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ohaa [14]
3 years ago
12

A constant volume calorimeter (bomb calorimeter) was calibrated by performing in it a reaction in which 5.23 kJ of heat energy w

as released, causing the calorimeter to rise by 7.33 °C. What is the heat capacity C of the calorimeter?
Chemistry
1 answer:
Serggg [28]3 years ago
5 0

Explanation:

Relation between heat energy and specific heat is as follows.

            q_{bomb} = c \times \Delta t

or,            c = \frac{q_{bomb}}{\Delta t}

where,     c = specific heat

           q_{bomb} = heat energy

           \Delta t = change in temperature

Putting the given values into the above formula we will calculate the specific heat as follows.

                c = \frac{q_{bomb}}{\Delta t}

                   = \frac{5.23 kJ}{7.33^{o}C}

                   = 0.713 kJ^{o}C

Thus, we can conclude that heat capacity C of the calorimeter is 0.713 kJ^{o}C.

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Answer:

An Arrhenius Base

Explanation:

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8 0
3 years ago
The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis
Dimas [21]
The solution is as follows:

K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106

The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol

Use the ICE approach

        borneol  →  isoborneol
I         0.0972           0.0486
C         -x                     +x
E     0.0972 - x        0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832 

Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
6 0
4 years ago
If we use 18 moles of H2 how many moles of H2O do we make? Use this chemical equation: 6H2 + O2 → 3H2O
Ilya [14]
<h3>Answer:</h3>

9 mol H₂O

<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
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<u>Atomic Structure</u>  

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<u>Stoichiometry</u>  

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 6H₂ + O₂ → 3H₂O

[Given] 18 mol H₂

[Solve] mol H₂O

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

[RxN] 6 mol H₂ → 3 mol H₂O

<u>Step 3: Stoich</u>

  1. [DA] Set up conversion:                                                                               \displaystyle 18 \ mol \ H_2(\frac{3 \ mol \ H_2O}{6 \ mol \ H_2})
  2. [DA] Simplify:                                                                                                 \displaystyle 18 \ mol \ H_2(\frac{1 \ mol \ H_2O}{2 \ mol \ H_2})
  3. [DA] Divide [Cancel out units]:                                                                     \displaystyle 9 \ mol \ H_2O
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