<h3>
Answer:</h3>
0.127 mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 25.0 g Au
[Solve] moles Au
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.126923 mol Au ≈ 0.127 mol Au
Carbon has 4 valence electrons, oxygen has 6
II. sulfur (S) and carbon (C)
and
III. fluorine (F) and oxygen (O)
will form covalent bonds, so the answer will be:
e. II and III
Explanation:
To know is what type of bond is formed between atoms we need to look at the electronegativity difference between the atoms.
If the electronegativity difference is less than 0.4 there is a nonpolar covalent bond.
If the electronegativity difference is between 0.4 and 1.8 there is a polar covalent bond. (if is a metal involved we consider the bond to be ionic)
If the electronegativity difference is greater then 1.8 there is an ionic bond.
We have the following cases:
I. lithium (Li) and sulfur (S)
electronegativity difference = 2.5 (S) - 1 (Li) = 1.5 but because there is a metal involved the bond will be ionic
II. sulfur (S) and carbon (C)
electronegativity difference = 2.5 (S) - 2.5 (C) = 0 so the bond will be nonpolar covalent
III. fluorine (F) and oxygen (O)
electronegativity difference = 4 (F) - 3.5 (O) = 0.5 so the bond will be polar covalent bond.
Learn more about:
covalent and ionic bonds
brainly.com/question/1802971
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3H + 3Br = HBr9 Organic chemistry mechanism
Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.
