C. The salt bridge maintains the flows of ions and allows electrons to move from the anode to the cathode.
Total in pot=28 L
400 mL in each bowl
16 bowls filled
1000mL=1L
16 bowls(400mL/1 bowl)=6400mL
6400mL(1L/1000mL)=6.4L
28L-6.4L=21.6 L
Answer:
78.96 g of NaC2H3O2
Explanation:
We are given;
- Volume of the solution as 350 mL
- Molarity of the solution as 2.75 M
- Molar mass of the solute NaC2H3O2 as 82.04 g/mol
We are required to determine the mass of the solute;
First we determine the number of moles;
Moles = Molarity × Volume
Therefore;
Moles of the solute = 2.75 M × 0.350 L
= 0.9625 moles
Second, we determine the mass
Mass = Moles × Molar mass
= 0.9625 moles × 82.04 g/mol
= 78.9635 g
= 78.96 g
Therefore, the mass of NaC2H3O2 needed is 78.96 g
I believe this property is vital since the sticky nature allows the wells to coat easily with the desired antigen by simply adding a small amount of it and allowing it time to incubate. Adhesion is the tendency of dissimilar particles or surfaces to cling to one another ( cohesion refers to the tendency of similar or identical particles/surfaces to cling to one another).
Iodine Strontium Silver...... ..I think.