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jeyben [28]
3 years ago
14

Do your answers to the previous questions satisfy

Chemistry
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

I put B&C

Explanation:

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How many moles are in 8.63 x 103 atoms of Li?
Ira Lisetskai [31]
<h3>Answer:</h3>

1.43 × 10⁻²⁰ mol Li

<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>Chemistry</u>

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

8.63 × 10³ atoms Li

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})
  2. Multiply/Divide:                \displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li

<u>Step 4: Check</u>

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

1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li

4 0
2 years ago
Someone please help me
ehidna [41]
I hop this will help u

8 0
2 years ago
During an investigation, similar glow sticks were placed in two beakers containing water at different temperatures. A record of
allochka39001 [22]

Answer:

i got you dawg just gimme one sec i'll get to you fr g

Explanation:

6 0
3 years ago
Read 2 more answers
What are the similarities and differences between a camera and a camera and the human eye?
Sunny_sXe [5.5K]
So a camera can zoom in and out and the human eye has peripheral vision
4 0
3 years ago
Read 2 more answers
a sample of an unknown gas has a mass of 0.582g. it's volume is 21.3 mL a a temp of 100 degrees Celsius and a pressure of 754 mm
Zielflug [23.3K]

Answer:

Molar mass = 0.09 × 10⁴ g/mol

Explanation:

Given data:

Mass = 0.582 g

Volume = 21.3 mL

Temperature = 100°C

Pressure = 754 mmHg

Molar mass = ?

Solution:

(21.3 /1000 = 0.0213 L)

(100+273= 373 K)

(754/760 = 0.99 atm)

PV = nRT

n = PV/RT

n = 0.99 atm × 0.0213 L / 0.0821 atm. L. mol⁻¹. k⁻¹ × 373 K

n =0.02 mol/ 30.6

n = 6.5 × 10⁻⁴ mol

Molar mass = Mass/ number of moles

Molar mass = 0.582 g / 6.5 × 10⁻⁴ mol

Molar mass = 0.09 × 10⁴ g/mol

6 0
3 years ago
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