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Umnica [9.8K]
3 years ago
13

In a coffee-cup calorimeter, 50.0ml of 0.100M AgNO3 and50.0ml of 0.100M HCl are mixed to yield the following reaction:

Chemistry
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

-66.88KJ/mol

Explanation:

It is possible to obtain the heat involved in a reaction using a calorimeter. Formula is:

q = -C×m×ΔT

<em>Where q is heat of reaction, C is specific heat capacity (4.18J/°Cg), m is mass of solution (100.0g) and ΔT is temperature change (23.40°C-22.60°C = 0.80°C)</em>

Replacing:

q = -4.18J/°Cg×100.0g×0.80°C

q = -334.4J

Now, in the reaction:

Ag⁺ + Cl⁻→ AgCl

<em>AgNO₃ as source of Ag⁺ and HCl as source of Cl⁻</em>

Moles that react are:

0.050L× (0.100mol /L) = 0.0050moles

If 0.0050 moles produce -334.4J. Heat of reaction is:

-334.4J / 0.0050moles = -66880J/mol = <em>-66.88KJ/mol</em>

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Answer:

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Solution:

Data Given:

                 Mass of MgS₂O₃  =  193 g

                 M.Mass of MgS₂O₃  =  136.43 g.mol⁻¹

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Formula Used:

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Putting values,

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3 years ago
How does Gregor Mendel's work with pea plants affect human life today? OA A. People continue to cross pea plants to fix Mendel's
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CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?
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<h3>Answer:</h3>

64 g O₂

<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>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   36 g H₂O

[Solve]   x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

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<u>Step 4: Check</u>

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

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