Answer: Matter
Explanation:
Matter is anything that has volume and/or mass.
It is a graph. It shows observations and then you record your results with any of the graph types.
Explanation:
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![NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}](https://tex.z-dn.net/?f=%20NaC6H5COO%20%5Crightarrow%20Na%7B%5E%7B%2B%7D%7D%20%2B%20C6H5COO%5E%7B-%7D%20)
Here the base is a benzoate ion, which is a weak base and reacts with water.
![C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)](https://tex.z-dn.net/?f=%20C6H5COO%5E%7B-%7D%28aq%29%20%2B%20H2O%20%28l%29%5Cleftrightarrow%20C6H5COOH%28aq%29%2B%20OH%5E%7B-%7D%28aq%29%20)
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 = ![\frac{Product}{Reactant}](https://tex.z-dn.net/?f=%20%5Cfrac%7BProduct%7D%7BReactant%7D%20)
![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 = ![1.6\times 10^{-10}](https://tex.z-dn.net/?f=%201.6%5Ctimes%2010%5E%7B-10%7D)
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-
![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)
![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)
or ![0.76\frac{mol}{L}](https://tex.z-dn.net/?f=%200.76%5Cfrac%7Bmol%7D%7BL%7D%20)
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 = ![0.76(\frac{mol}{L}) \times (0.50L)](https://tex.z-dn.net/?f=%200.76%28%5Cfrac%7Bmol%7D%7BL%7D%29%20%5Ctimes%20%280.50L%29%20)
Moles of NaC6H5COO (sodium benzoate) = 0.38 mol
5. The difference between mass and weight it that mass is the volume inside a object.