Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
Electronegativity measures how much an atom likes to pull electrons away from another one. Ionization energy measures how much an atom doesn't want to lose electrons. As an atom that wants to gain electrons will clearly not want to lose electrons, these trends are basically identical.
<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %
<u>Explanation:</u>
We are given:
A chemical compound having chemical formula of 
It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms
To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:
Mass of compound = ![[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol](https://tex.z-dn.net/?f=%5B%281%5Ctimes%2014%29%2B%285%5Ctimes%201%29%2B%281%5Ctimes%2012%29%2B%283%5Ctimes%2016%29%5D%3D79g%2Fmol)
Mass of hydrogen = 
Putting values in above equation, we get:
Hence, the percent composition by mass of hydrogen in given compound is 6.33 %
Answer:
there are 20 oxygen atoms in 4.00 moles of Dinitrogen pentoxide
Explanation:
there are 2 atoms in an oxygen molecule , so each oxygen molecules has at least 2. Dinitrogen pentoxide is N2O5, which has 7 atoms, 2 nitrogen and 5 oxygen. 1 molecule of N2O5 has 5 oxygen atoms, so 4 of then would be 20
