Write out the eqn of magnesium and oxygen. this should be under “metals” chapter. do revise.
next, find the mols of both oxygen and magnesium. compare the ratios and find the LIMITING REAGENT.
use the mols of the limiting reagent to compare with the mols of the product.
take the mols of the product/mr of the product.
this will give u the mass.
When air pressure is low it is Cold
Answer: chemical property
Explanation: This is a chemical property because it was combined with carbon, another molecule acting like an element. The answer to the question is chemical property. Hope this helps!
Answer:
MM = 680g / mol
Explanation:
Hello! To calculate the molar mass of the compound (g / mol), I first have to calculate the molarity.
Molarity can be calculated from the osmotic pressure equation.
op = M * R * T
op = osmotic pressure = 28.1mmHg * (1 atm / 760mmHg) = 0.037atm
M = molarity
R = gas constant
T = temperature (K) = 20 ° C + 273.15 = 293.15K
M (mol / L) = op / R * T
M = 0.037atm / ((0.082 (atm * L) / (K * mol)) * 293.15K) = 0.0015mol / L
As I have the volume = 100ml * (1L / 1000ml) = 0.1L
I can calculate the amount of moles
n = M * V = 0.0015 * 0.1 = 0.00015mol
n = m / MM
m = mass
MM = molar mass
MM = m / n = 0.102g / 0.00015mol
MM = 680g / mol
Answer:
25 mM Tris HCl and 0.1% w/v SDS
Explanation:
A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:
(250 mM Tris HCl)/10 = 25 mM Tris HCl
(1.92 M glycine)/10 = 0.192 M glycine
(1% w/v SDS)/10 = 0.1% w/v SDS
Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.
