Explanation:
Bayer process is industrial method of the refining of the bauxite to produce alumina which is aluminum oxide.
As alumina is amphoteric in nature, it exhibits a higher solubility at both the extremes of pH range, it is possible to dissolve alumina in low as well as in high pH solutions.
Dissolution of the alumina at high pH is well recognized in Bayer process. Bauxite is digested in very <u>high pH solution (> 13) of alkali</u> like sodium hydroxide at temperature of about 150–250°C and pressure at 20 atm. <u>This is done so that the dissolved alumina is separated from rest of insoluble bauxite minerals. </u>
Answer:
2.4mol
Explanation:
In order to calculate numbers of moles we have to divide the given mass by the molar mass of the substance (ibuprofen) n= m/M
To calculate the molar mass we have
C13= 12*13= 156
H18= 1*18=18
O2= 16*2 =32
Molar mass M = 153+18+32= 206g
Given the mass m= 250g per tablet
Hence number of moles in one tablet is n= 250/206= 1.2 mol
Therefore the total number of ibuprofen you took is 1.2*2 (for the two tablets) = 2.4 moles
For a general reaction,
General expression for rate law will be:
![r=k[A]^{a}[B]^{b}](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E%7Ba%7D%5BB%5D%5E%7Bb%7D)
Here, r is rate of the reaction, k is rate constant, a is order with respect to reactant A and b is order with respect to reactant B.
The reaction is first order with respect to 
, second order with respect to 
and zero order with respect to 
.
According to above information, expression for rate law will be:
![r=k[BrO_{3}^{-}]^{1}[Br^{-}]^{2}[H^{+}]^{0}](https://tex.z-dn.net/?f=r%3Dk%5BBrO_%7B3%7D%5E%7B-%7D%5D%5E%7B1%7D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%5BH%5E%7B%2B%7D%5D%5E%7B0%7D)
Or,
![r=k[BrO_{3}^{-}][Br^{-}]^{2}](https://tex.z-dn.net/?f=r%3Dk%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
...... (1)
- When concentration of get doubled, rate of the reaction becomes,
![r^{'}=2k[BrO_{3}^{-}][Br^{-}]^{2}](https://tex.z-dn.net/?f=r%5E%7B%27%7D%3D2k%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
...... (2)
Dividing (2) by (1)
![\frac{r^{'}}{r}=\frac{2k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=2](https://tex.z-dn.net/?f=%5Cfrac%7Br%5E%7B%27%7D%7D%7Br%7D%3D%5Cfrac%7B2k%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%7D%7Bk%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%7D%3D2)
Or,
Thus, rate of the reaction also get doubled.
- When the concentration of is halved, the rate of reaction becomes
^{2}](https://tex.z-dn.net/?f=r%5E%7B%22%7D%3Dk%5BBrO_%7B3%7D%5E%7B-%7D%5D%28%5BBr%5E%7B-%7D%5D%2F2%29%5E%7B2%7D)
Or,
![r^{"}=1/4k[BrO_{3}^{-}][Br^{-}]^{2}](https://tex.z-dn.net/?f=r%5E%7B%22%7D%3D1%2F4k%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D)
...... (3)
Dividing (3) by (1)
![\frac{r^{"}}{r}=\frac{1/4k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7Br%5E%7B%22%7D%7D%7Br%7D%3D%5Cfrac%7B1%2F4k%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%7D%7Bk%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B4%7D)
Or,
Thus, rate of reaction becomes 1/4th of the initial rate.
- When the concentration of is tripled:
Since, the rate expression does not have concentration of , it is independent of it. Thus, any change in the concentration will not affect the rate of reaction and rate of reaction remains the same as in equation (1).
Answer:
b. phosphoric acid & phosphorus acid
