We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl
Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.
To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH
Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH
Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH
Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH
I hope this helps.
Gregor Mendel
Hope this helped !
The term which is used is homogeneous.
when sugar is completely dissolved in the water, the mixture or solution homogeneous, both in same phase and same uniform texture that is liquid.
There two types of mixtures are heterogeneous and homogeneous in different phases.
If sugar is not completely dissolved in water and you see the crystals of sugar in water, then the solution will be heterogeneous.
Answer: Its average atomic mass is 114.9 amu
Explanation:
Mass of isotope 1 = 113 amu
% abundance of isotope 1 = 5% = 
Mass of isotope 2 = 115 amu
% abundance of isotope 2 = 95% = 
Formula used for average atomic mass of an element :
![A=\sum[(113\times 0.05)+(115\times 0.95)]](https://tex.z-dn.net/?f=A%3D%5Csum%5B%28113%5Ctimes%200.05%29%2B%28115%5Ctimes%200.95%29%5D)
Thus its average atomic mass is 114.9 amu
Answer:Switch off lighting in refrigeration areas when not in use?
Explanation:
