➯https://t.me/+QGFxHSeUJpU3OTJk

Qormaata Kutaa 6
#GUYYAAN_5_HAFE !

BBO- 20/09/2017 Qormaatni Xumura Barnoota Sad. 1ffaa Kutaa 6ffaa Naannoo Oromiyaa Guyyaa 26-27/09/2017 Kan Kennamu Ta'uun Ni Beekama.
Kanaafuu, Manneen Barnootaa
- Qormaatni keessatti kennamu,
- Barsiisotni qormaata kennan,
- Supparvaayizarootni qormaataa,
- Hogganaa Buufataalee qormaataa,
- Hoggansi haala qormaataa irratti ibsa (orientation) kennu,
- Hawaasni barnootaa hundi,
- Gumiin Tiraansifoormeeshinii Barnootaa (GTB) sadarkaan jiru,
- Maatiin Barattootaafi
- Abbaan CIDHAA, Barattootni keenya guyyaa 5 hafan kanneen keessatti xiyyeeffannoo lakk. 1 gochuun qormaatni Naannoo Kutaa 6ffaa akka milkaa'u tokko taanee gahee keenya haa baanu kan jedhu ergaa BBO ti.

#Guyyaa_5_Qofatu Hafe!
Xiyyeeffannoo Lakk. 1 !
#Qormaata_Kutaa_6 !!


💙Oromia Education Bureau 💠
💙Tolaa Bariisoo Gadaa 💠

Guys now we are mathematics quiz on group please join group and participate on it
https://t.me/education123444

Economics quiz is also there now guys

Guys be ready for biology quiz in our channel discussion group join now

https://t.me/education123444
https://t.me/education123444

⚗️🧪Here’s a mix of questions from grade 11 chemistry unit 3 and unit 5. Put reactions if you want the explanation!

1. An unknown gas diffuses at a rate of 1.3 × 10⁻⁶ mol/h. Ethane (C₂H₆) diffuses at a rate of 3.6 × 10⁻⁶ mol/h under the same conditions. Calculate the molar mass of the unknown gas.

2. Write the expressions for KC and KP for the following homogeneous gaseous reaction:
2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)

3. Write the equilibrium constant expression in terms of KC for the following heterogeneous reaction:
Fe³⁺ (aq) + 3OH⁻ (aq) ⇌ Fe(OH)₃ (s)

4. For the reaction 2NO (g) + Br₂ (g) ⇌ 2NOBr (g), an equilibrium mixture at a certain temperature is found to contain [NO] = 0.010 M, [Br₂] = 0.025 M, and [NOBr] = 0.15 M. Calculate the value of KC for this reaction at this temperature.

5. The reaction for the formation of nitrosyl chloride was studied at 25 °C:
2NO (g) + Cl₂ (g) ⇌ 2NOCl (g)
At equilibrium, the partial pressures were found to be 1.2 atm, 5.0 × 10⁻² atm, and 3.0 × 10⁻¹ atm for NOCl, NO, and Cl₂, respectively. Calculate the value of KP for this reaction at 25 °C.

6. The equilibrium constant, KC, for the reaction H₂ (g) + I₂ (g) ⇌ 2HI (g) is 54.3 at 430 °C. Suppose a mixture contains initial concentrations [H₂]₀ = 0.0218 M, [I₂]₀ = 0.0145 M, and [HI]₀ = 0.0783 M. Is the reaction at equilibrium? If not, in which direction will the reaction proceed to reach equilibrium?

7. For the equilibrium N₂O₄ (g) ⇌ 2NO₂ (g), KC = 0.36 at 100 °C. A sample of 0.25 mol of N₂O₄ is allowed to dissociate and come to equilibrium in a 1.5 L flask at 100 °C. What are the equilibrium concentrations of NO₂ and N₂O₄?

8. The equilibrium constant, KC, is 69 for the reaction H₂ (g) + I₂ (g) ⇌ 2HI (g). Given that 1 mole of H₂ and 2 moles of I₂ were added to a 500 mL reaction vessel, determine the molar concentrations in the mixture at equilibrium.

9. At 327 °C, the equilibrium concentrations are [CH₃OH] = 0.15 M, [CO] = 0.24 M, and [H₂] = 1.1 M for the reaction: CH₃OH (g) ⇌ CO (g) + 2H₂ (g). Calculate KP at this temperature.

10. Consider the exothermic reaction: N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g), ∆H°rxn = -92 kJ. According to Le Chatelier's principle, predict how the position of equilibrium will shift if the following changes are made:
a. Increasing the temperature
b. Increasing the pressure by decreasing the volume
c. Adding more N₂ gas
d. Removing NH₃ gas as it is formed
e. Adding an inert gas (like Argon) at constant volume
f. Adding a catalyst

11. Arrange the following reactions in order of their increasing tendency to proceed towards completion (least extent to greatest extent) based on their KC values:
a. CO (g) + Cl₂ (g) ⇌ COCl₂ (g), KC = 13.8
b. 2CO₂ (g) ⇌ 2CO (g) + O₂ (g), KC = 2.0 × 10⁻⁶
c. 2NOCl (g) ⇌ 2NO (g) + Cl₂ (g), KC = 4.7 × 10⁻⁴

1. Rate₁ / Rate₂ = √(M₂ / M₁)
Rate of unknown = 1.3 × 10⁻⁶ mol/h
Rate of ethane = 3.6 × 10⁻⁶ mol/h
Molar mass of ethane = 30 g/mol
=> (1.3×10⁻⁶ / 3.6×10⁻⁶) = √(30 / M₁)
(1.3 / 3.6)² ≈ 0.1304
=> 0.1304 = 30 / M₁ ⇒ M₁ = 30 / 0.1304 ≈ 230 g/mol

2. Reaction: 2SO₂ (g) + O₂ (g) ⇌ 2SO₃ (g)
KC expression:
KC = [SO₃]² / ([SO₂]² × [O₂])
KP expression:
KP = (P_SO₃)² / ((P_SO₂)² × P_O₂)

3. Equilibrium expression for
Fe³⁺ (aq) + 3OH⁻ (aq) ⇌ Fe(OH)₃ (s):
Kc = 1 / ([Fe³⁺][OH⁻]³)

4. Reaction: 2NO (g) + Br₂ (g) ⇌ 2NOBr (g)
[NO] = 0.010 M, [Br₂] = 0.025 M, [NOBr] = 0.15 M
Kc = [NOBr]² / ([NO]² × [Br₂])
= (0.15)² / ((0.010)² × 0.025)
= 0.0225 / (0.0001 × 0.025)
= 0.0225 / 0.0000025
= 9000

5.Reaction: 2NO (g) + Cl₂ (g) ⇌ 2NOCl (g)
P(NOCl) = 1.2 atm, P(NO) = 0.050 atm, P(Cl₂) = 0.30 atm
Kp = (P(NOCl))² / ((P(NO))² × P(Cl₂))
= (1.2)² / ((0.050)² × 0.30)
= 1.44 / (0.0025 × 0.30)
= 1.44 / 0.00075
= 1920
6. Reaction: H₂ (g) + I₂ (g) ⇌ 2HI (g), Kc = 54.3
[H₂] = 0.0218, [I₂] = 0.0145, [HI] = 0.0783
Q = [HI]² / ([H₂][I₂])
= (0.0783)² / (0.0218 × 0.0145)
= 0.00613 / 0.0003161
≈ 19.4

Since Q < Kc (19.4 < 54.3), the reaction is not at equilibrium and will proceed forward.

7. The equation for the reaction is:
N₂O₄(g) ⇌ 2NO₂(g)   
KC = [NO₂]² / [N₂O₄] = 0.36
[NO₂]initial = 0
[N₂O₄]initial = 0.25 mol ÷ 1.5 L = 0.167 M

Concentration N₂O₄ ⇌ 2NO₂

Initial 0.167 M 0 M
Change –x +2x
Equilibrium 0.167 – x 2x

KC = [NO₂]² / [N₂O₄]
0.36 = (2x)² / (0.167 – x)
0.36 = 4x² / (0.167 – x)
=> 4x² = 0.36(0.167 – x)
4x² = 0.06 – 0.36x
4x² + 0.36x – 0.06 = 0
By using quadratic formula:
x=(–0.36 ± √(0.36² –4×4×–0.06))/2×4
=(–0.36 ± √(1.0896))/2×4
= (-0.36±1.044)/8
x=0.0855 or x= -0.1755
But X cannot be negative.
Therefore,
[N₂O₄] = 0.167 – 0.0855 = 0.0815 M
[NO₂] = 2x = 2 × 0.0855 = 0.171 M

Check KC:
KC = (0.171)² / (0.0815) = 0.029241 / 0.0815 ≈ 0.36

8.[H₂] = 1 mol / 0.5 L = 2.0 M
[I₂] = 2 mol / 0.5 L = 4.0 M
[HI] = 0 (initially)

H₂ + I₂ ⇌ HI
Initial 2.0 M 4.0 M 0
Change –x –x +2x
Equilibrium 2.0 – x 4.0 – x 2x

KC = [HI]² / ([H₂][I₂])
  69 = (2x)² / [(2.0 – x)(4.0 – x)]
  69 = 4x² / (8.0 – 6x + x²)
4x²=552-414x+69x²
65x² - 414+ 552=0
By using quadratic formula:
x=(414 ± √(414² –4×65×552))/2×65
=(414 ± 167)/130
x=4.46 or x= 1.9

But X cannot be greater than 4M.
Therefore,
[H2] = 2 – 1.9 = 0.1M
[HI] = 2x = 2 ×1.9 = 3.8M
[I2]=4-1.9=2.1M

Check KC:

KC = (3.80)² / (0.10 × 2.10)
KC = 14.44 / 0.21 ≈ 68.76

9.CH₃OH ⇌ CO + 2H₂
[CH₃OH] = 0.15 M, [CO] = 0.24 M, [H₂] = 1.1 M
Δn = 3 - 1 = 2
Kc = [CO][H₂]² / [CH₃OH]
= 0.24 × (1.1)² / 0.15
= 0.24 × 1.21 / 0.15
= 0.2904 / 0.15 = 1.936
Use KP = Kc(RT)^Δn
R = 0.0821, T = 600 K (327 °C = 600 K)
KP = 1.936 × (0.0821 × 600)²
= 1.936 × (49.2)²
= 1.936 × 2,420.64 ≈ 4686.4

10. Reaction: N₂ + 3H₂ ⇌ 2NH₃, ∆H = -92 kJ (exothermic)
a. Increasing temperature → shifts left (endothermic direction)
b. Increasing pressure (by decreasing volume) → shifts right (fewer gas molecules)
c. Adding more N₂ → shifts right
d. Removing NH₃ → shifts right
e. Adding inert gas at constant volume → no effect
f. Adding a catalyst → no shift, only speeds up equilibrium

11.Arrange by KC value (lowest to highest):
b. 2CO₂ ⇌ 2CO + O₂ → KC = 2.0 × 10⁻⁶ (least tendency)
c. 2NOCl ⇌ 2NO + Cl₂ → KC = 4.7 × 10⁻⁴
a. CO + Cl₂ ⇌ COCl₂ → KC = 13.8 (greatest tendency)
Order: b < c < a