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3 years ago

Determine the number of grams of sodium carbonate needed to prepare 100.0 ml of a 2.5 m solution

Chemistry

1 answer:

SCORPION-xisa [38]3 years ago

6 0

Answer:

26.5 g

Explanation:

First we convert 100.0 mL to L:

100.0 mL / 1000 = 0.100 L

Now we calculate how many moles of sodium carbonate are needed, using the definition of molarity:

Molarity = moles / liters

moles = molarity * liters

2.5 M * 0.100 L = 0.25 mol

Finally we convert 0.25 moles of sodium carbonate into grams, using its molar mass:

0.25 mol * 106 g/mol = 26.5 g

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Explanation:

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The process in which Volume remains constant is called isochoric process.

The process in which pressure remains constant is called isobaric process.

The process in which no heat is gained or lost by the system, Hence heat remains constant is called as adiabatic process.

Thus option d) heat is correct

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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>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table
Writing Compounds
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

85 g AgNO₃ (silver nitrate)

Step 2: Identify Conversions

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

Step 3: Convert

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: $\displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

6 0

3 years ago