A 25.888 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 73.464 g of water. A 10.762 g aliquot of this solution is then titrated with 0.1039 M HCl . It required 31.89 mL of the HCl solution to reach the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.

Weight % of NH₃ in the aqueous waste  = 2.001 %

Explanation:

The chemical equation for the reaction

Moles of HCl = Molarity × Volume

= 0.1039 × 31.89 mL ×

= 0.0033 mole

Total mass of original sample = 25.888 g + 73.464 g

= 99.352 g

Total HCl taken for assay =

= 0.1083 g

Moles of NH₃ =

= 0.03047 moles

Mass of NH₃ = number of  moles × molar mass

Mass of NH₃ = 0.03047 moles × 17 g

Mass of NH₃  = 0.51799

Weight % of NH₃  = %

Weight % of NH₃ in the aqueous waste  = 2.001 %

it's filtration, but this is an assumption so my apologies if i got this

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.