Answer:
pH = 1.348
Explanation:
- H2SO3 + H20 ↔ H3O+ + HSO3-
∴ Ka1 = 1.7 E-2 = ( [ H3O+ ] * [ HSO3- ] ) / [ H2SO3 ]
∴ <em>C </em>H2SO3 (aq) = 0.163 M
- HSO3- + H2O ↔ H3O+ + SO32-
∴ Ka2 = 6.4 E-8 = ( [ H3O+ ] * [ SO32- ] ) / [ HSO3- ]
∴ Kw = 1 E-14 = [ H3O+ ] * [ OH- ]
since Ka1 >>> Ka2 ( difference between the constants is greater than the order of E3 ); then the most importan equilibrium is Ka1 and we can assume that [ SO32- ] <<< [ HSO3- ]
mass balance:
⇒ <em>C </em>H2SO3 = 0.163 = [ H2SO3 ] + [ HSO3- ]......(1)
charge balance:
⇒ [ H3O+ ] = [ HSO3- ]...........(2) where [ OH- ] and [ SO32- ] is neglected. the first comes from water and the second comes from Ka2.
(2) in (1):
⇒ [ H2SO3 ] = 0.163 - [ H3O+ ]......(3)
(2) and (3) in Ka1:
⇒ 1.7 E-2 = [ H3O+ ]² / ( 0.163 - [ H3O+ ] )
⇒ [ H3O+ ]² + 1.7 E-2 [ H3O+ ] - 2.771 E-3 = 0
⇒[ H3O+ ] = 0.045 M
⇒ pH = - Log [ H3O+ ] = - Log ( 0.045 )
⇒ pH = 1.348
Answer:
Interphase
Explanation:
I think you answered your own question chief? lol
I don't agree because ocean temperatures can get warm thanks to high sun temperature and warm fronts.
So there are three formulas that you must be aware of hen attempting these questions (by technicality, only two because the third is a transposition of the second).
These formulas are :
mole x L = # of atoms (where L = Avogadro's Constant)
mass / Mr = mole (where Mr = Molar Mass)
Mole x Mr = mass
So no we can easily ace these questions
1) a. 98.3 g of Hg
mass of Hg = 98.3g
Mr " " = 201 g / mol
mole = ![\frac{98.3 g}{201 g / mol}](https://tex.z-dn.net/?f=%20%5Cfrac%7B98.3%20g%7D%7B201%20g%20%2F%20mol%7D%20)
= 0.489 mol
∴ # of atoms = 0.489 mol * (6.03 * 10²³)
= 2.95 * 10²³ atoms
c. 10.7g of Li
mass of Li = 10.7 g
Mr " " = 7 g / mol
thus mole = ![\frac{10.7g}{7 g / mol}](https://tex.z-dn.net/?f=%20%5Cfrac%7B10.7g%7D%7B7%20g%20%2F%20mol%7D%20)
= 1.529 mol
# of atoms = 1.529 mol * (6.03 * 10²³)
= 9.22 * 10²³
2) a. 6.84 g of C₁₂H₂₂O₁₁
mass of sucrose = 6.84 g
Mr " " = ∑ (Mr of each element * amount of each element's atom)
= ∑ [(12 * 12)+(1 * 12)+(16 * 11)] g / mol
= 342 g / mol
∴ mole of sucrose = ![\frac{6.84 g}{342 g / mol}](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.84%20%20g%7D%7B342%20g%20%2F%20mol%7D%20)
= 0.02 mol
c) 68.0g of NH<span>₃</span>
<span> mass of NH₃ = 68.0 g</span>
<span> Mr " " = [(14 * 1) + ( 1 * 3)] g / mol</span>
<span> ∴ mole of NH₃ = </span>![\frac{68.0 g }{17 g / mol}](https://tex.z-dn.net/?f=%20%5Cfrac%7B68.0%20g%20%7D%7B17%20%20g%20%2F%20mol%7D%20)
= 4 mol
3) a. 3.52 mol of Si
mole of Si = 3.52 mol
Mr " " = 28 g / mol
Mass = 3.52 mol * 28 g / mol
= 98.56 g
c. 0.550 mol of F₂
mol of F₂ = 0.55 mol
Mr " " = (19 * 2 ) g / mol
Mass = 0.55 mol * 38 g / mol
= 20.9 g
Hope I as clear in my answers and that they helped