Its D. combustibillaty
A. density is how hard or densely the substances molecules are together
B.boiling point is the point at which bonds that hold molecules are broken and so the substances boils
C. malleability is how easily a substance can be stretched until it breaks
D. is a chemical property of how much chemical energy is stored in a substance for it to combust
hope that helps
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Answer:
0.6749 M is the concentration of B after 50 minutes.
Explanation:
A → B
Half life of the reaction = 
Rate constant of the reaction = k
For first order reaction, half life and half life are related by:
Initial concentration of A = ![[A]_o=0.900 M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.900%20M)
Final concentration of A after 50 minutes = ![[A]=?](https://tex.z-dn.net/?f=%5BA%5D%3D%3F)
t = 50 minute
![[A]=[A]_o\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_o%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}](https://tex.z-dn.net/?f=%5BA%5D%3D0.900%20M%5Ctimes%20e%5E%7B-0.02772%20min%5E%7B-1%7D%5Ctimes%2050%20minutes%7D)
[A] = 0.2251 M
The concentration of A after 50 minutes = 0.2251 M
The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M
0.6749 M is the concentration of B after 50 minutes.
If it loses an electron, it will become an ion.
The answer should be B: -3
Given the percentage composition of HC as C → 81.82 % and H → 18.18 %
So the ratio of number if atoms of C and H in its molecule can will be:
C : H = 81.82 12 : 18.18 1 C : H = 6.82 : 18.18 = 6.82 6.82 : 18.18 6.82 = 1 : 2.66 ≈ 3 : 8
So the Empirical Formula of hydrocarbon is:
C 3 H 8
As the mass of one litre of hydrocarbon is same as that of C O 2 The molar mass of the HC will be same as that of C O 2 i.e 44 g mol
Now let Molecular formula of the HC be ( C 3 H 8 ) n
Using molar mass of C and H the molar mass of the HC from its molecular formula is:
( 3 × 12 + 8 × 1 ) n = 44 n So 44 n = 44 ⇒ n = 1
Hence the molecular formula of HC is C 3 H 8
Does that help?
