Answer:
c) (12×0.9889) + (13×0.01108)
Explanation:
Given data:
Percentage of C-12 = 98.89%
Percentage of C-13 = 1.108%
Atomic mass = ?
Solution:
98.89/100 = 0.9889
1.108/ 100 = 0.01108
Atomic mass = (12×0.9889) + (13×0.01108)
Atomic mass = (11.8668 + 0.144034)
Atomic mass = 12.01084
Answer:6.94
Explanation:
Molar mass of CaCO3=40+12+16×3
=40+12+48=100g/mol
Moles=mass of substance/molar mass
=97mg/100g=0.097/100=0.00097moles/L.
PH=-log[CaCo3]=-log(0.00097)=6.94
P.s it's log to base e
the balanced equation for the formation of ammonia is
N₂ + 3H₂ ---> 2NH₃
molar ratio of N₂ to NH₃ is 1:2
mass of N₂ reacted is 8.0 g
therefore number of N₂ moles reacted is - 8.0 g / 28 g/mol = 0.286 mol
according to the molar ratio,
1 mol of N₂ will react to give 2 mol of NH₃, assuming nitrogen is the limiting reactant
therefore 0.286 mol of N₂ should give - 2 x 0.286 mol = 0.572 mol of NH₃
therefore mass of NH₃ formed is - 0.572 mol x 17 g/mol = 9.72 g
a mass of 9.72 mol of NH₃ is formed
Answer: See explanation
Explanation:
The negative effects of video games include:
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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.