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Finger [1]
3 years ago
8

Determine the enthalpy of the products of ideal combustion, i.e., no dissociation, resulting from the combustion of an isooctane

-air mixture
Chemistry
1 answer:
jeka943 years ago
8 0
The combustion of isooctane (C₈H₁₈) is written below:

2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (l)

The formula for heat of combustion is:

ΔHc = (∑Stoichiometric coefficient×ΔHf of products) - (∑Stoichiometric coefficient×ΔHf of reactants), where ΔHf is heat of formation.

ΔHf of isooctane = -259.2 kJ/mol
ΔHf of O₂ = 0 kJ/mol
ΔHf of CO₂ = -393.5 kJ/mol 
ΔHf of H₂O = <span>-285.8 kJ/mol
</span>
ΔHc = [(16 mol×-393.5 kJ/mol )+(18 mol×-285.8 kJ/mol)] - [(2 mol×-259.2 kJ/mol) + (25 mol*0 kJ/mol)]
ΔHc = -10,922 kJ
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The graph shows the solubility of several different salts in water, across a range of temperatures. According to the
sladkih [1.3K]

Answer:

  • <u><em>Sodium chloride</em></u>
  • See the graph attached

Explanation:

The attached graph with a green and a red arrow facilitates the understanding of this explanation.

To read the <em>solubility </em>on the <em>graph</em>, you can start with the temperature, on the x-axis.

The red vertical arrow shows how, departing from the <em>40ºC temperature</em> on the x-axis, you intersect the<em> solutibility curve </em>of sodium chloride at a height (y-axis) corresponding to <em>60 g/100cm³ of water</em> (follow the green horizontal arrow).

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4 0
3 years ago
Read 2 more answers
Can somewhere show me the dimensional analysis of this?
natita [175]

245 mm Hg = 32.6634 kPa

<u>Explanation:</u>

When the pressure inside the can is measured in mm Hg and it is needed to convert mm mercury (Hg) to kilo pascal, we have to multiply the pressure in mm Hg with  0.13332, so that the pressure is converted in kilo pascals.

1 mm Hg × 0.13332 = 1 kPa

245 mm Hg × 0.13332 = 32.6634 kPa

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6 0
3 years ago
How many molecules of H2O and O2 is present in 8.5g of H2O?​
Igoryamba

Answer:

2.8 x 10²³ molecules H₂O

1.4 x 10²³ molecules O₂

Explanation:

First, you will need the balanced chemical equation for the formation of water:

2H₂ + O₂ -> 2H₂O

This will help in determining the mole ratios between water and oxygen, which we will need later.

Let's first calculate the number of H₂O (water) molecules. This will require stoichiometry. We are also given the mass, so we must convert mass into moles, then moles into molecules. mass -> moles -> molecules

8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (6.02 x 10²³ molecules H₂O/1 mol H₂O) = 2.8404 x 10²³ molecules H₂O

Rounded to 2 significant digits: 2.8 x 10²³ molecules H₂O

Now, to find the molecules of water, we can begin with the same stoichiometric equation, but before we convert to molecules, we will have to convert moles of water to moles of oxygen. This is where we will use the mole ratio of water to oxygen we got from the balanced chemical equation earlier. 2H₂O:1O₂

8.5 g H₂O x (1 mol H₂O/18.01528 g H₂O) x (1 mol O₂/2 mol H₂O) x (6.02 x 10²³ molecules O₂/1 mol O₂) = 1.4202 x 10²³ molecules O₂

Rounded to 2 significant digits: 1.4 x 10²³ molecules O₂

3 0
3 years ago
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