<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
The atom positions in a general molecule of formula (not shape class) AXn that has shape square pyramidal at the corers of square and one at the above center of the square.
<h3>What is square pyramidal?</h3>
The square pyramidal is a shape geometry of the hybridization in which it consists of one lone pair and 5 bond pairs of electrons that repel each other and due to which the geometry changes from octahedral to square pyramidal.
As atoms are located at the four corners of the planer and one atom at the above center of the planner which is repelled by 4 atoms present at the corner of the planer.
Therefore, the atom positions in a general molecule of formula (not shape class) AXn that has a shape square pyramidal at the corners of the square and one at the above center of the square.
Learn more about square pyramidal, here;
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Answer:
ΔG° = -533.64 kJ
Explanation:
Let's consider the following reaction.
Hg₂Cl₂(s) ⇄ Hg₂²⁺(aq) + 2 Cl⁻(aq)
The standard Gibbs free energy (ΔG°) can be calculated using the following expression:
ΔG° = ∑np × ΔG°f(products) - ∑nr × ΔG°f(reactants)
where,
ni are the moles of reactants and products
ΔG°f(i) are the standard Gibbs free energies of formation of reactants and products
ΔG° = 1 mol × ΔG°f(Hg₂²⁺) + 2 mol × ΔG°f(Cl⁻) - 1 mol × ΔG°f(Hg₂Cl₂)
ΔG° = 1 mol × 148.85 kJ/mol + 2 mol × (-182.43 kJ/mol) - 1 mol × (-317.63 kJ/mol)
ΔG° = -533.64 kJ
Answer: 1.8124 g of 3-nitrophthalhydrazide were recovered.
Explanation:
The balanced chemical reaction will be :

moles of 3-nitrophthalic acid = 
As 1 mole of 3-nitrophthalic acid gives = 1 mole of 3-nitrophthalhydrazide
0.0110 moles of 3-nitrophthalic acid gives =
mole of 3-nitrophthalhydrazide
mass of 3-nitrophthalhydrazide = 
As the percentage yield is 93% , the mass of 3-nitrophthalhydrazide recovered = 
Therefore 1.8124 g of 3-nitrophthalhydrazide were recovered.
True. I think it's true but I could be wrong