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forsale [732]
3 years ago
15

How many CH4 molecules are in 14.8 g of CH4

Chemistry
2 answers:
murzikaleks [220]3 years ago
8 0
1.8021•1024 molecules
Alex73 [517]3 years ago
8 0

Answer:

\boxed {\boxed {\sf 5.56 \times 10^{23} \ molecules \ CH_4}}

Explanation:

We are asked to find how many molecules of methane are in 14.8 grams of the substance.

<h3>1. Convert Grams to Moles </h3>

First, we convert grams to moles. We use the molar mass, or the mass of 1 mole of a substance. These values are equivalent to the atomic masses found on the Periodic Table, however the units are grams per mole instead of atomic mass units.

We are given the compound methane, or CH₄. Look up the molar mass of the individual elements (carbon and hydrogen).

  • C: 12.011 g/mol
  • H: 1.008 g/mol

Check the formula for subscripts. Hydrogen (H) has a subscript of 4, so there are 4 moles of hydrogen in 1 mole of methane. We must multiply hydrogen's molar mass by 4, then add carbon's molar mass.

  • H₄: 1.008 * 4 = 4.032 g/mol
  • CH₄: 12.011 + 4.032 = 16.043 g/mol

Now we use dimensional analysis to convert. To do this, we set up a ratio using the molar mass.

\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}

Since we are converting 14.8 grams of methane to moles, we multiply by this value.

14.8 \ g \ CH_4 *\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}

Flip the ratio so the units of grams of methane cancel.

14.8 \ g \ CH_4 *\frac{ 1 \ mol \ CH_4} {16.043 \ g \ CH_4 }

14.8  *\frac{ 1 \ mol \ CH_4} {16.043}

\frac {14.8}{16.043} \ mol \ CH_4= 0.9225207256 \ mol \ CH_4

<h3>2. Moles to Molecules </h3>

Next, we convert moles to molecules. We use Avogadro's Number or 6.022  × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are moles of methane. Set up another ratio using Avogadro's Number.

\frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}

Multiply by the number of moles we calculated.

0.9225207256\ mol \ CH_4 * \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}

The units of moles of methane cancel.

0.9225207256* \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 }

5.55541981 \times 10^{23} \ molecules \ CH_4

<h3>3. Round </h3>

The original measurement of grams (14.8) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 5 in the hundredth place up to a 6.

5.56 \times 10^{23} \ molecules \ CH_4

14.8 grams of methane is equal to approximately <u>5.56 × 10²³ molecules of methane.</u>

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Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.
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The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₃PO₄ (nA) = 2
  • The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>
  • Molarity of acid, H₃PO₄ (Ma) = 0.390 M
  • Volume of acid, H₃PO₄ (Va) = 24.5 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
  • Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

9.555 / (0.279 × Vb) = 2/3

Cross multiply

2 × 0.279 × Vb = 9.555 × 3

0.558 × Vb = 28.665

Divide both side by 0.558

Vb = 28.665 / 0.558

Vb = 51.4 mL

Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL

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