To
determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample
of the compound with the given composition. Then, we calculate for the number
of moles of each element. We do as follows:<span>
mass moles
C 56.79 4.73
H 6.56 6.50
O 28.37 1.77
N 8.28 0.59
Dividing the number of moles of each element with
the smallest value, we will have the empirical formula:
</span> moles ratio
C 4.73 / 0.59 8
H 6.50 / 0.59 11
O 1.77 / 0.59 3
N 0.59 / 0.59 1<span>
</span><span>
The empirical formula would be C8H11O3N.</span>
Explanation:
your answer is Kelvin because it is the SI unit of temperature
Answer:
0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.
Answer:
(a)
(b)
Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
![\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3Dkt%20%2B%5Cfrac%7B1%7D%7B%5BNOBr%5D_0%7D%5C%5C%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3D%5Cfrac%7B0.8%7D%7BM%2As%7D%2A22s%2B%5Cfrac%7B1%7D%7B0.086M%7D%3D%5Cfrac%7B29.3%7D%7BM%7D%5C%5C)
![[NOBr]=\frac{1}{29.2/M}=0.0342M](https://tex.z-dn.net/?f=%5BNOBr%5D%3D%5Cfrac%7B1%7D%7B29.2%2FM%7D%3D0.0342M)
(b) Now, for a second-order reaction, the half-life is computed as shown below:
![t_{1/2}=\frac{1}{k[NOBr]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BNOBr%5D_0%7D)
Therefore, for the given initial concentrations one obtains:
Best regards.
Answer:
100.09 amu is the answer.
