Answer is: pH of hydroxylamine solution is 9,23.
Kb(NH₂OH) = 1,8·10⁻⁵<span>.
c</span>₀(NH₂OH)<span> = 0,0500 M =
0,05 mol/L.
c(NH</span>₂⁺) = c(OH⁻) = x.
c(NH₂OH<span>) = 0,05 mol/L - x.
Kb = c(NH</span>₂⁺) · c(OH⁻) / c(NH₂OH).
0,0000000066 = x² / (0,05 mol/L - x).
solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.<span>
pOH = -log(</span>0,000018 mol/L) = 4,74.<span>
pH = 14 - 4,74 = 9,23.</span>
Answer:
2%
Explanation:
oriented C-2, and (3) the minimizing of the number of ... (2) L. A. Mitscher, J. K. Paul, and L. Goldman,Experientia, 19, 195. (1963). ... SOzCeHiBr)3 in 147 ml. of anhydrous methanol containing 0.37 ... bicarbonate and saturated sodium chloride solution, and dried ... determined in 2% chloroform solution; infrared spectra on.
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
__________________________________________________
2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
Answer:
Explanation: A mixture of two partially miscible liquids
