Daryll

1.) Lim [√(x + 1) - (2)] / (x - 3) x -> 3 2.) Lim [ (1/ x + 4) - (1 / 4)] / (x) x -> 0

What are the molality and mole fraction of solute in a 30.1 percent by mass aqueous solution of formic acid (HCOOH) can you please explain it better? I still don't quite understand

9.0-17.8x-6.1x+6.1y=0 12.0-8.3y-6.1y+6.1x=0

Sam still has the right to the colt since NO DATE OR CONSIDERATION was agreed upon BEFORE the colt was born. This means that the OBLIGATION TO DELIVER the horse has NOT YET ARISEN and Ben has no right to the "fruits" of the object of the contract. HOWEVER...

1) since 1 kg = 1000 g then: 1000 g / 1 kg = x / 5.114 kg x = (5.114 kg * 1000 g) / 1 kg x = 5114 g 2) since 1 km = 1000 m then: 1000 m / 1 km = x / 0.5 km x = (0.5 km * 1000 m) / 1 km x = 500 m *** just do the same method for the remaining ones.. here's...

Pressure = height * density * gravity 101 Pa = height * (13600 kg / m3) * (9.81 m/s2) ** note that kg * m/s2 = N 101 Pa = height * 133416 N/m3 ** Pa = N/m2 therefore, (101 N/m2) / (133416 N/m3) = height height = 0.00075703 m = 0.75703mm ********* HOWEVER!...

The molality formula is mol solute/kg solvent. In this case, Na2SO4 is the solute and water is the solvent. So basically you have to find the moles of Na2SO4 by dividing 11.1 g by molar mass of the solute, which is about 142 g/mol (11.1g/ 142 g/mol) Then...

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