Answer:
Aye bro you from discovery too
Explanation:
According to the reaction equation:
and by using ICE table:
CN- + H2O ↔ HCN + OH-
initial 0.08 0 0
change -X +X +X
Equ (0.08-X) X X
so from the equilibrium equation, we can get Ka expression
when Ka = [HCN] [OH-]/[CN-]
when Ka = Kw/Kb
= (1 x 10^-14) / (4.9 x 10^-10)
= 2 x 10^-5
So, by substitution:
2 x 10^-5 = X^2 / (0.08 - X)
X= 0.0013
∴ [OH] = X = 0.0013
∴ POH = -㏒[OH]
= -㏒0.0013
= 2.886
∴ PH = 14 - POH
= 14 - 2.886 = 11.11
Answer:
Nomenclatura de sodio
nomenclatura de compuestos iónicos En compuesto químico: Compuestos iónicos que contienen iones poliatómicos ... el compuesto NaOH se llama hidróxido de sodio, ya que contiene el catión Na+ (sodio) y el anión OH (hidróxido).
Explanation:
The graph shouldn't change because a catalyst isn't supposed to have any type of affect
Answer: -
A 3.00-l flask is filled with gaseous ammonia, NH₃. the gas pressure measured at 26.0 ∘c is 1.55 atm . assuming ideal gas behavior, 3.23 grams of ammonia are in the flask
Explanation: -
Volume V = 3.00 L
Pressure P = 1.55 atm
Temperature T = 26.0 °C + 273 = 299 K
We know the value of Universal gas constant R = 0.082 L atm K−1
mol−1
We use the ideal gas equation
PV = nRT
Number of moles of Ammonia n = 
= 
= 0.19 mol
Molar mass of NH₃ = 14 x 1 + 1 x 3 = 17 g / mol
Mass of NH₃ = Molar mass of NH₃ x number of moles of NH₃
= 17 g / mol x 0.19 mol
= 3.23 g
