<u>Answer:</u> The hydroxide ion concentration and pOH of the solution is
and 2.88 respectively
<u>Explanation:</u>
We are given:
Concentration of barium hydroxide = 0.00066 M
The chemical equation for the dissociation of barium hydroxide follows:

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions
pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution
To calculate pOH of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
![[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%282%5Ctimes%200.00066%29%3D1.32%5Ctimes%2010%5E%7B-3%7DM)
Putting values in above equation, we get:

Hence, the hydroxide ion concentration and pOH of the solution is
and 2.88 respectively
1/5 + 1/4 =1/x
4x + 5x = 20
9x = 20
x = 2.22 hours Answer
Answer:
36.55 J
Explanation:
PE = Potential energy
KE = Kinetic energy
TE = Total energy
The following data were obtained from the question:
Position >> PE >>>>> KE >>>>>> TE
1 >>>>>>>> 72.26 >> 27.74 >>>> 100
2 >>>>>>>> 63.45 >> x >>>>>>>> 100
3 >>>>>>>> 58.09 >> 41.91 >>>>> 100
The kinetic energy of the pendulum at position 2 can be obtained as follow:
From the table above, at position 2,
Potential energy (PE) = 63.45 J
Kinetic energy (KE) = unknown = x
Total energy (TE) = 100 J
TE = PE + KE
100 = 63.45 + x
Collect like terms
100 – 63.45 = x
x = 36.55 J
Thus, the kinetic energy of the pendulum at position 2 is 36.55 J.
Reaction: 2K₍s₎ + 2H₂O₍l₎ → 2KOH₍aq₎ + H₂₍g₎.
K - potassium.
H₂O - water.
KOH - potassium-hydroxide.
H₂ - hydrogen.
s - solid phase.
l - liquid.
aq - disolves in water.
g - gas.
Reaction is exothermal (release of energy) and potassium burns a purple flame. H<span>ydrogen released during the reaction reacts with </span>oxygen<span> and ignites.</span><span>
</span>
Answer: a. 0.26mol
b. 0.000479mol
c. 1.12mol
Explanation: Please see attachment for explanation