Answer:
At equal concentration of HBCG and BCG^-, the colour is green. This colour first appears at pH = 3.8
Explanation:
HBCG is an indicator that is prepared by dissolving the solid in ethanol.
Since
Ka=[BCG−][H3O+][HBCG]When [BCG-] = [HBCG], then Ka = [H3O+].
If pH = 3.8
Ka= [H3O+] = -antilog pH = -antilog (3.8)
Ka= 1.58 ×10^-4
Those are the correct steps, young chemist. Don't be discouraged by an insane answer.
Answer:
Nsc=30
Explanation:
The solid angle subtended by the counter is
d=0.1mm2/(1cm)2=(10)3 sr
d=(10)10x10.5gcm3x10-4cm/108x1.6610x10-24gx0.510-24cm2x10-3=30
Gas particles spread out to fill a container evenly, unlike solids and liquids. Gas is a state of matter that has no fixed shape and no fixed volume. Gases have lower density than other states of matter, such as solids and liquids. Hence above statement is FALSE.
Answer: A. 4 unpaired electrons
B. Zero unpaired electrons
C. 1 unpaired electron
D. 5 unpaired electrons
E. Zero unpaired electrons
Explanation:
In A, Oxidation state of Co is +3 and Electronic configuration is [Ar]3d6
F is weak field ligand, causes no pairing of Electrons hence it has 4 unpaired electrons&2 are paired in t2g orbitals (dXY)
In B , Mn is in +3 with electronic configuration 3d4&CN is a strong field ligand hence causes pairing of Electrons hence it results 0 unpaired electron
In C, Mn is in +2 with electronic configuration 3d5 sinceCN is a strong field ligand hence it leaves one unpaired electron
In D, Mn is in+2 with 3d5& five unpaired electrons since cl is a weak field ligand causes no pairing.
In E, Rh is in +3, with d6 configuration and it is a low spin complex hence pairing of Electrons involved. So it leaves zero unpaired electrons.
