Answer:
83.8%
Explanation:
The balanced reaction equation is;
2Al(s) + 3Cl2(g) → 2AlCl3(s)
Now we have to obtain the limiting reactant as the reactant that produces the least amount of AlCl3
Amount of Al = 3.11g/27 g/mol = 0.115 moles
If 2 moles of Al yields 2 moles of AlCl3
Then 0.115 moles of Al yields 0.115 moles of AlCl3
For Cl2
Amount of Cl2 = 5.32 g/71 g/mol= 0.075 moles
If 3 moles of Cl2 yields 2 moles of AlCl3
0.075 moles of Cl2 yields 0.075 * 2/3 = 0.05 moles of AlCl3
Hence Cl2 is the limiting reactant
Theoretical yield of AlCl3 = 0.05 moles of AlCl3 * 133g/mol = 6.65 g
%yield = actual yield /theoretical yield * 100
%yield = 5.57 g/6.65 g * 100
%yield = 83.8%
Answer:
For carbon the most important forms of hybridization are the sp2- and sp3- hybridization. Besides these structures there are more possiblities to mix dif- ferent molecular orbitals to a hybrid orbital. An important one is the sp- hybridization, where one s- and one p-orbital are mixed together.
The answer is 200 for sure!!!!!
Answer:
pH of buffer =4.75
Explanation:
The pH of buffer solution is calculated using Henderson Hassalbalch's equation:
![pH=pKa+log[\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5B%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
Given:
pKa = 3.75
concentration of acid = concentration of formic acid = 1 M
concentration of salt = concentration of sodium formate = 10 M
![pH=3.75+log[\frac{10}{1}]=3.75+1=4.75](https://tex.z-dn.net/?f=pH%3D3.75%2Blog%5B%5Cfrac%7B10%7D%7B1%7D%5D%3D3.75%2B1%3D4.75)
pH of buffer =4.75
Answer:
9.8 × 10²⁴ molecules H₂O
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Organic</u>
<u>Stoichiometry</u>
- Analyzing reaction rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O
[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O
[Given] 130 g CH₄
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[RxN] 1 mol CH₄ → 2 mol H₂O
[PT] Molar Mass of C: 12.01 g/mol
[PT] Molar Mass of H: 1.01 g/mol
Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol
<u>Step 3: Stoichiometry</u>
- [DA] Set up conversion:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O