A water solution is found to have a molar oh- concentration of 3.2 x 10-5. the solution would be classified as neutral.
The concentration of hydroxide ions (OH-) is measured by pOH. It is a way of expressing how alkaline a solution is. At 25 degrees Celsius, aqueous solutions with pOH values of 7 or less are neutral, whereas those with pOH values of 7 or more are acidic. The hydrogen ion potential is known as pH. The potential of hydroxide ions is known as pOH. 2. It is a scale used to estimate the hydrogen ion (H+) concentration in the solution. The hydroxide ion (OH-) concentration of the solution is measured using this scale.
pH + pOH = 14
pOH = 3.2x 10-5
[OH-] = 10^(-pOH) =10^(- 3.2x 10-5)
= 0.99
Answer:

Explanation:
Hello there!
In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:
![HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=HA%5Crightleftharpoons%20H%5E%2B%2BA%5E-%5C%5C%5C%5CKa%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:
![Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7Bx%5E2%7D%7B%5BHA%5D_0-x%7D%20%3D10%5E%7B-4.74%7D%3D1.82x10%5E%7B-5%7D)
Thus, it is possible to find x given the pH as shown below:

So that we can calculate the initial concentration of the acid:
![\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B%281.82x10%5E%7B-5%7D%29%5E2%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1.82x10%5E%7B-5%7D%5C%5C%5C%5C%5Cfrac%7B1.82x10%5E%7B-5%7D%7D%7B%5BHA%5D_0-1.82x10%5E%7B-5%7D%7D%20%3D1%5C%5C%5C%5C)
![[HA]_0=3.64x10^{-5}M](https://tex.z-dn.net/?f=%5BHA%5D_0%3D3.64x10%5E%7B-5%7DM)
Therefore, the percent dissociation turns out to be:
![\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%](https://tex.z-dn.net/?f=%5C%25%20diss%3D%5Cfrac%7Bx%7D%7B%5BHA%5D_0%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%3D%5Cfrac%7B1.82x10%5E%7B-5%7DM%7D%7B3.64x10%5E%7B-5%7DM%7D%2A100%5C%25%20%5C%5C%5C%5C%5C%25%20diss%20%3D%2050%5C%25)
Best regards!
Answer:
4.68x10²⁵ ions of Na⁺
Explanation:
First of all, we dissociate the salt:
NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
An aqueous solution of NaCl dissociates in chlorides anions and sodium cations. Ratio is 1:1, per 1 mol of NaCl, we have 1 mol of Na⁺
We determine the moles of salt: 4543.3 g . 1mol / 58.45 g = 77.7 moles
77.7 moles are the amount of NaCl, therefore we have 77.7 moles of Na⁺.
We count the ions:
1 mol fo Na⁺ has 6.02x10²³ ions
77.7 moles of Na⁺ must have (77.7 . 6.02x10²³) / 1 = 4.68x10²⁵ ions of Na⁺
Answer:
The correct answer is "-268.667°C".
Explanation:
Given:
Temperature,
= 4.483 K (below)
Now,
The formula of temperature conversion will be:
⇒ 
By putting the values, we get
⇒ 
⇒ 
Thus the above is the correct answer.
Answer:
I think copper
Explanation:
Material IACS % Conductivity
Silver 105
Copper 100
Gold 70
Aluminum 61
Nickel 22
Zinc 27
Brass 28
Iron 17
Tin 15
Phosphor Bronze 15
Lead 7
Nickel Aluminum Bronze 7
Steel 3 to 15
the table might help- your indian brother