Autoionization Reactions are those reactions in which ions or molecules ionizes spontaneously without adding any external reagent.
For Example,
Autoionization of water.
H₂O + H₂O ⇆ H₃O⁺ + OH⁻
Autoionization reaction of Methanol is shown below,
Hey there!:
8) ΔTb = i*Kb*m
m is molality
Since same number of mol is added to same amount of water in both cases
m will be same for both
is 1 for glucose since it is covalent compound
is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻
So, ΔTb will be 4 times in aluminum nitrate case
So, boiling point will change by 4ºC
9) use Q = m* L
L = heat of vaporization so:
T1=T2=100ºC
5.40 * 1000 => 5400 cal/g
Q = 5400 / 540
Q = 10 grams
Hope that thlps!
Answer:
4 hydrogen atoms are needed to C5H8 to saturate it
Explanation: because C5H8 is an alkyne that contains a triple bond or alkene with 2 double bonds.
2 hydrogen atoms are needed to saturate one double bond and 4 hydrogen atoms are needed to fully Saturate a triple bond or two double bonds.
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = 
=74.8 %.
