Answer:- 335 kcal of heat energy is produced.
Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:
From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.
We could solve this using dimensional analysis as:
= 1401.5 kJ
Now, let's convert kJ to kcal.
We know that, 1kcal = 4.184kJ
So, 
= 335 kcal
Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.
This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:
Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰
Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻
Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.
In such a way, the correct choice is C.
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Fun FACT : An apple, potato, and onion all taste the same if you eat them with your nose plugged
Answer:
Dispersion Forces are found between n-Pentane (CH₃-CH₂-CH₂-CH₂-CH₃) and n-Hexane (CH₃-CH₂-CH₂-CH₂-CH₂-CH₃).
Explanation:
Dispersion Forces are present and developed by those compounds which are non-polar in nature. In given statement n-Pentane and n-Hexane both are non-polar in nature as the electronegativity difference between Hydrogen atoms and Carbon atoms is less than 0.4.
When non-polar molecules approaches each other, a Dipole is induced in one of them, this step is known as Instantaneous Dipole, This generated Dipole on approaching another non-polar molecule induces dipole in it and the process propagates. Hence, creating intermolecular interactions.
Answer:
0.0308 mol
Explanation:
In order to convert from grams of any given substance to moles, we need to use its molar mass:
- Molar mass of KAI(SO₂)₂ = MM of K + MM of Al + (MM of S + 2*MM of O)*2
- Molar mass of KAI(SO₂)₂ = 194 g/mol
Now we <u>calculate the number of moles of KAI(SO₂)₂ contained in 5.98 g</u>:
- 5.98 g ÷ 194 g/mol = 0.0308 mol
