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vichka [17]
3 years ago
11

D Which compound is insoluble in water?

Chemistry
1 answer:
densk [106]3 years ago
3 0

Answer:

Examples of substances insoluble in water: oil, acetone, ether

Explanation:

Such examples of substances are non-polar and do not dissolve in water (polar compound). The classic example is oil floating in water (this happens because oil has a lower density than water).

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tigry1 [53]
Atomic <span>mass He = 4.002 u.m.a

4.002 g --------------- 6.02x10</span>²³ atoms
50 g ------------------ ? atoms

50 x ( 6.02x10²³) / 4.002

= 3.01x10²⁵ / 4.002

= 7.52x10²⁴ atoms of He
6 0
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The half-life for the radioactive decay of calcium-47 is 4.5 days. how many half-lives have elapsed after 27 days?
Evgesh-ka [11]
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x half-lives   - 27 days

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6 <span>half-lives have elapsed after 27 days.</span>
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A flexible container holds 32.4L of gas at 55.0c. What will be the volume if the temperature falls to 17.0c
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5 0
3 years ago
How many molecules are in 85g of silver nitrate?
maksim [4K]
<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<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
  • Writing Compounds
  • 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>

85 g AgNO₃ (silver nitrate)

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

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

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

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

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})
  2. Multiply/Divide:                                                                                                \displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

6 0
2 years ago
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