Answer:
9.96*10^21
Explanation:
Molar mass of K2O=29*2+16
= 74g per mol
number of moles in the sample= 1.224/ 74
=0.1654
Number of particles in 1 mole=6.0221409*10^23
Number of particles= 0.01654*6.0221409*10^23
=9.96*10^21
I'm not sure about 2 but in 3 it would float
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.
31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag 6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ → ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻ = 107.9 g Ag
∅ | 63.5 g Cu | 1 mol Cu | 1 mol Ag 63.5
There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
Answer: The image from the question has the correct answers.
Explanation:
As summarized in the attached table.
Answer: 94.07%
Explanation:
Percentage yield can be calculated by the formula
%yield = Experimental yield/Theoretical yield x100
Experimental yield = 7.93g
Theoretical yield = 8.43
%yield = Experimental yield/Theoretical yield x100
%yield = 7.93/8.43 x 100 = 94.07%
