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Anastasy [175]
3 years ago
14

How many moles in 4.93 x 10E23 atoms of silver?

Chemistry
1 answer:
leva [86]3 years ago
8 0
<h3>Answer:</h3>

0.819 mol Ag

<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>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

4.93 × 10²³ atoms Ag

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 4.93 \cdot 10^{23} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})
  2. Divide:                              \displaystyle 0.818665 \ mol \ Ag

<u>Step 4: Check</u>

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

0.818665 mol Ag ≈ 0.819 mol Ag

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lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

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Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

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Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

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Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

5 0
3 years ago
If 3.0 moles of hydrogen react, how many grams of nitrogen will be used? Round your answer to the nearest whole number. N2 + 3H2
Iteru [2.4K]

Answer:

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Explanation:

Given parameters:

Number of moles of hydrogen  = 3moles

Unknown:

Mass of nitrogen used = ?

Solution:

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From the balanced reaction equation;

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Now,

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         Molar mass of N₂  = 2(14) = 28g/mol

   Mass of Nitrogen  = 1 mole x 28g/mole  = 28g

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