Answer:
Double displacement chemical reaction.
Explanation:
Hello,
In this case, for the given reaction we firstly balance it:
Then, since the lithium and potassium cations are being exchanged to each other from bromide and carbonate, we are talking about a double displacement chemical reaction.
Best regards.
U^235Number of protons = 92 pNumber of electrons = 92 eNumber of neutrons = 235 – 92 = 143n
U^238Number of protons = 92pNumber of electrons = 92eNumber of neutrons = 238 – 92 = 146n
Electron configuration of U atom U = 92 U = [Rn] 5f^6 6d^0 7s^0 U = [Rn] 5f^36d^17s^2, 7s is completely filled and others are less than half filled.
(_92^238)U Decays to (_90^234)ThIt loses 2 protons, 2 electrons and loses 2 neutrons Th = [Rn] 6d^2 7s^2 There is no electron in 5f subshell and 6d contains 2e^-, 7s completely filled
Hey there!:
Isotopes : abundance :
46 Ti 8.0%
47 Ti 7.8 %
48 Ti 73.4 %
49 Ti 5.5 %
50 Ti 5.3 %
Weighted average = ∑ Wa * % / 100
Therefore:
( 46 * 8.0) + (47 * 7.8 ) + (48 * 73.4 ) + ( 49 * 5.5 ) + ( 50*5.3 ) / 100 =
4792.3 / 100
= 47.923 a.m.u
Hope that helps!
Answer:
True
Explanation:
Pi bonds are formed using p orbitals. Atoms that have p orbitals have 3 of them. In an sp² hybridized atom, only 2 of the 3 p orbitals are being used in the sp² orbitals, leaving one p orbital free for pi bonding. Therefore, only one pi bond can be formed to an sp² hybridized atom.
Answer: The standard enthalpy change is -607kJ
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{C_6H_6}\times \Delta H_f^0_{(C_6H_6)}]-[n_{C_2H_2\times \Delta H_f^0_{(C_2H_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BC_6H_6%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_6H_6%29%7D%5D-%5Bn_%7BC_2H_2%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_2H_2%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(1\times 83)]-[(3\times 230)]=-607kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%2083%29%5D-%5B%283%5Ctimes%20230%29%5D%3D-607kJ)
The standard enthalpy change is -607kJ
