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Vladimir79 [104]
2 years ago
11

In what ways are final grade calculations and atomic mass calculations similar?

Chemistry
1 answer:
Gennadij [26K]2 years ago
3 0

Answer:

atomic mass calculations are similar to final grade calculations in how the two are calculated. In school, your given different kinds of work that has a different effect on your grade. Homework might be worth 20 percent while a tests are worth 80 percent. the percentage of these assignments are the weight they have on your grade. when you multiply the weight by the score, you get a weighted score that you can add with other weighted scores to reach a total, that total would be your grade. Atomic mass calculations are found the same way, first turn the percent abundance into a decimal (divide by 100, 97% becomes .97) multiply it by Mass (AMU) and you get that isotopes mass contribution. do this for the isotopes for whatever element, and add up the total. the total will be your element's atomic mass.

Explanation:

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What mass of Al contains the same number of atoms as 33.5g of Au
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A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.
dimulka [17.4K]

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution =  9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

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