Answer:
Products
Explanation:
During a chemical reaction, there are the reactants (left side), and the products (right side).
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
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.
<span>0.48 grams.
Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu(NO3)2, that the reaction will result in 1 atom of copper per molecule of Cu(NO3)2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced.
To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So
0.0075 mol * 63.546 g/mol = 0.476595 g.
Round the results to 2 significant figures, giving 0.48 grams.</span>
