Answer:
We need 2.933 L of 0.15 mg /mL of protein solution.
Explanation:
Concentration of given solution
1 mg = 0.001 g , 1 mL = 0.001 L
Molecular weight of protein = 22,000 Da =22,000 g/mol
Initial concentration in moles/liter:
Initial concentration in micromoles/mL :
1 L = 1000 mL
Initial concentration in micromoles/microLiter :
1 L = 1000,000 μL
Moles of protein required = 20 μmoles
n(Moles)=C(concentration) × V(Volume of solution)
We need 2.933 L of 0.15 mg /mL of protein solution.
Answer:
Option B. 25mL
Explanation:
Step 1:
Data obtained from the question:
Molarity of stock solution (M1) = 6M
Volume of stock solution needed (V1) =..?
Molarity of the diluted solution (M2) = 0.5M
Volume of diluted solution (V2) = 300mL
Step 2:
Determination of the volume of the stock solution needed.
Using the dilution formula, we can easily find the volume of the stock solution needed as follow:
M1V1 = M2V2
6 x V1 = 0.5 x 300
Divide both side by 6
V1 = (0.5 x 300) /6
V1 = 25mL
Therefore, 25mL of 6M NaOH is needed.
Answer:
Atom: The smallest particle of an element that has the properties of the element and can exist either alone or in combination.
Particle: One of the very small parts of matter (as a molecule, atom, or electron)
Kinetic: Of or relating to the motions of material bodies and the forces and energy associated with them
Explanation:
Answer:
Potassium (K) [First element in period 4]
