Answer:
130 Liters
Explanation:
if 1 mol is 22.4 L, then 5.8 mol is 130 L (129.92 but use sig figs)
Answer:

Explanation:
Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.
Hence, the rate law is:
![r=d[A]/dt=-k[A]](https://tex.z-dn.net/?f=r%3Dd%5BA%5D%2Fdt%3D-k%5BA%5D)
Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:
![[A]=[A]_0e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_0e%5E%7B-kt%7D)
You know [A]₀, k, and t, thus you can calculate [A].
![[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}](https://tex.z-dn.net/?f=%5BA%5D%3D0.548M%5Ctimes%20e%5E%7B-3.6%5Ccdot%2010%5E%7B-4%7D%2Fs%5Ctimes99.2s%7D)
![[A]=0.529M](https://tex.z-dn.net/?f=%5BA%5D%3D0.529M)
Oxygen carbon and hydrogen
Answer:
3.6 times 10^4
Explanation:
Scientific notation is between 1-9. So, we move 36000 to 4 decimal places. SO it would be 3.6 times 10^4. Scientific Notation always has the base of 10 . Enjoy :)
The proper name for the following alkyl side group where the main carbon chain is denoted with a squiggly line is isopropyl.
In natural chemistry, an alkyl substituent is an alkane missing one hydrogen. The term alkyl is intentionally unspecific to include many viable substitutions. An acyclic alkyl has the overall formulation of CₙH₂ₙ₊₁.
An alkyl is a purposeful institution of an organic chemical that includes only carbon and hydrogen atoms, that are organized in a chain. Examples include methyl CH3 (derived from methane) and butyl C2H5 (derived from butane). they may be now not located on their own however are discovered attached to different hydrocarbons.
what is an alkyl group? Alkyl group is shaped through putting off a hydrogen atom from the molecule of alkane. Alkanes are quite regularly represented as R-H and here R stands for alkyl group. the overall method of the alkyl organization is CₙH₂ₙ₊₁. The smallest alkyl organization is CH3 referred to as methyl.
Learn more about alkyl group here:- brainly.com/question/14272539
#SPJ4