Answer:
molarity= 0.238 mol L-
Explanation:
The idea here is that you need to use the fact that all the moles of sodium phosphate that you dissolve to make this solution will dissociate to produce sodium cations to calculate the concentration of the sodium cations.
Na 3 PO 4 (aq) → Na + (aq) + PO3−4 (aq)
Use the molar mass of sodium phosphate to calculate the number of moles of salt used to make this solution.
3.25g⋅1 mole N 3PO4 163.9g = 0.01983 moles Na3 PO 4
Now, notice that every
1 mole of sodium phosphate that you dissolve in water dissociates to produce
3bmoles of sodium cations in aqueous solution.
Answer: True the bicarbonate mixture can help save time and few routine.
Explanation:
For the purpose of making dialysate for hemodialysis patient therapies a bicarbonate mixing and delivering systems designed to prepare a liquid sodium bicarbonate formulation comes in handy.
Certain systems like the SDS unit also allow for the transfer and distribution of acid concentrate solutions. We also provide stand-alone acid concentrate delivery systems using a variety of holding tanks and delivery methods.
A challenge for hemodialysis providers is to properly provide bicarbonate solution in a cost effective manner. Preparation and disinfection can be time-consuming and labor intensive.
Bicarbonate however can corrode certain metals and painted surfaces leaving your preparation area encrusted and grimy.
Furthermore, if not mixed properly, bicarbonate can negatively affect the dialysate solution.
The answer to the above is true the bicarbonate mixture can help save time and few routine.
Answer:
The answer is D.) Beaker. Hope this helps! :)
Uranus. Its axis is tilted to almost 90 degrees.
Answer : The volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.
Explanation :
Let the volume of sodium benzoate (salt) be, x
So, the volume of benzoic acid (acid) will be, (100 - x)
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:
x = 29.0
The volume of sodium benzoate = x = 29.0 mL
The volume of benzoic acid (acid) = (100 - x) = (100 - 29.0) = 71 mL
Thus, the volume of sodium benzoate and benzoic acid solution mixed to prepare this buffer should be, 29.0 mL and 71 mL respectively.
