Answer:
We can solve this by the method of which i solved your one question earlier
so again here molar mass of C12H25NaSO4 is 288.372 and number of moles for 11900 gm of C12H25NaSO4 will be = 11900/288.372
which is almost = 41.26 moles
so to get one mole of C12H25NaSO4 we need one mole of C12H26O
so for 41.26 moles of C12H25NaSO4 it will require 41 26 moles of C12H26O
so the mass of C12H26O = 41.26× its molar mass
C12H26O = 41.26×186.34
= 7688.38 gm!!
so the conclusion is If you need 11900 g of C12H25NaSO4 (Sodium Lauryl Sulfate) you need C12H26O 7688.38 gm !!
Again i d k wether it's right or wrong but i tried my best hope it helped you!!
Lies just beyond the continental slope is the open ocean zone and it has three subzones namely the epipelagic(sunlit zone), mesopelagic(disphotic zone) and bathypelagic zones(aphotic zone). The oceanic zone covers 65% of the ocean's water and where different types of terrains can be found. From deep trenches, deep sea volcanoes and basins. A variety of sea creatures can also be found on each subzone.
Answer: (E) 300 bq
Explanation:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.
Half life is represented by
Half life of Thallium-208 = 3.053 min
Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. , after second half life, the activity would be reduced to half of 1200 i.e. , and after third half life, the activity would be reduced to half of 600 i.e. ,
Thus the activity 9 minutes later is 300 bq.
Answer:
its 303 Kelvin. (typing this for characters)
Answer:
10.64
Explanation:
Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.
C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻ pKb = 3.36
C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:
C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq) pKa
We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.
pKa + pKb = 14.00
pKa = 14.00 - pKb = 14.00 - 3.36 = 10.64