Answer:
H₂CO₃→H₂O + CO₂
Exphlanation:
Soda is carbonated water. It contains carbon dioxide dissolved in water , which is carbonic acid. On opening the soda, the carbonic acid undergoes the reverse reaction and releases carbon dioxide in the air . The soda is thus left with just water and loses it's fizz.
Answer:
The pH and pOH of a 2.2*10⁻³ HBr solution is 2.66 and 11.34 respectively.
Explanation:
pH - short for hydrogen potential - is a measure of the acidity or alkalinity of a solution. So the pH is a parameter that indicates the concentration of hydrogen ions [H]⁺ that exist in a solution.
The pH is expressed as the negative base 10 logarithm of the hydrogen ion concentration. This is represented by:
pH= - log [H⁺]
pH is measured on a scale of 0 to 14. On this scale, a pH value of 7 is neutral, which means that the substance or solution is neither acidic nor alkaline. A pH value of less than 7 means that it is more acidic, and a pH value of more than 7 means that it is more alkaline.
HBr is a strong acid. Then, in aqueous solution it will be totally dissociated. So the proton concentration is equal to the initial concentration of acid:
[H⁺]= [HBr]= 2.2*10⁻³ M
So:
pH= - log (2.2*10⁻³)
pH= 2.66
On the other hand, pOH is a measure of the concentration of hydroxyl ions in a solution. The sum of pH and pOH equals 14:
pH + pOH= 14
2.66 + pOH= 14
pOH= 14 - 2.66
pOH= 11.34
<u><em>The pH and pOH of a 2.2*10⁻³ HBr solution is 2.66 and 11.34 respectively.</em></u>
Here the base is a benzoate ion, which is a weak base and reacts with water.
The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.
Therefore [OH-] = [C6H5COOH]
In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]
pOH = 14 - pH
pH given = 9.04
pOH = 14-9.04 = 4.96
pOH = -log[OH-] or ![[OH^{-}] = 10^{^{-pOH}}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%2010%5E%7B%5E%7B-pOH%7D%7D%20)
![[OH^{-}] = 10^{^{-4.96}}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%2010%5E%7B%5E%7B-4.96%7D%7D%20)
![[OH^{-}] = 1.1\times 10^{-5}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%201.1%5Ctimes%2010%5E%7B-5%7D%20)
The base dissociation equation kb = 
![kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}](https://tex.z-dn.net/?f=%20kb%20%3D%5Cfrac%7B%5BC6H5COOH%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BC6H5COO%5E%7B-%7D%5D%7D)
H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.
Value of Kb is given = 
And value of [OH-] we have calculated as 
and value of C6H5COOH is equal to OH-
Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-
![1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}](https://tex.z-dn.net/?f=%201.6%5Ctimes%2010%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5BC6H5COO%5E%7B-%7D%5D%7D%20)
![[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}](https://tex.z-dn.net/?f=%20%5BC6H5COO%5E%7B-%7D%5D%20%3D%20%5Cfrac%7B%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B1.6%5Ctimes%2010%5E%7B-10%7D%7D%20)
![[C6H5COO^{-}] = 0.76 M](https://tex.z-dn.net/?f=%20%5BC6H5COO%5E%7B-%7D%5D%20%3D%200.76%20M%20)
or 
So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L
Moles of NaC6H5COO would be = 
Moles of NaC6H5COO (sodium benzoate) = 0.38 mol
Density is equal to mass divided by volume so the densest object will be the object that has the largest mass in the smallest area.
In this case object A is the densest with a density of 10g/cm^3.
I hope this helps. Let me know if anything is unclear.
