Answer:
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
Explanation:
Step 1: Data given
Mass of Al2O3 = 3.50 grams
Molar mass of Al2O3 = 101.96 g/mol
Number of Avogadro = 6.022 * 10^23 /mol
Step 2: Calculate moles Al2O3
Moles Al2O3 = mass Al2O3 / molar mass Al2O3
Moles Al2O3 = 3.50 grams / 101.96 g/mol
Moles Al2O3 = 0.0343 moles
Step 3: Calculate moles Aluminium
In 1 mol Al2O3 we have 2 moles Al
in 0.0343 moles Al2O3 we have 2*0.0343 = 0.0686 moles Al
Step 4: Calculate aluminium atoms
Aluminium atoms = moles aluminium * Number of Avogadro
Aluminium atoms = 0.0686 * 6.022 * 10^23
Aluminium atoms = 4.13 *10^22 aluminium atoms
The correct answer is E
Answer:
H2O<en<phen
Explanation:
The degree of d- splitting is observed from the intensity of colour. The order of d splitting from least to greatest is H2O<en<phen. Phen shows the greatest d-splitting. The degree of splitting of d- orbitals by ligands depends on their relative positions in the spectrochemical series. The spectrochemical series is an experimentally determined series. The series separates the ligands into strong field and weak field ligands. Strong field ligands are found towards the end of the series. Strong field ligands such as en and phen can participate in metal to ligand or ligand to metal pi-bonding. Hence they cause more d-splitting. Ethylendiamine and phenanthroline occur towards the end of the spectrochemical series hence the higher order of d-splitting.
Answer:
Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]
Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]
Explanation:
An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:
HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)
And Ka, the acid dissociation constant is:
<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />
When base:
HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)
And kb, base dissociation constant is:
<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>
Answer:
28.52 L
Explanation:
First, let's calculate the density of the ocean, which is the mass divided by the volume:
d = m/V
d = 35.06/1
d = 35.06 g/L
So, for a mass of 1.00 kg = 1000.00 g
d = m/V
35.06 = 1000.00/V
V = 1000.00/35.06
V = 28.52 L
How all the data are expressed with two significant figures, the volume must also be expressed with two.
Answer:
Four moles of the cation
Explanation:
2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)
Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.
The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.
This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.