Please someone solve this and tell me how you solve it

Chemistry

1 answer:

lianna [129]3 years ago

8 0

Answer:

Supersaturated.

Explanation:

Hello there!

In this case, according to this solubility chart, we infer that for NH3, the solubility starts at 90 grams of NH3 that are soluble in 100 g of water at 0 °C and ends in about 8 g in 100 g of water at 100 °C for a saturated solution.

However, since we are asked for the solubility of NH3 at 20 °C, we can see that, according to the table and the curve for NH3, about 52 g of NH3 are soluble in 100 g of water; thus, for the given 60 g of NH3, we will say that 8 grams will remain undissolved, and therefore, this solution will be supersaturated.

Regards!

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3 years ago

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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>

<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

<u>Stoichiometry</u>

Using Dimensional Analysis
Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

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

[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

<u>Step 3: Stoich</u>

[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]: $\displaystyle 1.68873 \ g \ Mg_3N_2$