Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
Answer:
His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.
Explanation:
The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.
Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.
Suppose the mass of the solute is m.
Originally, the density is = 
Now after adding extra 10 mL , the density becomes 
.
Therefore, 
So the density decreases when we add more solution.
Answer:
The correct answer is -
1. a) The bubbles will shrink, some may vanish.
2. a) Can A will make a louder and stronger fizz than can B.
Explanation:
In the first question, it is given that the bottle is not opened and therefore, squeezing the bottle filled with a carbonated drink will increase the pressure on the carbonated liquid which forces the bubbles to dissolve or displace or vanish as it moves to empty space.
Thus, the correct answer would be - The bubbles will shrink, some may vanish
In the second question, there are two different conditions for two different unopened cans of carbonated water that are different temperatures one at the garage with higher temperature and one in the fridge at low temperature. As it is known that higher the temperature less will be solubility of gas in liquid so gas in can A will be less soluble which means it has more gas and it will make louder and stronger fizz than B which was stored at low temperature.
thus, the correct answer would be - Can A will make a louder and stronger fizz than can B.
