You should try to strt a conversation with him or say hi introduce yourself or ask him how hes doing just talk to him i guess
Answer:
Water outside the cell will flow inwards by osmosis to attain equilibrium
Explanation:
In the hypotonic environment, the concentration of water is greater outside the cell and the concentration of solute is higher inside. A solution outside of a cell has a lower concentration of solutes relative to the cytosol.
If concentrations of dissolved solutes are greater inside the cell, the concentration of water inside the cell is correspondingly lower. As a result, water outside the cell will flow inwards by osmosis to attain equilibrium.
Osmosis is a process by which molecules of a solvent tend to pass from a less concentrated solution into a more concentrated one through a semipermeable membrane.
Answer:
d) cut the large sized Cu solid into smaller sized pieces
Explanation:
The aim of the question is to select the right condition for that would increases the rate of the reaction.
a) use a large sized piece of the solid Cu
This option is wrong. Reducing the surface area decreases the reaction rate.
b) lower the initial temperature below 25 °C for the liquid reactant, HNO3
Hugher temperatures leads to faster reactions hence this option is wrong.
c) use a 0.5 M HNO3 instead of 2.0 M HNO3
Higher concentration leads to increased rate of reaction. Hence this option is wrong.
d) cut the large sized Cu solid into smaller sized pieces
This leads to an increased surface area of the reactants, which leads to an increased rate of the reaction. This is the correct option.
Answer:
The answer is Denaturation is reversible
Explanation:
Denaturation of proteins is the process whereby there is a disruption in the tertiary, quaternary or secondary structure of proteins which causes a conformation change in its action
denaturation is reversible, that is, the proteins can regain their native state when the denaturing influence is removed. This process can be called renaturation.
Answer:
625 mL
Explanation:
From the question given above, the following data were obtained:
Volume of stock solution (V₁) = 250 mL
Molarity of stock solution (M₁) = 5 M
Molarity of diluted solution (M₂) = 2 M
Volume of diluted solution (V₂) =?
The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5 × 250 = 2 × V₂
1250 = 2 × V₂
Divide both side by 2
V₂ = 1250 / 2
V₂ = 625 mL
Therefore, the volume of the diluted solution is 625 mL.
