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2 years ago

Match the following terms:

Chemistry

1 answer:

AURORKA [14]2 years ago

3 0

Answer:

See below ~

Explanation:

A to B

⇒ Solid Phase

==========================================================

B to C

⇒ Melting

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C to D

⇒ Liquid Phase

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D to E

⇒ Vaporizing

============================================================

E to F and beyond

⇒ Gas Phase

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What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N

Citrus2011 [14]

<h3>Answer:</h3>

1.69 g Mg₃N₂

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis
Reactions RxN

<h3>Explanation:</h3>

Step 1: Define

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

Step 2: Identify Conversions

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

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

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

Step 3: Stoich

[DA] Set up: $\displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 1.68873 \ g \ Mg_3N_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

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