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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.
Molar mass of 13c = 13 grams
number of moles = mass / molar mass
therefore,
number of moles = 7 / 13
To know the number of atoms in 7/13 moles, we simply multiply the number of moles by Avogadro's number as follows:
number of atoms = (7/13) x 6.022 x 10^23 = 3.2426 x 10^23 atoms
Molar mass Li2CO3 = 73.89 g/mol
Molar mass Li = 6.94g/mol Li = 6.94*2 = 13.88g
% LI = 13.88/73.89*100 = 18.78% perfectly correct.
