B) As water temp increases solubility increases
Answer:
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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
Answer:
(a) C5H8O2
(b) C2H2Cl2
(c) CH2
(d) CH
Explanation:
We need to find the proportion of the atoms in whole numbers. Given the percentages we can calculate the number of moles and find their proportions.
Assume 100 g and given the atomic weights the moles are calculated.
(a) C = 59.9/ 12.01 = 4.98 ≈ 5.00
H = 8.06/1.007 = 8.00
O = 32/15.999 = 2.00
C5H8O2
(b) C= 24.8/12.01 = 2.06≈ 2.00
H = 2.0/1.007 = 1.99 ≈ 2.00
Cl = 73.1/ 35.453 = 2.06 ≈ 2.00
C2H2Cl2
(c) C = 86/12.01 = 7.16
H= 14/1.007 = 13.90
7.16:13.90 ≈ 1:2
CH2
(d) C = 92.30/12.01 = 7.68
H = 7.7 / 1.007 = 7.65
7.68:7.65 ≈ 1:1
CH
