If the O-Zone Tropo increased to a point, like it has, it can cause a huge variety of health problems and even death! Some of the problems include, but are not limited to: Asthma, E<span>mphysema, Chest Pain, Inflamed Lungs and Lung Scarring from tissue damage. </span>
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.
Answer:
Correct answer is: "Cold Water"
Explanation:
EDGE 2020
Answer:
% yield = 82.5%
Explanation:
HgO + 2Cl₂ → HgCl₂ + Cl₂O
Our reactants are:
Our products are:
We do not have information about moles of reactants, but we do know the theoretical yield and the grams of product, in this case Cl₂O, we have produced.
Percent yield = (Yield produced / Theoretical yield) . 100
Theoretical yield is the mass of product which is produced by sufficent reactant. We replace data:
% yield = (0.71 g/0.86g) . 100 = 82.5%
