Asaturated solution of nano3 is prepared at 60.°c using 100. grams of water. as this solution is cooled to 10.°c, nano3 precipitates (settles)
out of the solution. the resulting solution is saturated. approximately how many grams of nano3 settled out of the original solution?
(1) 46 g (3) 85 g
(2) 61 g (4) 126 g

The answer is 46 g

The explanation :

According to the attached curve of the relation between the temperature and the grams in solutions:

-we can see from the curve that the grams of NaNO3 at 60°C = 126 g

-and the grams of NaNO3 at 10°C = 80 g

-so, To get the grams of NaNO3 settled out of the original solution, we will subtract the grams of NaNO3 at 10°C - the grams of NaNO3 at 60°C

126 g - 80 g = 46 g

∴ the correct answer is 46 g

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

air

in a tire you fill a tire with air

then a couple days later you check the air pressure in the tire

hope this mark me brainliest