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Daryll
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What are the molality and mole fraction of solute in a 30.1 percent by mass aqueous solution of formic acid (HCOOH)
can you
2 answers
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What are the molality and mole fraction of solute in a 30.1 percent by mass aqueous solution of formic acid (HCOOH)
1 answer
518 views
What are the molality and mole fraction of solute in a 30.1 percent by mass aqueous solution of formic acid (HCOOH)
1 answer
800 views
If you have 1911.9 g of a bleach solution. The percent by mass of the solute sodium hypochlorite NaOCl is 3.62. How many grams
1 answer
568 views
What volume of a 4.50 M KI stock solution would you use to make 0.455 L of a 1.25 M KI solution?
2 answers
522 views
1.) Lim [√(x + 1) - (2)] / (x - 3)
x -> 3 2.) Lim [ (1/ x + 4) - (1 / 4)] / (x) x -> 0
1 answer
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Answers (6)
126
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
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 *********
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
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,
9.0-17.8x-6.1x+6.1y=0 12.0-8.3y-6.1y+6.1x=0
