Answer:
okay let me know when i can help, ill be happy too help
The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
<span>The volume in liters is </span>v=3.7xL
The image of the bonds are missing, so i have attached it.
Answer:
A) - Sigma bond
-Sp³ and Sp³
- None
B) - Sigma and pi bond
- Sp² of C and p of O
- p of C and P of O
Explanation:
A) For compound 1;
- the molecular orbital type is sigma bond due to the end-to-end overlapping.
- Atomic orbitals in the sigma bond will be Sp³ and Sp³
- Atomic orbitals in the pi bond would be nil because there is no pi bond.
B) For compound 2;
- the molecular orbital type is sigma and pi bond
-Atomic orbitals in the sigma bond would be Sp² of C and p of O
- The Atomic orbitals in the pi bond will be; p of C and p of O
Answer:
Type of hybridisation
Explanation:
For example , saturated hydrocarbons have sp3 hybridisation while unsaturated hydrocarbons have either sp2 or sp hybridisation.
Example: Ethane( C2H6) has sp3 hybridisation and ethene(C2H4) has sp2 hybridisation and Ethyne(C2H2) has sp hybridization.
