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[6 pointe] Perform the following conversions and show allyour work for fufl credit. You may need look up conversion factors in yourbook online . 13,500 mg of NaCl t...

Question

[6 pointe] Perform the following conversions and show allyour work for fufl credit. You may need look up conversion factors in yourbook online . 13,500 mg of NaCl to grams NaCl 25 8 0f He to moles of Ile 0.3265 moles of CH_COOH grams of CH;COOH 1.5 g of water ML of water

[6 pointe] Perform the following conversions and show allyour work for fufl credit. You may need look up conversion factors in yourbook online . 13,500 mg of NaCl to grams NaCl 25 8 0f He to moles of Ile 0.3265 moles of CH_COOH grams of CH;COOH 1.5 g of water ML of water



Answers

A certain brine has $3.87 \%$ NaCl by mass. A $75.0 \mathrm{mL}$ sample weighs 76.9 g. How many liters of this solution should be evaporated to dryness to obtain $725 \mathrm{kg}$ $\mathrm{NaCl} $?

In this problem, we have to calculate the quantity off. Sodium chloride are common started in the seawater. We know that mass is equal toe volume multiplied by density density. Given here of the seawater is 1.0 t gram. For Emily, the volume off seawater is 1.5, multiplied by 10 to part 21 little. The mass of the seawater, therefore, can be killed. Created by in putting these values which comes out toe 1.55 multiplied by 10 to part 24 grams. It is known back so it employed as a person. T is 3.1% off the total weight off See voto. Therefore sort employed will be 1.55 multiplied by 10 24 gram multiplied by three point fun times 100 is a call toe 4.8 multiplied by 10 to 2 part 22 grounds. To convert this into Katie, we know that one kg physical 2000 grams in putting that as a conversion factor here vigor The final answer. In kilograms, the quantity off sodium chloride is equal to 4.8 multiplied by 10 to part 19 k g. Also, we know that one ton is equal to £2000 on £1 is equal to 4 53 points exclaims. Therefore, the months to become a leader, in turn is 4.8, multiplied by 10 to part 22 graham, multiplied by £1 by 4 53.6 gram multiplied by one turn, divided by £2000. This is achieved by in putting all the known value. So the mask off sodium chloride in seawater in tons is five country in tow. Temper 16 tons.

Going to calculate clarity. We have polarity speak for two moles of solid, divided by liters of solution. So in the first example, we have 2.92 moles divided by 1.45 liters. Gives us 2.0 mola. Next we have another example. So we have 7.2 times 10 minus five. Most divide that by no point Not not 165 liters. We get no point. Not for mola The next example We have 9.955 moles divided by 9.125 liters. We get 9.44 mola and the last example in Part D. We have 9.78 miles. Divide that by 7.25 liters. That God does not 0.11 Mola, Yeah.

If you recognize what to do, this is pretty much just a conversion calculation. We'll start with our middle leaders, converted toe leaders. When we have leaders, we can now use Mill Arat. AEA's a conversion factor to go from leaders. Two moles by multiplying by the polarity. Then we have moles. We can convert the moles into grams by multiplying by the Moeller math. So there are 2.7 g and a C. L and 35 mil leaders of the solution.

May be the hardest part of answering. This question is just the set up. So let me help you with a set up first. If this is our definition of mass percent. Graham solid over Graham solution times 100 equals R percentage. Then if we divide, will sides by percent and multiply both sides by Graham solution. We get this equation. So if I know the Graham cell you'd in the percent, I can easily calculate the Gramps solution. So if we know the masses 1.5 g of sodium chloride that's desired, we divide it by the percent of the solution. 0.58 Multiply it by 100. That'll be the master of the solution 2.6 kg, or 2600 g. The next 11.5 is still is desired, divided by the different percentage multiplied by 100. It's 100 g, and then the 1.5 g of sodium chlorate is desired. From an 8.44% solution. Times 100 would get 18 g


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