b. 70 kg of ammonia is equivalent to: kmol c. 70 kg/h of ammonia is equivalent t

Question

b. 70 kg of ammonia is equivalent to: kmol c. 70 kg/h of ammonia is equivalent to: kmol/h 4 20 g of sugar (CisHzOn) and 10 g of table salt (NaCI) are dissolved in 250 g of a. what is the mass fraction and mass% for each component in the b, what is the mol fractions and mol% for each component in the water. mixture? mixture? Molar masses of elements: Na = 23 g/mol, Cl = 35.5 g/mol, C = 12 g/mol, H-1 g/mol, 0-16 g/mol 5.A gold leaching plant processes gold ore slurry (a mixture of solids and liquid) containing 30 wt.% rocks and the remaining water, within the solid rock phase, gold particles are present at a concentration of 4 ppm by mass. a. How much gold is present in 1 tonne of slurry? b. What is the concentration of gold in the rock in units of g/tonne? c. If the market price for pure gold is approximately R 510/g how many tonnes of slurry would have to be processed to extract the equivalent of R1 million worth of gold? [Upload answer to SAKAI quiz 6.)The Food Defect Action Levels Handbook is a document written by the United States Food and Drug Administration (FDA) that describes the acceptable level of food contamination from sources such as maggots, mold, insect fragments and faeces. According to the handbook, chocolate can contain no more than 60 insect fragments per 100 g sample for it still to be edible. Assuming the density of an insect is 0.22 g/cm, and the average size of an insect fragment is 1 mm2, what is the acceptable concentration of insects in chocolate as a mass percentage? upload answer to SAKAI quiz] oswer circled questi ion Page 4 of 4

Explanation / Answer

Mass of sugar = 20 g

Molar mass of sugar = 22 x12 + 1x22+11x16 = 462 g/mol

moles of sugar = 20 g / 462 = 0.0433 moles

Mass of salt = 10 g

molar mass of salt = 23+ 35.5 = 58.5 g/mol

moles of salt = 10 g / 58.5 g/mol = 0.171 moles

Mass of water = 250 g

moles of water = 250 / 18 = 13.9 moles

Mass fraction of sugar = mass of sugar / total mass

= 20 g / 280 g

= 0.071

mass percentage of sugar = mass fracation x 100 = 7.1%

Mass fraction of salt= mass of salt / total mass

= 10 g / 280 g

= 0.0355

mass percentage of salt = mass fracation x 100 = 3.55 %

Mass fraction of water = mass of sugar / total mass

= 250 g / 280 g

= 0.892

mass percentage of water = mass fracation x 100 = 8.9%

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(b)

Mole fraction of sugar = mole of sugar / total moles

= 0.0433 moles / (0.0433 moles + 13.9 +0.171 moles)

= 0.0030

mole percentage of sugar = mole fracation x 100 = 0.3 %

Mole fraction of salt= mole of salt / total moles

= 0.171 moles / (0.0433 moles + 13.9 +0.171 moles)

= 0.012

mole percentage of salt = mole fracation x 100 = 1.2 %

Mole fraction of water = mole of water / total moles

= 13.9 moles / (0.0433 moles + 13.9 +0.171 moles)

= 0.985

mole percentage of salt = mole fracation x 100 = 98.5 %

pls post remaining question separely only one question should be answered

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