Answer:
The minimum molecular weight of the enzyme is 29.82 g/mol
Explanation:
<u>Step 1:</u> Given data
The volume of the solution = 10 ml = 10*10^-3L
Molarity of the solution = 1.3 mg/ml
moles of AgNO3 added = 0.436 µmol = 0.436 * 10^-3 mmol
<u>Step 2:</u> Calculate the mass
Density = mass/ volume
1.3mg/mL = mass/ 10.0 mL
mass = 1.3mg/mL *10.0 mL = 13mg
<u>Step 3:</u> Calculate minimum molecular weight
Molecular weight = mass of the enzyme / number of moles
Molecular weight of the enzyme = 13mg/ 0.436 * 10^-3 mmol
Molecular weight = 29.82 g/mole
The minimum molecular weight of the enzyme is 29.82 g/mol
Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %
Explanation :
In the given case different law related to gas is given. The attached figure shows the required solution.
Boyle's law states that the pressure is inversely proportional to the volume of the gas i.e.


k is a constant.
Charle's law states that the volume of directly proportional to the temperature of the gas.


Combined gas law is the combination of the pressure, volume and the temperature of the gas i.e.

Hence, this is the required solution.
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>