Answer:
Explanation:
No not always but it is more likely they will
20600Cal
Explanation:
Given parameters:
Mass of water = 319.5g
Initial temperature = 35.7°C
Final temperature = 100°C
Unknown:
Calories needed to heat the water = ?
Solution:
The calories is the amount of heat added to the water. This can be determined using;
H = m c Ф
c = specific heat capacity of water = 4.186J/g°C
H is the amount of heat
Ф is the change in temperature
H = m c (Ф₂ - Ф₁)
H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J
Now;
1kilocalorie = 4184J
85996.56J to kCal;
= 20.6kCal = 20600Cal
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Answer : The correct option is, (C) 0.675 M
Explanation :
Using neutralization law,

where,
= concentration of
= 13.5 M
= concentration of diluted solution = ?
= volume of
= 25.0 ml = 0.0250 L
conversion used : (1 L = 1000 mL)
= volume of diluted solution = 0.500 L
Now put all the given values in the above law, we get the concentration of the diluted solution.


Therefore, the concentration of the diluted solution is 0.675 M
The hours taken for concentration to decrease from 0 to 74 min. to 0.21 m is 91.7 hours.
<h3>What is the rate law of a reaction?</h3>
Rate law depicts the rate of a chemical reaction depend on the concentration of the reactant.
The given reaction is second order reaction
Thus, the hours taken for concentration to decrease from 0 to 74 min. to 0.21 m is 91.7 hours.
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