Compute the freezing point of this solution:
25.5 g ch. no: s
(4-nitro-2-toluenesulfonoic acid dihydrate) in 1.00 x 10
g h: 0 (nonionizing solute)

Kc = 0.040 for the system below at 450oc. if a reaction is initiated with 0.40 mole of cl2 and 0.40 mole of pcl3 in a 2.0 liter container, what is the equilibrium concentration of cl2 in the same system? pcl5(g) ⇄ pcl3(g) + cl2(g)

Astudent is given a sample of a blue copper sulfate hydrate. he weighs the sample in a dry covered porcelain crucible and got a mass of 23.875 g for the crucible, lid, and sample. the mass of the empty crucible and lid was found earlier to be 22.652 g. he then heats the crucible to expel the water of hydration, keeping the crucible at red heat for 10 minutes with the lid slightly ajar. on colling, he finds the mass of crucible, lid, and contents to be 23.403 g. the sample was changed in the process to very light clue anhydrous cuso4. if there are again 100.0 g of hydrate, how many grams of cuso4 are in it? how many moles of cuso4? (hint: molar mass of cuso4 = 159.6 g / mole. what per cent of the hydrate is cuso4? you may convert the mass of cuso4 to moles.)

Imagine that a new planet is discovered with two moons of equal mass: moon a and moon b. the mass of the new planet is greater than the combined mass of its moons. moon a is farther away from the new planet than moon b. what is the planet's gravitational pull on moon a compared to the planet's gravitational pull on moon b? the planet's gravity repels moon a with a greater force than it repels moon b, which is why moon a is farther away. the gravitational pull on moon b is greater than on moon a because moon b is closer to the new planet than moon a. the gravitational pull on moon b is greater than on moon a because moon b is farther away from the new planet than moon a. the gravitational pull on moon a is the same as the gravitational pull on moon b because distance does not affect the planet's gravity.