Answer:
A. Coefficients
Explanation:
that's the number in front of the molecules
The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.
Answer :
The basic rules for naming of hydrocarbons are :
First select the longest possible carbon chain.
The longest possible carbon chain should include the carbons of double or triple bonds.
The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne.
The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.
The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.
If two or more similar alkyl groups are present in a compound, the prefixes di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.
Benefit you and your life in many ways!
Answer:
D) 0.86 M
Explanation:
Given that:
The rate constant, k = 6.7×10⁻⁴ s⁻¹
Initial concentration [A₀] = 1.33 M
Time = 644 s
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
So,
![[A_t]=1.33\times e^{-6.7\times 10^{-4}\times 644}](https://tex.z-dn.net/?f=%5BA_t%5D%3D1.33%5Ctimes%20e%5E%7B-6.7%5Ctimes%2010%5E%7B-4%7D%5Ctimes%20644%7D)
![[A_t]=0.86 M](https://tex.z-dn.net/?f=%5BA_t%5D%3D0.86%20M)
