7. A 28.4 g sample of aluminum is heated to 39.4 °C, then is placed in a calorimeter
1 answer:
Answer:
The specific heat of aluminum is 0.214 calorie /gram °C
Explanation:
Specific heat:The heat requires to increase 1°C temperature of that object per unit gram.
When the aluminum comes in contact of of water then
The heat lose by aluminum = The heat gain by the water.
Q₁=Q₂
Where,
Q = m ×s× ΔT
m= mass of an object
s= specific heat of the object
ΔT= change of temperature.
The mass of aluminum(m₁)= 28.4 gram,
ΔT for aluminum=(initial temperature - final temperature)= (39.4-23)°C=16.4°C
s₁=?
The mass of water(m₂) = 50 gram
ΔT for water =( final temperature-initial temperature)=(23-21)°C= 2°C
S₂=1 calorie /gram °C
Q₁=Q₂
⇒m₁s₁ΔT=m₂s₂ΔT
⇒(28.4×s₁×16.4)= 50×1×2
calorie /gram °C
The specific heat of aluminum is 0.214 calorie /gram °C
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Answer : The pH of the solution is, 5.24
Explanation :
First we have to calculate the volume of
Formula used :
where,
are the initial molarity and volume of
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are the final molarity and volume of
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We are given:
Putting values in above equation, we get:
Now we have to calculate the total volume of solution.
Total volume of solution = Volume of + Volume of
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Total volume of solution = 1246.79 mL
Now we have to calculate the Concentration of salt.
Now we have to calculate the pH of the solution.
At equivalence point,
Thus, the pH of the solution is, 5.24