Answer:
cells like bacteria are bisexual so they split and their offspring is 100 percent like the parent and this process happens over and over
Explanation:
Kinetic
because the boy is walking in the street soo it has force to use
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
Answer:
632.32 mmHg
Explanation:
Millimetre mercury:
It is the monometric unit of pressure. It is define as "The pressure exerted by the column pf mercury at the height of 1 millimetre.
It is represented as mmHg.
It can also be written as mm Hg.
Atmosphere (atm):
It is barometric pressure, define as "The force exerted by atmospheric column on per unit area".
It is written as "atm".
Conversion of atm to mmHg:
0.832 atm × 760 mmHg / 1 atm
632.32 mmHg
Answer:
Unsaturated
Explanation:
In order to successfully answer this question, we need to think about the solubility of solutes in specific solvents, typically water.
- A solution is considered to be unsaturated if at a given temperature and volume of water we may still add more solute and it will dissolve;
- A solution is considered to be saturated if at a given temperature and volume of water we have a maximum amount of solute dissolved and trying to add more solute results in undissolved crystals that can be seen in the solution;
- A solution is considered to be oversaturated (or supersaturated) i at a given temperature and volume of water we exceeded the maximum amount of a solute that could possibly dissolve.
In this case, if we can continue to add more solute to a solution and the solute dissolves, we may state that we are still at a point in which we have an unsaturated solution.
