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3 years ago

Assuming that each cubic centimeter of water has a mass of exactly 1 g, find the mass of 1.27 cubic meter of water.

Chemistry

1 answer:

iren [92.7K]3 years ago

7 0

Answer:

1270g or 1.27kg

Explanation:

This is a case of conversions. Firstly, we will need to know the amount or number of cubic centimeters in one cubic meter. By conversion, we have 1000 of such.

This practically means before we could form one cubic meter of water, 1000 units of water having a volume of one cubic centimeter each would have come together.

Now, we know that one cubic centimeter has a mass of 1g, this means 1000 cubic centimeter would have a mass of 1000 * 1 = 1000g or 1kg

Hence 1.27 cubic meter of water will thus have a mass of 1.27 * 1000g = 1270g or simply 1.27kg

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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

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