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mr Goodwill [35]
2 years ago
14

1.20 x 10^22 molecules NaOH to gram

Chemistry
1 answer:
Trava [24]2 years ago
4 0

Answer:

\boxed {\boxed {\sf 0.797 \ g \ NaOH}}

Explanation:

<u>1. Convert Molecules to Moles</u>

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}

Multiply by the given number of molecules.

1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}

Flip the fraction so the molecules cancel out.

1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}

1.20*10^{22}  *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}

\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}

0.0199269345732 \ mol \ NaOH

<u>2. Convert Moles to Grams</u>

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

  • Na: 22.9897693 g/mol
  • O: 15.999 g/mol
  • H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

  • NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

\frac {39.9967693 \ g  \ NaOH }{1 \ mol \ NaOH}

Multiply by the number of moles we calculated.

0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g  \ NaOH }{1 \ mol \ NaOH}

The moles of sodium hydroxide cancel.

0.0199269345732 *\frac {39.9967693 \ g  \ NaOH }{1}

0.0199269345732 *39.9967693 \ g  \ NaOH

0.79701300498 \ g \ NaOH

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

0.797 \ g \ NaOH

1.20*10²² molecules of sodium hydroxide is approximately 0.797 grams.

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Answer:

1.386 KJ

Explanation:

From the question given above, the following data were obtained:

Mass (M) of copper = 45 g

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

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Next, we shall determine the change in temperature. This can be obtained as follow:

Initial temperature (T1) = 20.0°C

Final temperature (T2) = 100.0°C

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Suppose a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O). The total pressure of
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Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.

<h3>Definition of mole fraction</h3>

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:

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In this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).

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Learn more about mole fraction:

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