Answer:
The information that can be used to determine which mixture has the higher proportion of KCl IS INFORMATION ABOUT THE MASS OF CHLORINE IN EACH MIXTURE, THIS INFORMATION CAN BE OBTAINED BY USING THE LAW OF DEFINITE PROPORTION.
Explanation:
The law of definite proportion states that the chemical composition by mass of a chemical compound is always constant. For instance, a chemical compound that is made up of two elements will always contain the same proportions of the constituent elements regardless of the quantity of chemical that was used.
Using the law of definite proportion, we can determine the proportion of sodium and chlorine in NaCl and the proportion of potassium and chlorine in KCl if the mass of chlorine that was used is known. Based on the results obtained, one can easily determine the mixtures that has higher proportion of KCl.
We find the weight of the empirical formula:
12.0107 + 2 x 1.00794 + 15.9994
= 30.03
Now, we divide the molecular weight by the weight of the empirical formula to find the number of times the empirical formula repeats:
90.09 / 30.03
= 3
The formula is 3(CH₂O)
C₃H₆O₃
Atomic mass Boron ( B ) = 10.811 u.m.a
10.811 g -------------- 6.02x10²³ atoms
5.40 g ----------------- ?? atoms
5.40 x ( 6.02x10²³) / 10.811 =
3.0069x10²³ atoms
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.
That statement is true
it was easy to organize 'all' the elements in a group of threes because back then they only knew about 15 - 20 elements
hope this helps
