Answer:
- 1.602 x 10⁻¹⁹coulombs
Explanation:
Charge on individual oil droplet would be multiple of charge on one electron . So we will find out the minimum common factor of given individual charges that is the LCM of all the charges given.
LCM of given charges like 3.204 , 4.806 ,8.01 and 14.42 . We have neglected the power of ten( 10⁻¹⁹) because it is already a common factor to all.
The LCM is 1.602 . So charge on electron is 1.602 x 10⁻¹⁹.
Moles of Zn: 26 / 65 = 0.4
Moles of S: 12.8 / 32 = 0.4
Molar ratio of Zn : S = 1 : 1
Empircal formula: ZnS
The answer is C
Answer:
The chemist would require to use 43.43 grams.
Explanation:
In order to solve this problem we need to know<u> how much do 0.550 moles of selenium weigh</u>. To do that we use selenium's<em> molar mass </em>and multiply it by the given number of moles:
- 0.550 mol * 78.96 g/mol = 43.43 g
The chemist would require to use 43.43 grams.
Answer: silicon Si, Germanium GE
Explanation:
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
