Explanation:
When carbon atom tends to form single bonds then its hybridization is 
, when carbon atom tends to form double bond then its hybridization is 
and when a carbon atom is attached to a triple bond or with two double bonds then its hydridization is sp.
For example, in HCN molecule there is a triple existing between the carbon and nitrogen atom.
So, hybridization of carbon in this molecules is sp. Moreover, nitrogen atom is also attached via triple bond and it also has a lone pair of electrons. Hence, the hybridization of nitrogen atom is also sp.
Thus, we can conclude that s and p type of orbitals overlap to form the sigma bond between C and N in H−C≡N:
Answer:
* No precipitate: 
* Precipitate: 
* Precipitate: 
Explanation:
Hello!
In this case, since these all are double displacement reactions, in which the cations and anions are exchanged, we can write the resulting chemical reactions as follows:
a. LiOH and NaCl: No precipitate is formed since LiOH and NaOH are both largely soluble in water:
b. BaCl2 and Na3PO4: barium phosphate precipitate is formed because it has a large molar mass which makes it insoluble in water:
c. MgSO4 and KOH: magnesium hydroxide "milky" precipitate is formed because it is not soluble in water:
Moreover, we can relate the solubility of a substance by considering its polarity, molar mass and nature; usually, heavy substances tend to be insoluble in water as well as nonpolar compounds.
Best regards!
Answer:
52.9 KJmol-1
Explanation:
From;
log(k2/k1) = Ea/2.303 * R (1/T1 - 1/T2)
The temperatures must be converted to Kelvin;
T1 = 25° C + 273 = 298 K
T2= 35°C + 273 = 308 K
R= gas constant = 8.314 JK-1mol-1
Substituting values;
log 2 = Ea/2.303 * 8.314 (1/298 - 1/308)
Ea = 52.9 KJmol-1
