Am I doing this right?
What is [OH-] of a solution
with a pH of 4.0?
Answer in M
Ya
Answer:
25.6 %.
Explanation:
- To calculate the % of Cl in KClO₄:
<em>% Cl = [Atomic mass of (Cl)/molar mass of KClO₄] x 100</em> = [(35.5 g/mol)/(138.55 g/mol)] x 100 = <em>25.62 % ≅ 25.6 %.</em>
Answer:
Explanation:
The <em>dimensional analysis</em> helps you by focusing your attention into the units: you must set up the equation starting with the available units and add the factors with the units such that the cancelation of numerators and denominators ends up with the desired units: grams of aluminium in this case.
In this case, you start with a mass of 2.56 grams per penny, which is 
, whose value is $200.00 and want to end with the grams of
Aluminium needed.
Then, you must use the mole ratios derived from the chemical equation in a way that the units lead from grams of copper to grams of aluminium.
The equation is:
From which, the mole ratio between copper and aluminium is:
You must also use the atomic masses of Cu and Al which are 63.5g/mol and 27g/mol respectively.
The equation is:
You can check that afeter multiplying all the factors, the most units will cancel and the final units will be grams of Al.
<span>83.9% is the weight percent of Al(C6H5)3 in the original 1.25 g sample.
First, look up the atomic weights of all elements involved.
Atomic weight of Aluminum = 26.981539
Atomic weight of Carbon = 12.0107
Atomic weight of Chlorine = 35.453
Atomic weight of Hydrogen = 1.00794
Now calculate the molar mass of Al(C6H5)3, and C6H6
Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 15.999
= 258.293239 g/mol
Molar mass of C6H6 = 6 * 12.0107 + 6 * 1.00794
= 78.11184 g/mol
Determine how many moles of C6H6 was produced
0.951 g / 78.11184 g/mol = 0.012174851 mol
Since the balanced formula indicates that 3 moles of C6H6 is produced for each mole of Al(C6H5)3 used, divide by 3 to get the number of moles of Al(C6H5)3 that was present.
0.012174851 mol / 3 = 0.004058284 mol
Now multiply by the molar mass of Al(C6H5)3 to get the mass of Al(C6H5)3 originally present.
0.004058284 mol * 258.293239 g/mol = 1.048227218 g
Finally, divide the mass of Al(C6H5)3 by the total mass of the original sample to get the weight percentage.
1.048227218 g / 1.25 g = 0.838582
Since all our measurements had 3 significant figures, round the result to 3 significant figures, giving 0.839 = 83.9%</span>
