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DENIUS [597]
3 years ago
15

If it takes 38.70cm³ of 1.90 M NaOH to neutralize 10.30cm³ of H2SO4 in a battery, What's the Molarity of the H2SO4 ?​

Chemistry
2 answers:
meriva3 years ago
8 0

Answer:

Volume of acid Va = 38.70cm3,

Concentration of acid Ca = 1.90M

Volume of Base Vb = 10.30cm3

Concentration of base Cb = ?

Equation of the reaction

H2SO4 + 2NaOH = Na2SO4 + 2H2O

No of mole of acid Na = 1

No of mole of base Nb = 2

CaVa/CbVb = Na/Nb

Cb = CaVaNb/VbNa

Cb = 1.9*38.7*2/10.30*1

Cb = 14.2777M

Explanation:

arlik [135]3 years ago
3 0

Answer: 3.57M

Explanation:Please see attachment for explanation

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<h3>Answer:</h3>

7.3 × 10⁻⁷ g Ni

<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
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<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

7.5 × 10¹⁵ atoms Ni

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

Avogadro's Number

Molar Mass of Ni - 58.69 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 7.5 \cdot 10^{15} \ atoms \ Ni(\frac{1 \ mol \ Ni}{6.022 \cdot 10^{23} \ atoms \ Ni})(\frac{58.69 \ g \ Ni}{1 \ mol \ Ni})
  2. Multiply:                                                                                                           \displaystyle 7.30945 \cdot 10^{-7} \ g \ Ni

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

7.30945 × 10⁻⁷ g Ni ≈ 7.3 × 10⁻⁷ g Ni

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victus00 [196]

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Answer:

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Explanation:

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