Sulfuric acid is the answer
The food raises the temperature of the calorimeter by 1.29 °C. The amount of calorie produced by the combustion would be: 1.29°C * <span> 6.15 kJ/°C= 7.9335 kJ.
If the food weight is </span>0.521g, then the food value would be:
7.9335 kJ/ 0.521g / (4.184kcal/kJ) =
(15.227 kJ/g) / (4.184kcal/kJ)= 3.64 kilo calorie/gram
If I understand the question correctly, I think the answer is true. My reasoning is that each formula for the molecules used in a reaction are set and can't be modified to make it easier to balance or else the reaction would change fundamentally.
an example of this is H₂+1/2O₂⇒H₂O. You can't change the H₂O to H₂O₂ to make the reaction cleaner. Water and hydrogen peroxide are two different chemicals and therefore can't be interchanged with out changing the fundamental qualities of the reaction
Answer:
28.0mL of the 0.0500M NaOH solution
Explanation:
<em>0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.</em>
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The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:
H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O
<em>Where 1 mole of the acid reacts per mole of the base.</em>
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You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.
the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:
0.126g ₓ (1mol / 90.08g) = <em>1.40x10⁻³ moles of H₃C-CH(OH)-COOH</em>
To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:
1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =
<h3>28.0mL of the 0.0500M NaOH solution</h3>