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Sonbull [250]
3 years ago
5

what is molarity of a sodium hydroxide solution made by combining 2.0 L of 0.60 M NaOH with 495 mL of 3.0 M NaOH? Assume the vol

umes of the two solutions to be additive
Chemistry
1 answer:
hammer [34]3 years ago
8 0

Answer:

Molarity of the sodium hydroxide solution is 1.443 M/L

Explanation:

Given;

0.60 M concentration of NaOH contains 2.0 L

3.0 M concentration of NaOH contains 495 mL

Molarity is given as concentration of the solute per liters of the solvent.

If the volumes of the two solutions are additive, then;

the total volume of NaOH = 2 L + 0.495 L = 2.495 L

the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M

Molarity of NaOH solution = 3.6 / 2.495

Molarity of NaOH solution = 1.443 M/L

Therefore, molarity of the sodium hydroxide solution is 1.443 M/L

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When 2.50 mol of mg3n2 are allowed to react how many moles of h2o also react?
spin [16.1K]

Notice that in the equation given, there is 1 mol of Mg3N2 and 6 mol of H2O. Consequently, the question is requesting, if we took 2.5 mol of Mg3N2, how many mol of H2O would be essential to complete a similar reaction -- in quintessence, multiplying everything by 2.5: 

the solution would be: 6 * 2.5 = 15 mol

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4 years ago
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astraxan [27]
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3 years ago
2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2
miv72 [106K]
<h3>Answer:</h3>

4 g AgCl

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN]   2AgNO₃ + BaCl₂ → 2AgCl + Ba(NO₃)₂

[Given]   5.0 g AgNO₃

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

[Reaction - Stoich] 2AgNO₃ → 2AgCl

Molar Mass of Ag - 107.87 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

Molar Mass of AgCl - 107.87 + 35.45 = 143.32 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                              \displaystyle 5.0 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{2 \ mol \ AgCl}{2 \ mol \ AgNO_3})(\frac{143.22 \ g \ AgCl}{1 \ mol \ AgCl})
  2. Multiply/Divide:                                                                                                  \displaystyle 4.21533 \ g \ AgCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

4.21533 g AgCl ≈ 4 g AgCl

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