Dextrose 25% 1000ml is ordered. you have 70% dextrose on hand. how much of the dextrose 70% sol. and?
how much sterile water will you use to complete this order? a. 250ml of dex. and 750 ml of water b. 357ml of dex. and 643ml of water c.424 ml of dex. and 576 ml of water d. none of the above plz show me how u solve it. thanks
Asked by Nu Lueschen 1 year ago.
You have dextrose at 70%. 50%-50% of dextrose & water will give 35% 50% dextrose *0.7 * 100 = 35% x% dextrose * 0.7 * 100 = 25 % x * 0.7 = 0.25 x=0.25 / 0.7 x=0.357 or 35.7% of 1000 ml b) 357 ml of dextrose solution at 70% topped with 643 ml of water (in the first approximation) will yield 1000 ml of solution of 25% dextrose. Answered by Gearldine Bailleu 1 year ago.
A doctor needs a liter of 25% dextrose solution,you have 70% dextrose and sterile water.?
How much of each fluid will you need?
Asked by Rudy Yamkosumpa 1 year ago.
A liter of 25% dextrose solution contain .25 * 1000cc or 250 cc of dextrose The question is how many cc of 70% solution will give you 250 cc. x*.70 = 250 By division. x = 357.14. So to 357.14 cc of the 70% solution add 1000 - 357.14 distilled water. (642.86 cc) Answered by Rosemary Piantanida 1 year ago.
I believe you can use the old CV=C1V1 equation: C=25% V=1L C1=70% V1=x therefore: x = 25/70 x= 5/14 L 70%, now bring up to volume (up to 1L) with saline add 9/14L saline Answered by Zenobia Felske 1 year ago.
A solution of anhydrous dextrose (m.w. 180) contains 25.1 g in 500 mL of water. Calculate the freezing point o?
Asked by Kimberely Shetz 1 year ago.
For a non-ionic solute like dextrose, the freezing point depression equation is delta Tf = Kf x molality where delta Tf is the amount that the freezing point is lowered (degrees C) Kf = freezing point constant for H2O = 1.86 C/m molality = moles of solute / kg of solvent 25.1 g dextrose x (1 mole dextrose / 180 g dextrose) = 0.139 moles dextrose molality dextrose = 0.139 moles dextrose / 0.500 L water = 0.278 molal delta Tf = Kf x m = (1.86 C/m)(0.278 m) = 0.517 C So the freezing point of water is lowered by 0.517 C. Since the normal freezing point of water is 0.000 C, the new freezing point is -0.517 C. Answered by Clarence Seanger 1 year ago.
Anhydrous Dextrose Answered by Beaulah Cantell 1 year ago.
If a physician orders 25% dextrose 1000 mL and all you have is...?
If a physician orders 25% dextrose 1000 mL and all you have is 70% dextrose 1000 mL, how much 70% dextrose and how much sterile water will be used? I am ashamed that I cannot answer this question. -_-
Asked by Kristofer Vancamp 1 year ago.
Set up a simple ratio. You have a 70% dextrose solution 1000 mL, but you don't know how much of it you should use, so you have a 0.70 * 1000 * x solution. You want a 25% 1000 mL solution, so 0.25 * 1000. Set them equal to each other to get: 0.70 * 1000 * x = 0.25 * 1000 x = 0.357 = 357 mL dextrose 1000 mL - 357 mL = 643 mL water So you need 357 mL 70% dextrose and 643 mL water. Answered by Jani Nunoz 1 year ago.
Doctor needs a liter of 15% dextrose.You have 25% Dextrose and sterile water. How much fluid will you need?
Asked by Nicholle Zidek 1 year ago.
That depends on how much (what volume) of the 15% dextrose you want. For example, if you desire to make 100 mls of 15% dextrose then... (x ml)(25%0 = (100 ml)(15%) and x = 60 mls of 15% dextrose + 40 mls sterile water. You can use this equation for any volume, by substituting that volume for the 100 mls and solving for x Answered by Illa Erker 1 year ago.
Dr.is requesting a liter of 15% dextrose solution. You have 25% dex and sterile H20, how much of ea. do u need?
Asked by Columbus Streifel 1 year ago.
You can figure this out using alligation. [Ignore the dots] 25..........15 (parts of 25% [15-0]) .......15........Final concentration 0............10 (parts of sterile water [25-15]) ..............--- ..............25 (total # of parts) You can reduce the parts by dividing each by 5 3 parts D25, 2 parts water, 5 parts total Since 1000ml = 5 parts then 1 part is 1000ml/5parts or 200ml How much D25: 3 parts * 200ml/part = 600ml How much Water: 2 parts * 200ml/part = 400ml Check: 1000ml = 600ml + 400ml EDIT: You can also use the C1V1 = C2V2 formula 25% * Xml = 15% * 1000ml Xml = (15% * 1000ml) / 25% Xml = 600ml that's how much D25 is needed How much Sterile water: 1000ml - 600ml = 400ml PS: This formula doesn't always work but the alligation procedure always will. Answered by Joanna Schwimmer 1 year ago.
For a 15 % solution you need 15 grams dextrose/100 ml of H2O. If you have a 25% solution you have 25 Gm of dextrose /100 ml of H2O therefore you need to dilute your solution. Or you need to add 150g of dextrose to 1000 ml of sterile H2O. You didn't state volume of 25% dextrose in sterile H20 which you already have. You may not have enough of a 25% solution to bring to a final volume of 1000 ml or 1 liter at 15% concentration. Answered by Quincy Mojica 1 year ago.
Dextrose 15 Answered by Misha Sawyer 1 year ago.
Exactly 100lbs of dextrose were discharged in a small stream saturated with oxygen from the air at 25 celsius.?
How many liters of this water could be contaminated to the extent of removing all the dissolved oxygen by biodegradation?
Asked by Raymundo Jelden 1 year ago.
Xiaoyu, First, we have to find the balanced equation: C6H12O6 + 6O2 ---> 6CO2 + 6H2O Now we need to find the moles of dextrose: 100 lb * (453.59 g/lb) / (180.156 g/mole) = 251.8 moles According to the balanced equation, this would absorb 6 times that number of moles of oxygen: Moles O2 = (6 moles O2/mole dextrose) * 251.8 moles dextrose = 1511 moles Gms O2 = 1511 moles * 32.00 g/mole = 4.834 x 10^4 grams Now, we need to know the solubility of oxygen in water at 25C; if they did not give this to you, you can find it on-line. The reference below gives it as 0.003931 gm oxygen / 100 gm water, or 0.03931 gm/L. Therefore, Volume of contaminated water = (4.834 x 10^4 g)/(3.931 x 10^-2 g/L) = 1.23 x 10^6 L. Hope that helped! Answered by Josefine Propst 1 year ago.