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3 years ago

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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How many atoms is 50 g of He?

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

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3 years ago

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]

1 half-life - 4.5 days
x half-lives - 27 days

x=(1*27)/4.5=6
6 <span>half-lives have elapsed after 27 days.</span>

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3 years ago

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What is the lewis dot structure of phosphorus decasulfide

Gnoma [55]

Https://www.youtube.com/watch?v=7ZtLaCucV98

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3 years ago

A flexible container holds 32.4L of gas at 55.0c. What will be the volume if the temperature falls to 17.0c

mash [69]

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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>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
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>

Set up: $\displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})$
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