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choli [55]
3 years ago
5

Air has a mass of 1.2g and a volume of 4,555ml. What is the density

Chemistry
1 answer:
coldgirl [10]3 years ago
3 0

Answer:

<h2>Density = 0.00026 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>Density(\rho) =  \frac{mass}{volume}</h3>

From the question

mass of air = 1.2 g

volume = 4,555 mL

Substitute the values into the above formula and solve for the density

That's

<h3>Density =  \frac{1.2}{4555}</h3>

= 0.0002634

We have the final answer as

<h3>Density = 0.00026 g/mL</h3>

Hope this helps you

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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

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1) To find the change in enthalpy, determine the difference between the potential energy of the products and the potential energy of the reactants. (on this diagram, C-A) To find the activation energy, find the difference between the potential energy of the reactants and the "peak" of the curve (on this diagram, B-A). For this diagram, both the enthalpy and activation energy are positive.

2) If the reaction was exothermic, enthalpy would be negative, and the potential energy of the reactants would be greater than the potential energy of the products.

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