Answer:
A1+3
Explanation:
A1+3 is the most common ion of aluminum.
First we assume that the compound containing only C,H,and
O is combusted completely in the presence of excess oxygen, so that the only
things that can be produced are water and carbon dioxide.
From there we should back calculate the amount of
Hydrogen that is in the original sample by taking all of the hydrogen in the 0.239g
to came from the organic compound.
And since we know that the original mass of the sample
was .100g, we can also easily get a mass % H by taking the mass Hydrogen
calculated over the total original mass (.100 g)
So that:
0.239g H2O / (18.01 g/mol) = .01327 moles H20
.01327 Moles H20 * 2.02g H (per every mole H2O) = .0268g
H initially present in the sample
.0268g H / .100g sample = 26.8% H by mass
When there are 14c-lable uracil that are added to the growth medium of cells, the macromolecules that will be labled are RNA. Uracil is a nucleobase that make up the DNA or the RNA. In RNA, uracil binds with other nucleobase (adenine) through hydrogen bonds.
Answer:
Explanation:
![[A]_0](https://tex.z-dn.net/?f=%5BA%5D_0)
= Initial concentration = 1.28 M
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= Final concentration = ![0.17[A]_0](https://tex.z-dn.net/?f=0.17%5BA%5D_0)
k = Rate constant = 0.0632 s
t = Time taken
For first order reaction we have the relation
![kt=\ln\dfrac{[A]_0}{[A]}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{[A]}}{k}\\\Rightarrow t=\dfrac{\ln\dfrac{[A]_0}{0.17[A]_0}}{0.0632}\\\Rightarrow t=28.037\ \text{s}](https://tex.z-dn.net/?f=kt%3D%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B%5BA%5D%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B%5BA%5D%7D%7D%7Bk%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B%5Cln%5Cdfrac%7B%5BA%5D_0%7D%7B0.17%5BA%5D_0%7D%7D%7B0.0632%7D%5C%5C%5CRightarrow%20t%3D28.037%5C%20%5Ctext%7Bs%7D)
Time taken to reach the required concentration would be .
the answer to the first one is D and the second one is the first one
