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Chemistry

2 answers:

Tcecarenko [31]3 years ago

7 0

The first question should be B as the correct answer

Bad White [126]3 years ago

3 0

The answers to your questions:

1: B.

2: True

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A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.

Elodia [21]

Answer:

<em>Hydration number:</em> 4

Explanation:

<u>1) Mass of water in the hydrated compound</u>

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

<u>2) Number of moles of water</u>

Number of moles = mass in grams / molar mass

molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>

The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).

Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol

<u>4) Ratio</u>

0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

4 0

3 years ago

Read 2 more answers

4. How many moles of gold (Au) are in a pure gold nugget having a mass of 25.0 grams.

ikadub [295]

<h3>Answer:</h3>

0.127 mol Au

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

<u>Math</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 25.0 g Au

[Solve] moles Au

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

[PT] Molar Mass of Au - 196.97 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 25.0 \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
[DA] Multiply/Divide [Cancel out units: $\displaystyle 0.126923 \ mol \ Au$