Answer: 66.2 g
Explanation:
1) The ratio of Al in the molecule is 1 mol to 1 mol .
2) The mass of 1 mol of molecules of Al (CH2H3O2)3 is the molar mass of the compound.
3) You calculate the molar mass of the compound using the atomic masses of each atom, in this way:
Al: 27 g/mol
C: 2 * 3 * 12 g/mol = 72 g/mol
H: 3 * 3 * 1 g/mol = 9 g/mol
O: 2 * 3 * 16 g/mol = 96 g/mol
Molar mass = 27 g/mol + 72 g/mol + 9 g/mol + 96 g/mol = 204 g/mol
4) Set a proportion:
27 g/mol x
-------------------- = ----------
204 g/mol 500 g
5) Solve for x:
x = 500 g * 27 g/mol / 204 g/mol = 66.2 g
= 24.3
The average atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its % abundance).
Thus,
0.790 × 24 u = 18.96 u
0.100 × 25 u = 2.50 u
0.110 × 26 u = <u>2.86 u</u>
TOTAL = 24.3 u
∴ The relative atomic mass of X is 24.3.
This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.
1 mole is equal to 6.022x10²³ particles as given, so:
<h3>
Answer:</h3>
2.49 mol
Let me know if you have any questions.
Answer:
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