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3 years ago

Convert 6.7 x 1024 molecules of nitrogen dioxide into grams.

Chemistry

1 answer:

BlackZzzverrR [31]3 years ago

6 0

Answer:

510 g NO₂

General Formulas and Concepts:

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Reading the Periodic Table
Writing Compounds
Using Dimensional Analysis

Explanation:

Step 1: Define

6.7 × 10²⁴ molecules NO₂ (Nitrogen dioxide)

Step 2: Define conversions

Avogadro's Number

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of NO₂ - 14.01 + 2(16.00) = 46.01 g/mol

Step 3: Use Dimensional Analysis

$6.7 \cdot 10^{24} \ molecules \ NO_2(\frac{1 \ mol \ NO_2}{6.022 \cdot 10^{23} \ molecules \ NO_2} )(\frac{46.01 \ g \ NO_2}{1 \ mol \ NO_2} )$ = 511.901 g NO₂

Step 4: Check

We are given 2 sig figs. Follow sig fig rules.

511.901 g NO₂ ≈ 510 g NO₂

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The following information was provided by the question:

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To solve this problem, we need to understand what happens when NaOH and HCl react. A neutralization reaction occurs between a strong acid, such as HCl, and a strong base, such as NaOH, where the amount of H+ and OH- ions in solution are equal. We can write their reaction as:

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When the base is completely neutralized by the acid, it means that:

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The equality above is what we'll use to start our calculations.

Another important information to this question is that, by definition, the molar concentration is the number of moles of a compound divided by the volume of the solution:

$\text{molarity = }\frac{number\text{ of moles (mol)}}{\text{volume (L)}}\to\text{ M=}\frac{n}{V}$

Since we have the equality between the number of moles of acid and base, we can rewrite the equation above as:

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and use this to calculate the molar concentration (M) of NaOH.

Thus, so far we have that:

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Since the volume of NaOH, molarity of HCl and volume of HCl were provided by the question, we can rearrange the equation above to calculate the molarity of NaOH:

$M_{NaOH}=\frac{M_{HCl}\times V_{HCl}}{V_{NaOH}}$

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$M_{NaOH}=\frac{(0.950\text{ mol/L)}\times(80.0\text{ mL)}}{(50.0\text{ mL)}}=1.52\text{ mol/L}$

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