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ELEN [110]
3 years ago
10

What is the mass of 1.24 x 10^23 molecules of N2O2

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
7 0
<h3>Answer:</h3>

12.4 g N₂O₂

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.24 × 10²³ molecules N₂O₂

[Solve] grams N₂O₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of N₂O₂ - 2(14.01) + 2(16.00) = 60.02 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                        \displaystyle 1.24 \cdot 10^{23} \ molecules \ N_2O_2(\frac{1 \ mol \ N_2O_2}{6.022 \cdot 10^{23} \ molecules \ N_2O_2})(\frac{60.02 \ g \ N_2O_2}{1 \ mol \ N_2O_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                          \displaystyle 12.3588 \ g \ N_2O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

12.3588 g N₂O₂ ≈ 12.4 g N₂O₂

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The bigger the atomic radius the easier it is to oxidise the atom. Remember that an atom is oxidized by the loss of an electron.

Explanation:

The bigger the atomic radius the further away the valence electron are from the attractive force of the atomic nucleus. This means that the energy required to remove an electron from the valence shell is easier compared to an atom with a smaller atomic radius. This is because you need to overcome the attractive force of the nucleus on the electron for you to oxidize the atom.

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3 years ago
How many atoms of each element are present in the compound
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THIS IS A THREE PART QUESTION IF YOU CAN HELP IT WOULD BE REALLY APPRECIATED SO I DONT FAIL.
Nostrana [21]

Answer:

1. B = 1.13M

2. A. 8.46%

3. D = 0.0199

Explanation:

1. Molarity of a a solution = number of moles of solute/ volume of solution in L

Number of moles of solute = mass of solute/molar mass of solute

Molar mass of NaC₂H₃O₂ = 82 g/mol, mass of NaC₂H₃O₂ = 245 g

Number of moles of NaC₂H₃O₂ = 245 g / 82 g/mol = 2.988 moles

Molarity of solution = 2.988 mols/ 2.65 L = 1.13 M

2. Percentage by mass of a substance = mass of substance /mass of solution × 100%

Mass of 2.65 L of water = density × volume

Density of water = 1 Kg/L = 1000 g/L; volume of waterb= 2.65 L

Mass of water = 1000 g/L × 2.65 L = 2650 g

Mass of solution = mass of water + mass of solute = 245 + 2650 =2895 g

Percentage by mass of NaC₂H₃O₂ = 245/2895 × 100% = 8.46%

3. Mole fraction of NaC₂H₃O₂ = moles of NaC₂H₃O₂/moles of solution

Moles of water = mass /molar mass

Mass of water = 2650 g; molar mass of water = 18 g/mol

Moles of water = 2650 g / 18 g/mol = 147.222 moles

Moles of solution = moles of solute + moles of water = 147.222 + 2.988 = 150.21

Moles of NaC₂H₃O₂ =2.988 moles

Moles fraction of NaC₂H₃O₂ =2.988/150.21 = 0.0199

6 0
3 years ago
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
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Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

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Answer:

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Explanation:

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