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Sonbull [250]
3 years ago
5

12. How many moles of atoms are present in 154 g of Na2O?

Chemistry
1 answer:
tia_tia [17]3 years ago
6 0

Answer:

\boxed {\boxed {\sf A. \ 2.48 \ mol \ Na_2O}}

Explanation:

We are asked to convert an amount in grams to moles. To do this, we use the molar mass. This is the number of grams in one mole of a substance. It is the same value numerically as the atomic mass on the Periodic Table, however the units are grams per mole, not atomic mass units.

Look up the molar masses for the individual elements.

  • Sodium (Na): 22.9897693 g/mol
  • Oxygen (O): 15.999 g/mol

Look back at the formula: Na₂O. Notice there is a subscript of 2 after sodium. This means there are 2 atoms of sodium in every molecule, so we have to multiply sodium's molar mass by 2 before adding oxygen's.

  • Na₂O: 2(22.9897693 g/mol)+ 15.999 g/mol = 61.9785386 g/mol

Set up a ratio using the molar mass.

\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}

Multiply by the given number of grams.

154 \ g \ Na_2O*\frac {61.9785386 \ g \ Na_2O}{1 \ mol \ Na_2O}

Flip the ratio so the grams of sodium oxide can cancel each other out.

154 \ g \ Na_2O*\frac {1  \ mol \ Na_2O}{61.9785386 \ g \ Na_2O}

154 *\frac {1  \ mol \ Na_2O}{61.9785386 }

\frac {154}{61.9785386 }  \ mol \ Na_2O

2.48473106141 \ mol \ Na_2O

The original measurement of grams given has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place.

  • 2.48<u>4</u>73106141

The 4 in the thousandth place tells us to leave the 8.

2.48 \ mol \ Na_2O

There are <u>2.48 moles of sodium oxide</u> in 154 grams, so choice A is correct.

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Answer:

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2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

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With this in mind we can <u>add the electrons</u>:

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Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

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1.66 V - 0.69 V = <u>0.97 V</u>

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Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

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