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Sam wished to investigate how fertilizer run-off affects the growth of algae in freshwater lakes and streams. He set his experim
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The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Hence, the volume of HCl neutralized is 1.25 mL
Answer:
38.3958 °C
Explanation:
As,
1 gram of carbohydrates on burning gives 4 kilocalories of energy
1 gram of protein on burning gives 4 kilocalories of energy
1 gram of fat on burning gives 9 kilocalories of energy
Thus,
27 g of fat on burning gives 9*27 = 243 kilocalories of energy
20 g of protein on burning gives 4*20 = 80 kilocalories of energy
48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories of energy
Total energy = 515 kilocalories
Using,
Given: Volume of water = 23 L = 23×10⁻³ m³
Density of water= 1000 kg/m³
So, mass of the water:
Mass of water = 23 kg
Initial temperature = 16°C
Specific heat of water = 0.9998 kcal/kg°C
Solving for final temperature as:
<u>Final temperature = 38.3958 °C </u>