Balance each one by adding electrons to make the charges on both sides the same:
Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag
Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:
2 Ag+ + 2 e- --> 2 Ag
Now, just add the two half reactions together, cancelling anything that's the same on both sides:
2 Ag+ + Sn --> Sn2+ + 2 Ag
And you're done.
You would know that the variable is quantitative if it shows any number to express the quantity. For example, quantitative variables are 50°C, 5 atm, 2 moles, 100 L and so on. A variable is qualitative if it expresses a relative quantity but not expressing a number. Examples would be: few, too hot, several, or even describing the characteristics of a variable. Hence, when the variable is in grams, then that would be quantitative.
Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
When a wave is moving from gamma rays to radio waves, this means that the frequency is decreasing whereas the wavelength is increasing in value.Thank you for your question. Please don't hesitate to ask in Brainly your queries.
Where’s the problem I know how to do it
