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Ad libitum [116K]
3 years ago
11

How many atoms are in a 158.69 g sample of nickel?

Chemistry
1 answer:
Alchen [17]3 years ago
8 0
<h3>Answer:</h3>

1.6283 × 10²⁴ atoms Ni

<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
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

158.69 g Ni

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

Avogadro's Number

Molar Mass of Ni - 58.69 g/mol

<u>Step 3: Convert</u>

  1. Set up:                      \displaystyle 158.69 \ g \ Ni(\frac{1 \ mol \ Ni}{58.69 \ g \ Ni})(\frac{6.022 \cdot 10^{23} \ atoms \ Ni}{1 \ mol \ Ni})
  2. Multiply/Divide:        \displaystyle 1.62827 \cdot 10^{24} \ atoms \ Ni

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

1.62827 × 10²⁴ atoms Ni ≈ 1.6283 × 10²⁴ atoms Ni

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3 years ago
Which of the following elements would require the most energy to
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<h3>What does the term “ionization energy” mean? </h3>

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8 0
1 year ago
After performing a dilution calculation, you determine you need 25.0 milliliters of an aqueous stock solution to make 100.0 mill
IRINA_888 [86]
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.

D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.

C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.

B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.

I would add the final step of stirring, but B is the only answer that can be correct.
8 0
3 years ago
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In the reaction of 6 moles of CH4 with excess oxygen how many moles of water could be produced
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6 0
3 years ago
When 231. mg of a certain molecular compound X are dissolved in 65. g of benzene (CH), the freezing point of the solution is mea
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Answer:

Molar mass X = 18.2 g/mol

Explanation:

Step 1: Data given

Mass of compound X = 231 mg = 0.231 grams

Mass of benzene = 65.0 grams

The freezing point of the solution is measured to be 4.5 °C.

Freezing point of pure benzene = 5.5 °C

The freezing point constant for benzene is 5.12 °C/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C

⇒i = the Van't hoff factor = 1

⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m

⇒m = the molality = moles X / mass benzene

m = 1.0 / 5.12 °C/m

m = 0.1953 molal

Step 3: Calculate moles X

Moles X = molality * mass benzene

Moles X = 0.1953 molal * 0.065 kg

Moles X = 0.0127 moles

Step 4: Calculate molar mass X

Molar mass X = mass / moles

Molar mass X = 0.231 grams / 0.0127 moles

Molar mass X = 18.2 g/mol

5 0
3 years ago
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