Mass percentage of a solution is the amount of solute present in 100 g of the solution.
Given data:
Mass of solute H2SO4 = 571.3 g
Volume of the solution = 1 lit = 1000 ml
Density of solution = 1.329 g/cm3 = 1.329 g/ml
Calculations:
Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g
Therefore we have:
571.3 g of H2SO4 in 1329 g of the solution
Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987
Mass percentage of H2SO4 (%w/w) is 42.99 %
Answer:
A. Sublimation of camphor
Answer:
The temperature is always lower.
Explanation:
The temperature is always lower at the end of the state as compared to beginning of the state. We can see in the given data, the temperature is higher at the beginning i. e. 140 degree Celsius but with the passage of time, the temperature of a state decreases constantly and the temperature at the end is lower i. e. 20 degree Celsius. So we can conclude that the temperature is always lower.
The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J
The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:
7560J= 124g * (100-26)* specific heat
specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C
88 Grams because the principal of conservation of mass states total mass in equals total out so 200 in 122+88=200 out
