Answer:
pent-3-ene-1-yne
Explanation:
1 2 3 4 5
CH ≡ C - CH = CH - CH3
IUPAC name : Pent-3-ene-1-yne
ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹
According to the balanced equation,
ΣδΗ(bond breaking) = N ≡ N x 1 + Cl - Cl x 3
= 945 + 3(242)
= 1671 kJ mol⁻¹
ΣδΗ(bond making) = N - Cl x 3 x 2
= 192 x 6
= 1152 kJ mol⁻¹
δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
= 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
= 519 kJ mol⁻¹
Answer:
Initial rate of the reaction when concentration of hydrogen gas is doubled will be
.
Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
Initial rate of the reaction = R = 
![R = k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=R%20%3D%20k%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
![4.0\times 10^5 M/s=k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=4.0%5Ctimes%2010%5E5%20M%2Fs%3Dk%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)
The initial rate of the reaction when concentration of hydrogen gas is doubled : R'
![[H_2]'=2[H_2]](https://tex.z-dn.net/?f=%5BH_2%5D%27%3D2%5BH_2%5D)
![R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3](https://tex.z-dn.net/?f=R%27%3Dk%5Ctimes%20%5BN_2%5D%5BH_2%5D%27%5E3%3Dk%5Ctimes%20%5BN_2%5D%5B2H_2%5D%5E3)
![R'=8\times k\times [N_2][H_2]^3](https://tex.z-dn.net/?f=R%27%3D8%5Ctimes%20k%5Ctimes%20%5BN_2%5D%5BH_2%5D%5E3)

Initial rate of the reaction when concentration of hydrogen gas is doubled will be
.
Answer: Thomson
Explanation: It verified J. J. Thomson's work on the atomic structure.
Answer:
Yes
Explanation: Had a question like this and I said yes and got it right