The photoelectric effect occurs when light shines on a metal. ... Light of any frequency will cause electrons to be emitted.
Answer:
6.4 × 10^-10 M
Explanation:
The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).
CaF2 will dissociate as follows:
CaF2 ⇌Ca2+ + 2F-
1 mole of Calcium ion (x)
2 moles of fluorine ion (2x)
NaF will also dissociate as follows:
NaF ⇌ Na+ + F-
Where Na+ = 0.25M
F- = 0.25M
The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M
Ksp = {Ca2+}{F-}^2
Ksp = {x}{0.25}^2
4.0 × 10^-11 = 0.25^2 × x
4.0 × 10^-11 = 0.0625x
x = 4.0 × 10^-11 ÷ 6.25 × 10^-2
x = 4/6.25 × 10^ (-11+2)
x = 0.64 × 10^-9
x = 6.4 × 10^-10
Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M
Answer:
what can you speak English
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
