Answer:
D.
Explanation:
-log(1.0x10^-5) = pH
pH + pOH = 14 (rearrange it)
OH- = 10^-pOH = 1.0 x 10^-9
- Hope that helped! Let me know if you need further explantion.
Answer:
false,soil damage and nutrient loss can be a major problem so it can not gain nutrients quickly
Answer:
4.823 x 10^-19 J
Explanation:
Energy is calculated by E = hv where h - Planck's constant in joule.s
v - frequency.
in this particular question the wave length is 4.12 x 10^-7 m. to exhaustively use this we need a relation between wave length & frequency. c=wv where C is approximately 3 x 10^8m/s
-v = c/w = 3x10^8m/s / 4.12 x 10^-7m = 7.28 x 10^14 Hz or 1/sec
now we can simply use Planck's constant in E=hv =
(6.626 x 10^-34) x (7.28 x 10^14Hz) = 4.823 x 10^-19 J.
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.
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as
pH = pKa + log [A]/[HA]
where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid
Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol
Solution:
pKa = - log ( 1.8x10^-5) = 4.74
[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles
pH = 4.74 + log (.115/0.085)
pH = 4.87
