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.
70.306 would be your answer.
A compound is a pure substance formed by the chemical combination of two or more different elements.
A compound may be splitted into simpler substances by chemical reactions, and has different properties to those of the elements that form it.
The composition of a compound is fixed: every piece of a compound has the same kind of atoms, bonded in the same way and proportion.
Some examples of compounds are H₂O, NaCl, H₂O₂, CH₃COOH. As you see, they have a chemical formula which states the kind and number of the atoms that form them.
They are different to mixtures, which are formed by two or more compounds, in a variable proportion, and can be separated by physical media. Some examples of mixtures are the solutions (e.g. NaCl dissolved in H₂O), and some solid mixtures (e.g. a mixture of marbles and sand).
7.86 is the pOH of water at this temperature of 100 degrees celsius.
Option E is the right answer.
Explanation:
Data given:
Kw = 51.3 x 
pOH = ?
we know that pure water is neutral and will have pH pf 7.
The equation for relation between Kw and H+ and OH- ion is given by:
Kw = [H+] [OH-}
here the concentration of H+ ion and OH- ion is equal
so, [H+]= [OH-]
Putting the values in the equation of Kw
pKw = -log[Kw]
pKw = -log [51.3 x ]
pKw = 12.28
since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14
Now, pOH is calculated by using the equation:
14 = pOH + pH
14- 6.14 = pOH
pOH = 7.86
The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol / 0.250 mol/L = 40.0 mL
40.0 mL of 0.250 M HCl is required
