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it has less water from x then y

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Q: Why was more water lost from vial x then from y?
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When 1 gram of liquid water at 0 Celsius freezes to form ice how many total Joules of heat are lost by the water?

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What do you do after you get the vial on Poptropica?

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A 48.26g sample of aluminum at 100.0C is dropped into 34.47g of water at 25.0C What is the final temperature?

To find the final temperature, you can use the principle of conservation of energy, Q lost = Q gained. The heat lost by the aluminum will be equal to the heat gained by the water. Use this formula: (mass of aluminum) x (specific heat capacity of aluminum) x (change in temperature) = (mass of water) x (specific heat capacity of water) x (change in temperature). You can then solve for the final temperature.


What is the equation to use to find mass of x lost by hydrate?

The equation to find the mass of water lost by a hydrate during heating is: Mass of water lost = Mass of original hydrate - Mass of anhydrous compound left


How are the amount of heat transferred and change in temperature of water related?

Hi, heat transferred = mass x specific heat capacity x rise/fall in temperature If heat is lost then fall in temperature If heat is gained then rise in temperature. More the transfer then greater the difference in temperature.


How are the amount of heat transferred and the changes in temperature of water related?

Hi, heat transferred = mass x specific heat capacity x rise/fall in temperature If heat is lost then fall in temperature If heat is gained then rise in temperature. More the transfer then greater the difference in temperature.


How are the amount of heat transferred and the change in temperature of water related?

Hi, heat transferred = mass x specific heat capacity x rise/fall in temperature If heat is lost then fall in temperature If heat is gained then rise in temperature. More the transfer then greater the difference in temperature.


An unknown hydrate ac xH2O has a mass of 1.632 g and the anhydrous compound AC has a mass of 1.008 g What is the experimental percentage of water in the hydrate?

To find the experimental percentage of water in the hydrate, we need to calculate the mass of water lost during dehydration. Mass of water lost = 1.632 g - 1.008 g = 0.624 g Experimental percentage of water = (mass of water lost / initial mass of hydrate) x 100% = (0.624 g / 1.632 g) x 100% ≈ 38.24%


A 34.44-g sample of metal is heated to 98.6 degrees Celsius in a hot water bath until thermal equilibrium is reached What is the specific heat of the metal?

Heat lost by the metal = heat gained by the water. Heat gained by the water = 50.0 g x 4.184 J/g K x (28.3-22.2) = 1276 J Heat lost by metal = 1276 J = 34.44 g x Sp Heat x (98.6 - 28.3) Specific Heat = 1276 J / 2421 g K = 0.527 J/g K