Answer:
The common ion will be di-positive ion.
Explanation:
The ionization energy is defined as the amount of energy needed for removal of most loosely bound electron from an isolated atom in gaseous state.
The low ionization energy shows that the atom is able to give electron easily as after losing electron it may attain noble gas configuration or half filled stability.
Here the first and second ionization energy, both are low suggesting that the element is ready to give two electrons easily to form a di-positive ion however the third ionization energy is high which shows that it will not form tri-positive ion commonly.
Answer:
Explanation:
We must do the conversions
mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 180.16
C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O
m/g: 24.5
(a) Moles of C₆H₁₂O₆
(b) Moles of CO₂
(c) Volume of CO₂
We can use the Ideal Gas Law.
pV = nRT
Data:
p = 0.960 atm
n = 0.8159 mol
T = 37 °C
(i) Convert the temperature to kelvins
T = (37 + 273.15) K= 310.15 K
(ii) Calculate the volume
Answer is: 7,826 kg of cryolite.
Chemical reaction: Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂<span>O.
m(</span>Al₂O₃) = 12,1 kg = 12100 g.
n(Al₂O₃) = m(Al₂O₃) ÷ M(Al₂O₃).
n(Al₂O₃) = 12100 g ÷ 101,96 g/mol = 111,86 mol; limiting reactant.
m(NaOH) = 60,4 kg = 60400 g.
n(NaOH) = 60400 g ÷ 40 g/mol.
n(NaOH) = 1510 mol.
m(HF) = 60,4 kg = 60400 g.
n(HF) = 60400 g ÷ 20 g/mol = 3020 mol.
From chemical reaction: n(Al₂O₃) : n(Na₃AlF₆) = 6 : 2.
n(Na₃AlF₆) = 2 ·111,86 mol ÷ 6 = 37,28 mol.
m(Na₃AlF₆) = 37,28 mol · 209,94 g/mol.
m(Na₃AlF₆) = 7826,56 g = 7,826 kg.
Over all the size of the crystal depends on how fast magma will harden over time if the magma cools slowly the crystal will be larger than they would if the magma cooled faster
