Answer:
Equilibrium constant is equal to 
Explanation:
If the reaction is of this form
aA(g) + bB(g) ⇄ cC(g) + dD(g)
Then , the Kp of this equation will be equal to
Given
MPa and c is equal to two
MPa
MPa
Here a and b is equal to zero.
Substituting the given values we get -
Equilibrium constant is equal to
Answer:
- g) 0.20 mol
- h) 2.0 mol/dm³
- i) 2.8 g
Explanation:
<h2>a) </h2>
1. Data:
2. Solution:
i) Formulae:
- number of moles = mass in grams/Ar
- number of atoms = number of moles × Avogadro constant
- Avogadro constant = 6.022 × 10²³ atoms/mol
ii) Calculations:
- number of moles = 71g / 35.5(g/mol) = 2 mol
- number of atoms = 2mol × 6.022 × 10²³ atoms/mol = 1.20 × 10²⁴ atoms
<h2>b) </h2>
1. Data:
- 0.2 mol KOH
- Ar: H = 1, O = 16, k = 39
2. Solution
i)Formula:
- mass = number of moles × molar mass
ii) Molar mass:
- 1×1g/mol + 1×39g/mol + 1×16g/mol = 56g/mol
iii) Calculations:
- mass = 0.2 mol × 56g/mol = 11.2 g/mol ≈ 10g/mol (rounded to 1 significant figure)
<h2>c) </h2>
1. Data:
2. Solution
i) Formula:
- number of moles = mass in grams / Ar
ii) Calculations:
- number of moles = 1.4g / 7 gmol = 0.7 mol
<h2>d) </h2>
1. Data:
2. Solution
i) Formula:
- mass = number of moles × molar mass
ii) Calculations:
- molar mass = 8 × 32g/mol = 256 g/mol
- mass = 4 mol × 256 g/mol = 1,024 g ≈ 1,000 g (rounded to 1 significant figure)
<h2>e)</h2>
1. Data:
- Mr: Ca(NO₃)₂ . 2H₂O = 200
2. Solution
i) Formulae:
- number of moles = mass in grams / molar mass
ii) Calculations:
- number of moles = 50 g / 200 g/mol = 0.25 mol of Ca(NO₃)₂ . 2H₂O
The number of moles of water molecules is 2 times the number of moles of Ca(NO₃)₂ . 2H₂O.
- number of moles of water molecules = 2 × 0.25 mol = 0.5 mol
<h2>f) </h2>
1. Data:
2. Solution:
i) Formulae:
- number of atoms = number of moles × Avogadro constant
- number of moles = mass in grams / molar mass
ii) Calculations
- number of moles = 4.9g / 98g/mol = 0.050 mol
- number of atoms = 0.050 mol × 6.022 × 10²³ atoms/mol = 3.0×10²² atoms
<h2>g) </h2>
1. Data:
- V = 24 dm³ (air)
- 20% oxygen
- r.t.p ⇒ T = 298K, p = 1 atm
2. Solution
i) Formula:
ii) Calculation:
- R = 0.08206 (atm.dm³/K.mol)
- n = [ 1 atm × 24 dm³] / (0.08206atm.dm³/K.mol × 298K) = 0.98mol of air
- moles of oxygen = 20% × 0.98 mol ≈ 0.20 mol
<h2>h) </h2>
1. Data:
2. Solution
i) Formula:
- Molarity = moles of solute / volume of solution in dm³
ii) Calculations:
25cm³ × (0.1cm/dm)³ = 0.125 dm³
- Molarity = 0.25 mol / 0.125 dm³ = 2.0 mol/dm³
<h2>i) </h2>
1. Data:
- V = 35cm³
- M = 2 mol/dm³
- Mr NaOH = 40
2. Solution
i) Formulae:
- number of moles = M × V (in dm³)
- mass = number of moles × mola rmass
ii) Calculations:
- V = 35cm³ × (0.1dm/cm)³ = 0.035dm³
- number of moles = 2 mol/dm³ × 0.035 dm³ = 0.070 mol
- mass = 0.070 mol × 40 g/mol = 2.8 g
Answer:
a) in pure water:
⇒ S (25°C) Ni(OH)2 = 2.0 E-5 M
b) in 0.013 M NaOH:
⇒ S = 9.467 E-11 M
Explanation:
a) Ni(OH)2 ↔ Ni2+ + 2OH-
S S 2S
⇒ Ksp = [ Ni2+ ] * [ OH- ]² = 1.6 E-14.....Ksp value comes from the literature
⇒ Ksp = S * 2S² = 1.6 E-14
⇒ 2S³ = 1.6 E-14
⇒ S = ∛( 1.6 E-14 / 2 )
⇒ S = 2.0 E-5 M
b) NaOH ↔ Na+ + OH-
0.0130 M 0.013 0.013
Ni(OH)2 ↔ Ni2+ + 2OH-
S S 2S + 0.0130
⇒ Ksp = [ Ni2+ ] * [ OH- ]²
⇒ Ksp = S * (2S + 0.0130 )² = 1.6 E-14
Due to the small value of the Ksp it can be assumed that "s" is a very small number, so "2s" can be neglected when added to 0.0130.
⇒ 1.6 E-14 = S * ( 0.0130 )²
⇒ S = 9.467 E-11 M
That would be the exosphere. The reason it feels cold, despite high temperatures, is due to lower air density.
Hope it helps!
