50mm x 24mm or 54mm x 25mm
shut de routes
The answer depends on what X is meant to represent.
There is more than one vial (not called that) on Poptropica islands:Astro Knights IslandGet the replacement green fuel rod from the robot invader in the dungeon and replace the damaged one in the saucer Excalibur. Get the list of planets from the Queen, but set the saucer dials to X-56 Y-52 (the Pewter Moon).Steamworks IslandThe vial (beaker) from the Museum will be used to mix a herbicide in the lab (production zone) on the far side of the drawbridge.
L x w x avg depth x 7.5
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.
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
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.
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.
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.
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%
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