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:
Average atomic mass = 63.553 amu.
Explanation:
Given data:
Abundance of Y-63 = 69.17%
Abundance of Y-65 = 100 - 69.17 = 30.83%
Atomic mass of Y-63 = 62.940 amu
Atomic mass of Y-65 = 64.928 amu
Atomic mass of Y = ?
Solution:
Average atomic mass= (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass= (62.940×69.17)+(64.928×30.83) /100
Average atomic mass = 4353.560 + 2001.730 / 100
Average atomic mass = 6355.29 / 100
Average atomic mass = 63.553 amu.
<span>I think we can use rovers, landers, orbiters, probes and satellites.
Hoped this helped</span>
<u>Answer:</u> For the given reaction, the value of 
is greater than 1
<u>Explanation:</u>
For the given chemical equation:
The expression of for above equation follows:
![K_c=\frac{[Ca^{2+}]\times [CO_3^{2-}]}{[CaCO_3}]\\\\K_c=[Ca^{2+}]\times [CO_3^{2-}]](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCa%5E%7B2%2B%7D%5D%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BCaCO_3%7D%5D%5C%5C%5C%5CK_c%3D%5BCa%5E%7B2%2B%7D%5D%5Ctimes%20%5BCO_3%5E%7B2-%7D%5D)
The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression
As, the denominator is missing and the numerator is the only part left in the expression. So, the value of will be greater than 1.
Hence, for the given reaction, the value of is greater than 1
