seaweed having 50 g salt in 1 L water. The bucket contains 150 g of salt in 2 L of water
amount of water present in bucket is twice to amount of water in weed
At equilibrium, volume of water in weed is x and volume in bucket is y but concentration remain same as follows:
At equilibrium, weed loose z L from 1 L water to bucket containing 2 L as follows:
Thus, Weed will loose 0.25 L of water
Oh wow , what grade are you in ukliblo ?
Answer:
1.26 × 10¹⁵ s⁻¹
Explanation:
Work function is the minimum energy required to remove an electron from the surface of metal
energy of the electron = hf - Φ
Φ = work function = hf₀ where f₀ = threshold frequency
f₀ = Φ / h where h ( Planck constant = 6.626 × 10⁻³⁴ Js)
Φ = 5.22eV = 5.22 × 1 eV where 1 eV = 1.60217662 × 10⁻¹⁹ J
Φ = 5.22 × 1.60217662 × 10⁻19 J = 8.363362 × 10⁻¹⁹ J
f₀ = (8.363362 ×10⁻¹⁹ J) / (6.626× 10⁻³⁴ Js) = 1.26 × 10¹⁵ s⁻¹
The frequency must be greater than the 1.26 × 10¹⁵ s⁻¹ to observe the emission
Answer:
The electron in xenon are dropping more energy levels than helium
The combustion reaction of octane is as follow,
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O
According to balance equation,
8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted
So,
6.20 moles of CO₂ will release when = X g of Octane is reacted
Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol
X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.
