Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Answer:
A neutral particle made of an electron and hole
Explanation:
Exciton
It is the combination of an electron and a hole ( hole refers to the vacancy of an electron ) . And , as both the electron and the hole have the same charge but the polarity is opposite , the combination will lead to a neutral compound , i.e. , Exciton have no charge and so neutral .
It is free to move in the nonmetallic crystal and since it charge less , it is difficult to detect it directly .
Combined gas law is
PV/T = K (constant)
P = Pressure
V = Volume
T = Temperature in Kelvin
For two situations, the combined gas law can be applied as,
P₁V₁ / T₁ = P₂V₂ / T₂
P₁ = 3.00 atm P₂ = standard pressure = 1 atm
V₁ = 720.0 mL T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
By substituting,
3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
V₂ = 2012.6 mL
hence the volume of gas at stp is 2012.6 mL
