For example...
You have solution with [H+] = 0,01M
>>>> pH = -log[H+] = -log0,01 = 2
And you increase the [H+] by 10x ---> 0,01×10 = 0,1M
>>>> pH = -log0,1 = 1
○ pH decrease by 2x
○ pH is more acidic
Answer:
Radiation is energy. It can come from unstable atoms that undergo radioactive decay, or it can be produced by machines. Radiation travels from its source in the form of energy waves or energized particles. There are different forms of radiation and they have different properties and effects.
Explanation:
Answer:

Explanation:
We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.
Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.
We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.
- Na: 22.9897693 g/mol
- C: 12.011 g/mol
- O: 15.999 g/mol
Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.
- Na₂ = 22.9897693 * 2= 45.9795386 g/mol
- O₃ = 15.999 * 3= 47.997 g/mol
- Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol
We will convert using dimensional analysis. Set up a ratio using the molar mass.

We are converting 57.3 grams to moles, so we multiply by this value.

Flip the ratio so the units of grams of sodium carbonate cancel.




The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.
Answer:
11.9 is the pOH of a 0.150 M solution of potassium nitrite.
Explanation:
Solution : Given,
Concentration (c) = 0.150 M
Acid dissociation constant = 
The equilibrium reaction for dissociation of
(weak acid) is,

initially conc. c 0 0
At eqm.

First we have to calculate the concentration of value of dissociation constant
.
Formula used :

Now put all the given values in this formula ,we get the value of dissociation constant
.



By solving the terms, we get

No we have to calculate the concentration of hydronium ion or hydrogen ion.
![[H^+]=c\alpha=0.150\times 0.0533=0.007995 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Calpha%3D0.150%5Ctimes%200.0533%3D0.007995%20M)
Now we have to calculate the pH.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


pH + pOH = 14
pOH =14 -2.1 = 11.9
Therefore, the pOH of the solution is 11.9
Solute=potassium chloride, solvent=water