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crimeas [40]
2 years ago
10

A student prepares a weak acid solution by dissolving 0.2000 g HZ to give 100. mL solution. The titration requires 33.5 mL of 0.

1025 M NaOH. Calculate the molar mass of the acid. WebAssign will check your answer for the correct number of significant figures. g/mol (a) Would the molar mass be too high, too low, or unaffected if the student accidentally used 0.1000 g in the calculation
Chemistry
1 answer:
nexus9112 [7]2 years ago
3 0

Answer:

a)M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88 g/mol

b) if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94  hence the molar mass will be too low

Explanation:

0.2000 gHZ gives  100ml acid solution

33.5 ml of 0.1025 M NaOH  is required to prepare it

the moles = mass / molar mass

mass = 0.200 gHZ

moles = 0.0335*100 * 0.1025 = 0.034

therefore molar mass = mass / moles

                                  M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88

if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94  hence the molar mass will be too low

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Answer:

The % yield of CO2 is 85.05 %

Explanation:

Step 1: Data given

Mass of octane = 702 grams

Molar mass octane = 114.23 g/mol

Mass CO2 =1.84 kg = 1840 grams

Molar mass of CO2

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles of octane

Moles octane = mass octane / molar mass octane

Moles octane = 702.0 grams / 114.23 g/mol

Moles octane = 6.145 moles

Step 4: Calculate moles of CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 6.145 moles octane we'll have 8*6.145 moles =49.16 moles

Step 5: Calculate mass of CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 49.16 moles * 44.01 g/mol

Mass CO2 = 2163.5 grams

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3 years ago
Which of the following terms describes a piece of pure gold?
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Explanation:

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<h3>Answer:</h3>

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<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

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<u>Stoichiometry</u>

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[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

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<u>Step 3: Stoich</u>

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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

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