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Natasha_Volkova [10]
3 years ago
8

Determine the number of grams of carbon 3.14x 1023 atoms of carbon

Chemistry
2 answers:
tiny-mole [99]3 years ago
7 0
<h3>Answer:</h3>

6.26 g C

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<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] 3.14 × 10²³ atoms C

[Solve] grams C

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

Avogadro's Number

[PT] Molar Mass of C - 12.01 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 3.14 \cdot 10^{23} \ atoms \ C(\frac{1 \ mol \ C}{6.022 \cdot 10^{23} \ atoms \ C})(\frac{12.01 \ g \ C}{1 \ mol \ C})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 6.26227 \ g \ C

<u>Step 4: Check</u>

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

6.26227 g C ≈ 6.26 g C

pishuonlain [190]3 years ago
5 0
I think it would be 6.26 g
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3 years ago
A mixture of carbon dioxide and helium gases is maintained in a 7.91 L flask at a pressure of 1.42 atm and a temperature of 33 °
Crank

Answer:

The gas mixture contains 1.038 grams of helium

Explanation:

<u>Step 1:</u> Data given

Volume of the flask = 7.91 L

Total Pressure = 1.42 atm

Temperature = 33 °C

Mass of CO2 = 8.25 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of He = 4 g/mol

<u>Step 2</u>: Calculate total number moles of gas

p*V = n*R*T

⇒ p = the pressure = 1.42 atm

⇒ V = the volume = 7.91 L

⇒ n= the number of moles = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L* atm/K*mol

⇒ T = the temperature = 33 °C = 306 Kelvin

n = (p*V)/(R*T)

n = (1.42*7.91)/(0.08206 * 306)

n = 0.447 moles

<u>Step 3</u>: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 8.25 grams / 44.01 g/mol

Moles CO2 = 0.1875 moles

<u>Step 4:</u> Calculate moles of Helium

Moles Helium = total moles of gas - moles of CO2

Moles Helium = 0.447 - 0.1875 = 0.2595 moles of helium

<u>Step 5: </u> Calculate mass of helium

Mass of helium = moles of helium * molar mass of helium

Mass of helium = 0.2595 moles * 4 g/mol

Mass of helium = 1.038 grams

The gas mixture contains 1.038 grams of helium

8 0
3 years ago
An element has ccp packing with a face-centered cubic unit cell. its density is 8920 kg/m3 and the unit cell volume is 4.72 x 10
ziro4ka [17]
<span>63.4 g/mol
   First, let's determine how many atoms per unit cell in face-centered cubic. There is 8 corners, each of which has 1 atom, and each of those atoms is shared between 8 other unit cells. So 8*1/8 = 1 atom per unit cell. Additionally, there are 6 faces, each of which has 1 atom that's shared between 2 unit cells. So 6*1/2 = 3 atoms per unit cell. So each unit cell has the mass of 1+3 = 4 atoms. Since there is 1000 liters per cubic meter, the mass per liter is 8920 kg/1000 = 8.920 kg/L. Now the mass per unit cell is 8920 g * 4.72x10^-26 = 4.21024x10^-22 g per unit cell. The mass per atom is 4.21024x10^-22 g / 4 = 1.05256x10^-22 g/atom, Finally, multiply by Avogadro's number, getting 1.05256x10^-22 g/atom * 6.0221409x10^23 atom/mol = 63.38664625704 g/mol.
   Rounding to 3 significant digits gives 63.4 g/mol.</span>
5 0
3 years ago
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