Answer:
di ko po Alam sorry
Explanation:
sorry sorry sorry sorry sorry
Answer:
76.5g KCl/74.55 grams per mole Kcl = x
molality= x/.085 kg H2O
Explanation:
well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.
Hbro dissociate as follows
HBro---> H+ + BrO-
Ka= (H+)(BrO-) / HBro
PH = -log (H+)
therefore (H+) = 10^-4.48= 3.31 x10^-5
ka is therefore= ( 3.31 x 10^-5)^2/0.55=1.99 x10^-9
Answer:
When the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy
Explanation:
The answer that "the entropy will is increases" is correct as:
The water at 90° C i.e at higher temperature is mixed with the water at 10° C i.e the water at the lower temperature.
The water at lower temperature will have molecules with lower energy while the water with higher temperature will have molecules undergoing high thermal collisions. Thereby, when the water is mixed with water at lower temperature the effective temperature of the system (i.e the water at lower temperature) will increase, thereby increasing it's entropy.
Therefore, the answer is correct with respect to the water at lower temperature.
Meanwhile, for the water at higher temperature , the temperature of the system will decrease. Thus, the entropy of the water at higher level will decrease.
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>
