How many ml of 50 percent dextrose solution and how many ml of water are needed to prepare 100ml of 15 percent dextrose solution?

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30 ml 50% dextrose + 70 ml water
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What are indications for using 5 percent dextrose in water?

D5W or Dextrose 5% Water is indicated for:. Rehydration . Provides calories for some metabolic needs (100ml provides 5g dextrose) -- useful in traumas when there's a shock related calorie burn. . By adding carbohydrates, reduces metabolism of proteins for calorie uptake. . May produce diuresis. ( Full Answer )

To make a 10 percent solution of salt in water how many grams of salt should you dissolve into 100 ml of water?

10% salt solution is the equivalent of adding 100gr salt in a 900 ml (1000ml -100ml) of water. you now have one liter of a 10% solution. . Similarly 10 gr of salt in 90 ml water will give you 100 ml of a 10% salt solution. . However if you must use 100 ml of water you need to add 11 gr of salt to ( Full Answer )

How much pure acid should be added to 50 ml of a 10 percent acid solution to give a 50 percent acid solution?

A 50 ml solution that is 10% acid will consist of 5 ml of acid (10% of the volume) and 45 ml of water (90% of the volume). You're not adding any water, but you want to add enough acid to make a solution that is 50% acid and 50% water. You will need to have a total of 45 ml of acid in the mixture to ( Full Answer )

What solution is the 5 percent Dextrose in Lactated Ringer's?

Dextrose is a synonym of D-glucose (also known as grape sugar, corn sugar, and when it's present in blood, blood sugar).In 2013, Dextrose 5 percent in lactated Ringer's injection wasrecalled. This recall stemmed from allegations of the producthaving mold in it.

What is dextrose solution?

Dextrose solution is a medication in an IV form used to supply water and calories to the body. It is also used as a mixing solution (diluent) for other IV medications

Find the amount of 14 percent acid solution that should be added to 6 percent acid solution to obtain 50 mL of a 12 percent solution?

Write eqn for weight; x = 14% sol & y = 6% sol which is x + y = 50. Write eqn for solution: .14x + .06y = .12*50 = 6; 2 equations 2 unknowns. Solve: Mult 1st eqn by -.14; now -.14x-.14y = -.14*50 or -7 -.14x-.14y=-7 .14x+.06y=6 0-.08y=-1 or y=12.5 or 12.5mL of 6% solution x+12.5=50 or x=37.5 or 37.5 ( Full Answer )

How many grams of NaOH are required to prepare 200 ml of a 0.450 M solution?

First you have to find the molar mass of NaOH which is,23+16+1=40g/mol. Then you take 0.450M X 200mL=90 mM. which can be converted to .090moles just move the decimal three decimal places. Then you take youmolar mass which is 40g/mol X .090moles= 3.6grams. (The molescancelled themselves out and your ( Full Answer )

How many liters of a 10 percent alcohol solution must be mixed with 80 liters of a 80 percent solution to get a 50 percent solution?

The following solution is just an approximation, in practical circumstances sufficient to use. The addendum explains why. . x = number of litres (10%) to be added. You have 64 litres of alcohol in your original 80 litres. . 64 + x/10 = (80 + x)/2 . 640 + x = 400 + 5x . 4x = 240 . x = 60 and a ( Full Answer )

50 percent dextrose solution and you need a 1 percent dextrose soultion. How do you figure that out?

If the percents given are by weight or mass, this is very straightforward: The ratio between the desired percentage and the initial percentage is 1/50. Therefore, a given mass of initial solution must be diluted to 50 times its original mass to obtain the desired lower concentration, or in other wor ( Full Answer )

How much of a 37 percent formaldehyde solution should be put into 900 ml of water to make a 10 percent formaledhyde solution?

If the percentages specified are by mass and the 900 ml of water are assumed to be at standard temperature and pressure, this problem can be solved as follows: The 900 ml of water will have a mass of 900 grams. Call the unknown mass of the 37 percent solution to be added m. this will contain 0.37m g ( Full Answer )

How many ounces of pure water must be added to 50 ounces at a 15 percent saline solution to make a saline solution that is 10 percent salt?

Add 25 oz of pure water Let X be the volume of pure water to add. Then the total amount of pure water you add plus the total amount of water in the 15% solution will equal the total amount of water in the final 10% solution. So, X + 50*(1-0.15) = (50 + X) * (1-0.10) X + 50*(0.85) = (50 + X)*(0.9 ( Full Answer )

How many grams of Na2SO4 are needed to prepare 750 ml of a 0.375 M solution?

Let us find moles first. Molarity = moles of solute/Liters of solution ( 750 ml = 0.750 Liters ) 0.375 M Na 2 SO 4 = moles Na 2 SO 4 /0.750 Liters = 0.28125 moles Na 2 SO 4 === 0.28125 moles Na 2 SO 4 (142.05 grams/1 mole Na 2 SO 4 ) = 39.95 grams Na 2 SO 4 needed ------------- ( Full Answer )

How many milliliters of a 50 percent acid solution and how many milliliters of a 20 percent acid solution must be mixed to produce 36 mL of a 30 percent acid solution?

A nswer:. 12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solution. Let x = 50% acid solution. y = 20% acid solution. Equations:. x + y = 36mL ----equation (1). 0.5x + 0.2y = 0.3 * 36. 0.5x + 0.2y = 10.8. multiplying by 10. 5x + 2y = 108 ----equatio ( Full Answer )

How many grams are KCl are required to prepare 250 ml of a 4M solution?

By definition, 1 liter of a 4 M solution must contain 4 moles of the solute. Because solutions are homogeneous mixtures and 250 ml is one quarter of a liter, the amount of KCl required is 1 mole. The gram molar mass* of KCl is 74.55 grams; this is therefore the amount of KCl required.** ____________ ( Full Answer )