Explanation:
The required concentration of 
M1 =0.222 M.
The required volume of is V1 =225 mL.
The standard solution of is M2 =16 M.
The volume of standard solution required can be calculated as shown below:
Since the number of moles of solute does not change on dilution.
The number of moles 
Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.
Answer is: amount of sugar in milk chocolade is 3333 mg.
To solve this question, make proportion: if 12,00 grams of milk chocolate contain 8,00 grams of sugar, than 5,00 grams contain:
12,000 g : 8,000 g = 5,000 g : m(sugar).
12,000 g · m(sugar) = 8,000 g · 5,000 g.
m(sugar) = 40,000 ÷ 12,000.
m(sugar) = 3,333 g = 3333 mg.
Answer:
I feel like it could lead to flooding
Answer:
![[SO_2Cl_2]=0.0175M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%3D0.0175M)
Explanation:
Hello!
In this case, considering that the decomposition reaction of SO2Cl2 is first-order, we can write the rate law shown below:
![r=-k[SO_2Cl_2]](https://tex.z-dn.net/?f=r%3D-k%5BSO_2Cl_2%5D)
We also consider that the integrated rate law has been already reported as:
![[SO_2Cl_2]=[SO_2Cl_2]_0exp(-kt)](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%3D%5BSO_2Cl_2%5D_0exp%28-kt%29)
Thus, by plugging in the initial concentration, rate constant and elapsed time we obtain:
![[SO_2Cl_2]=0.0225Mexp(-2.90x10^{-4}s^{-1}*865s)](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%3D0.0225Mexp%28-2.90x10%5E%7B-4%7Ds%5E%7B-1%7D%2A865s%29)
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No mountains in the world are increasing in size
