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Sidana [21]
3 years ago
11

Aluminum costs about 85 cents per pound (453.592 g). How much is one atom of aluminum?

Chemistry
1 answer:
kakasveta [241]3 years ago
5 0

One aluminum atom will cost 0.84 × 10⁻²³ cents.

Explanation:

Aluminium costs 85 cents per 453.592 grams.

number of moles = mass / atomic weight

number of moles of aluminium = 453.592 / 27 = 16.8 moles

To calculate the number of aluminum atoms we use the Avogadro's number:

if         1 mole of aluminum contains 6.022 × 10²³ aluminum atoms

then    16.8 moles of aluminum contains X aluminum atoms

X = (16.8 × 6.022 × 10²³) / 1 = 101.17 × 10²³ aluminum atoms

Now, taking in account the aluminium price, we devise the following reasoning:

if         101.17 × 10²³ aluminum atoms costs 85 cents

then   1 aluminum atom costs Y cents

Y = (1 × 85) / (101.17 × 10²³) = 0.84 × 10⁻²³ cents

Learn more about:

Avogadro's number

brainly.com/question/1445383

brainly.com/question/13272803

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Answer:

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Explanation:

To convert from moles to grams, we must use the molar mass.

Recall that water's molecular formula is H₂O. It contains hydrogen and oxygen. Look up the two elements masses on the Periodic Table.

  • Hydrogen (H): 1.008 g/mol
  • Oxygen (O): 15.999 g/mol

Now, use these masses to find water's mass. The subscript of 2 tells us there are 2 atoms of hydrogen, so we multiply hydrogen's mass by 2 and add oxygen's.

  • H₂O= 2(1.008 g/mol) + 15.999 g/mol = 18.015 g/mol

Use the molar mass as a ratio.

\frac{18.015 \ g \ H_2 O}{ 1 \ mol \ H_2 O}

Multiply by the given number of moles.

3 \ mol \ H_2O*\frac{18.015 \ g \ H_2 O}{ 1 \ mol \ H_2 O}

The moles of water will cancel.

3 *\frac{18.015 \ g \ H_2 O}{ 1 }

3 *{18.015 \ g \ H_2 O}

54.045 \ g \ H_2O

Round to the nearest whole number. The 0 in the tenth place tells us to leave the number as is.

54 \ g \ H_2O

There are about <u>54 grams</u> of water in 3 moles.

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if a gas sample in a balloon occupies 1.5 L at atmospheric pressure, what would be the pressure (in mmHg) if the volume was redu
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1425 mmHg.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 1.5 L

Initial pressure (P1) = 1 atm

Final volume (V2) = 0.8 L

Final pressure (P2) =?

Next, we shall determine the final pressure of the gas by using the Boyle's law equation as follow:

P1V1 = P2V2

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Divide both side by 0.8

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