moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
[H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is [H2SO4] = 0.08955 M
Answer:
The smell of a chocolate is from the presence of volatile compounds present in the chocolate bar which at room temperature readily changes phase from solid to liquid to vapor or gas
Explanation:
There are nearly 600 identified compounds present in a chocolate bar and out of these, there are volatile components which gives the chocolate bar its distinctive aroma.
These volatile chocolate contents readily change phase from solid to vapor, with very short duration liquid phase.
For example, 3 methylbutanal, vanillin, and several organic compounds which are known to be readily volatile.
As of now, the nuclear fission is the most feasible energy source for human use. All the nuclear power plants are based on the controlled nuclear fission reaction, where the unstable nucleus is bombarded with high speed neutrons, thus, splitting the nucleus into stable ones and releasing huge amount of energy. The nuclear fusion requires very high temperature, the temperature equal's to that of the sun. Hence, it is not feasible right now. As the technology advances, we will see advancement in other form of energies.
Answer:
the normality of the given solution is 0.0755 N
Explanation:
The computation of the normality of the given solution is shown below:
Here we have to realize the two sodiums ions per carbonate ion i.e.
N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)
= 0.1886eq ÷ 0.2500L
= 0.0755 N
Hence, the normality of the given solution is 0.0755 N
Explanation:
(a) As the given chemical reaction equation is as follows.
So, when we double the amount of hypochlorite or iodine then the rate of the reaction will also get double. And, this reaction is "first order" with respect to hypochlorite and iodine.
Hence, equation for rate law of reaction will be as follows.
Rate = ![K \times [OCl^{-}] \times [l^{-}]](https://tex.z-dn.net/?f=K%20%5Ctimes%20%5BOCl%5E%7B-%7D%5D%20%5Ctimes%20%5Bl%5E%7B-%7D%5D)
(b) Since, the rate equation is as follows.
Rate = ![K [OCl^{-}][l^{-}]](https://tex.z-dn.net/?f=K%20%5BOCl%5E%7B-%7D%5D%5Bl%5E%7B-%7D%5D)
Let us assume that (![[OCl^{-}] = [l^{-}]](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%20%5Bl%5E%7B-%7D%5D)
)
Putting the given values into the above equation as follows.
K = 
= 
Hence, the value of rate constant for the given reaction is
.
(c) Now, we will calculate the rate as follows.
Rate =
= 
= 
Therefore, rate when ![[OCl^{-}] = 1.8 \times 10^{3}](https://tex.z-dn.net/?f=%5BOCl%5E%7B-%7D%5D%20%3D%201.8%20%5Ctimes%2010%5E%7B3%7D)
M and ![[I^{-}]= 6.0 \times 10^{4}](https://tex.z-dn.net/?f=%5BI%5E%7B-%7D%5D%3D%206.0%20%5Ctimes%2010%5E%7B4%7D)
M is .
