Answer:
0.416 mol CaBr₂
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
83.1 g CaBr₂
<u>Step 2: Identify Conversions</u>
Molar Mass of Ca - 40.08 g/mol
Molar mass of Br - 79.90 g/mol
Molar Mass of CaBr₂ - 40.08 + 2(79.90) = 199.88 g/mol
<u>Step 3: Convert</u>
<u />
= 0.415749 mol CaBr₂
<u>Step 4: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.415749 mol CaBr₂ ≈ 0.416 mol CaBr₂
Balanced chemical equation:
2 H2 + 1 O2 = 2 H2O
4 g H2 -------> 32 g O2 -----------> 36 g H2O
↓ ↓ ↓
14.0 g ---------> 2.0 g O2 ----------> mass H2O ?
32 * mass H2O = 2.0 * 36
32 * mass H2O = 72
mass of H2O = 72 / 32
mass of H2O = 2.25 g
hope this helps!.
Answer:
KCIO
Explanation:
the only reason is because I looked online and the meaning for that is potassium hypochlorite my guy so if this is not right then I'm sorry but good luck.
For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
To know more about hybrisisation:
brainly.com/question/23038117
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