u>
- Analyzing reactions rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[rxn] 2KClO₃ → 2KCl + 3O₂
[Given] 58.3 g O₂
[Solve] mol KClO₃
<u>Step 2: Identify Conversions</u>
[rxn] 2 mol KClO₃ → 3 mol O₂
[PT] Molar Mass of O: 16.00 g/mol
Molar Mass of O₂: 2(16.00) = 32.00 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
![\displaystyle 58.3 \ g \ O_2(\frac{1 \ mol \ O_2}{32.00 \ g \ O_2})(\frac{2 \ mol \ KClO_3}{3 \ mol \ O_2})](https://tex.z-dn.net/?f=%5Cdisplaystyle%2058.3%20%5C%20g%20%5C%20O_2%28%5Cfrac%7B1%20%5C%20mol%20%5C%20O_2%7D%7B32.00%20%5C%20g%20%5C%20O_2%7D%29%28%5Cfrac%7B2%20%5C%20mol%20%5C%20KClO_3%7D%7B3%20%5C%20mol%20%5C%20O_2%7D%29)
- [DA] Divide/Multiply [Cancel out units]:
![\displaystyle 1.21458 \ mol \ KClO_3](https://tex.z-dn.net/?f=%5Cdisplaystyle%201.21458%20%5C%20mol%20%5C%20KClO_3)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.21458 mol KClO₃ ≈ 1.21 mol KClO₃
Soil
profile
Soil
profile is the vertical sequence of the layers of soil. Furthermore, there are
actually six layers of the soil which involves the organic matter, -where most
humus is present-, surface soil, the subsoil, the parent rock, and the bedrock
as the innermost and core layer of the soil. In addition each soil layer has
three to four soil horizon. These horizons are the physical features of the soil,
mainly the texture, color and composition.
<u>Given:</u>
Rate of appearance of O2 = 0.250 M/s
Time period = 5.50 s
<u>To determine:</u>
The concentration of O2 formed
<u>Explanation:</u>
2O3 (g) ↔ 3O2 (g)
Rate of appearance of O2 = 1/3 * Δ[O2]/Δt
Based on the given data:
0.250 M/s = 1/3 * [O2]/5.50 s
[O2 ] = 0.250 Ms⁻¹ * 3 * 5.50 s = 4.125 M
Ans: Amount of oxygen formed is 4.13 M
The unknown of this problem is the experimental percent of water in the compound in order to remove the water of hydrogen, given the following:
Mass of crucible, cover and contents before heating 23.54 g
Mass of empty crucible and cover 18.82 g
Mass of crucible, cover, and contents after heating to constant mass 20.94 g
In order to get the answer, determine the following:
Mass of hydrated salt used = 23.54 g – 18.82 g = 4.72 g
Mass of dehydrated salt after heating = 20.94 g – 18.82 g = 2.12 g
Mass of water liberated from salt = 4.72 g – 2.12 g = 2.60 g
Then solve the percent of water in the hydrated salt by:
% water = (mass of water / mass of hydrated salt) x 100
% water = 2.60 g / 4.72 g x 100
% water = 55.08 % in the compound