Answer:
a. Xm = 0.0229
b. 0.0234 moles
c. 354.1 g/mol
Explanation:
ΔP = P° . Xm
ΔP = P° - P', where P° is vapor pressure of pure solvent and P', vapor pressure of solution-
This is the formula for lowering vapor pressure.
If we apply the data given: 523 Torr - 511 Torr = 523 . Xm
Xm = ( 523 Torr - 511 Torr) / 523 Torr → 0.0229
Xm = Mole fraction of solute → Moles of solute / Total moles (sv + solute)
We can make this equation to determine moles of solute
0.0229 = Moles of solute / Moles of solute + 1
0.0229 (Moles of solute + 1) = Moles of solute
0.0229 = Moles of solute - 0.0229 moles of solute
0.0229 = 0.9771 moles of solute → 0.0229 / 0.9971 = 0.0234 moles
Molecular mass of solute → g/mol → 8.3 g / 0.0234 mol = 354.1 g/mol
Answer:
She should take one
Explanation:
because it is the closest option to 1.04 and if you dont have the right dosage then you always want to take the smaller dose as opose to a bigger dose
Explanation:
When the reaction takes place,2 grams of carbon dioxide will be made but vaporized into air.
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
Answer:
1) chemical indicators won't work above it's pH range so therefore it probably won't change colour.
2) the solution should be clear and colourless to see colour change.
3) indicators tend to be of low accuracy so it's not 100% reliable.
