Given :
0.00072 M solution of 
at 
.
To Find :
The concentration of 
and pOH .
Solution :
1 mole of gives 2 moles of ions .
So , 0.00072 M mole of gives :
![[OH^-]=2 \times 0.00072\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2%20%5Ctimes%200.00072%5C%20M)
![[OH^-]=0.00144\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00144%5C%20M)
![[OH^-]=1.44\times 10^{-3}\ M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.44%5Ctimes%2010%5E%7B-3%7D%5C%20M)
Now , pOH is given by :
![pOH=-log[OH^-]\\\\pOH=-log[1.44\times 10^{-3}]\\\\pOH=2.84](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D%5C%5C%5C%5CpOH%3D-log%5B1.44%5Ctimes%2010%5E%7B-3%7D%5D%5C%5C%5C%5CpOH%3D2.84)
Hence , this is the required solution .
Correct, it is a start of a experiment, something that is asked first.
Temperature of the same cup of water will rise by 6 °C unless it boils.
<h3>Explanation</h3>
.
However, neither 
nor 
is given.
Adding 
to this cup of water of mass rises its temperature by 
.
In other words,
.
Both and are constant for the same cup of water unless the water boils. It's possible to reuse the value of 
in the second calculation. Here's how:
.
The flow rate of water differs from honey due to the texture and thickness of the product. Honey’s thickness doesn’t allow it to move as quickly as the water, therefore affecting the flow rate.
When a liquid is frozen, the atoms in the liquid slow their movement due to lack of energy so the substance becomes hard.
