Answer:
0.500 mol/dm³
Explanation:
Using the formula below;
CaVa = CbVb
Where;
Ca = concentration of acid (mol/dm³)
Cb = concentration of base (mol/dm³)
Va = volume of acid (cm³)
Vb = volume of base (cm³)
In accordance to the information provided in this question is;
Va = 5cm³
Vb = 250 cm³
Ca = 12 mol/dm³
Cb = ?
Using CaVa = CbVb
12 × 5 = Cb × 250
60 = 120Cb
Cb = 60/120
Cb = 0.500 mol/dm³
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.
Answer:
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Answer:
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