Answer: The approximate molecular mass of the polypeptide is 856 g/mol
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
Or,
where,
= osmotic pressure of the solution = 4.19 torr
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (polypeptide) = 0.327 g
Volume of solution = 1.70 L
R = Gas constant = 
T = temperature of the solution = ![26^oC=[273+26]K=299K](https://tex.z-dn.net/?f=26%5EoC%3D%5B273%2B26%5DK%3D299K)
Putting values in above equation, we get:
Hence, the molar mass of the polypeptide is 856 g/mol
Answer:
54.99% yield
Explanation:
percent yield is just the amount you obtained over the amount expected times 100%.
(experimental value/theoretical value) x 100%
= (107.9 g/196.2 g) x 100%
=54.99% yield
Answer:
For H3O concentration you do 10^-pH so if pH is 5 then H3O+ is 10^-5= 1*10^-5 H3O+ ions
For OH is one extra step. First find H3o+ ions using equation above then you have to use that to divide 1*10^-14
So if pH is 5....the H3O+ is 1*10^-5 then OH- = (1*10^-14)/(1*10^-5) = 1*10^-9 OH ions
as far as acid/base pH 0-6 is Acid 8-14 is Base. pH of 7 is neutral. Recheck your work *hint* *hint* water is neutral. Spit is above 7 so is base.
Answer:
7.16x10⁻⁸M = [Ag+]
Explanation:
Using the equation:
E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)
<em>Where E</em>⁰<em>= 0.4249V</em>
<em>E(Cell) = -(-0.0019V) -Measured value-</em>
<em>[Cu2+] = 1M</em>
<em />
Replacing:
0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)
-0.423V = - 0.0296 • log (1M/[Ag+]²)
14.29 = log (1M/[Ag+]²)
1.95x10¹⁴ = 1M / [Ag+]²
[Ag+]² = 5.12x10⁻¹⁵M
7.16x10⁻⁸M = [Ag+]
