Question

При электролизе ZnSO4 была осуществлена работа 10 кВт*час. Определите количество полученного цинка, если напряжение на зажимах ванны 4 В***На дифракционную решетку падает нормально свет. При этом максимум другого порядка для линии 0,65 мкм отвечает углу 45 градусов. Найдите угол, отвечающий максимуму третьего порядка для линии 0,5 мкм

Answer 1

m=M/(ne N[A]) I dt

I dt= A/U

 

m=M/(ne N[A]) A/U

m=0,16147/ (2 1.6*10^-19 6.02*10 ^23) 10^4 /4 =0.002 кг

 

 

d sin y = k lambda

 

sin y= k lambda /d

 

d=2*0.65*10^-6 * sqrt(2)

 

sin y=3 * 0.5*10^-6/ (2 *0.65*10^-6 *sqrt(2))=3*10/(2*11*sqrt(2))=30sqrt(2)/44~~0.9643

Answer 2

d sin y = k lambda

 

sin y= k lambda /d

 

d=2*0.65*10^-6 * sqrt(2)

 

sin y=3 * 0.5*10^-6/ (2 *0.65*10^-6 *sqrt(2))=3*10/(2*11*sqrt(2))=30sqrt(2)/44~~0.9643

Answer 3

Electrolysis of ZnSO4

To determine the amount of zinc obtained during the electrolysis of ZnSO4, we need to consider the work done and the voltage applied.

The work done during the electrolysis can be calculated using the formula:

Work (W) = Power (P) × Time (t)

In this case, the power is given as 10 kW (kilowatts) and the time is not provided. Therefore, we cannot determine the work done without knowing the time.

However, we can calculate the amount of zinc obtained using Faraday's laws of electrolysis. According to Faraday's laws, the amount of substance produced or consumed during electrolysis is directly proportional to the amount of electricity passed through the electrolyte.

The amount of electricity passed through the electrolyte can be calculated using the formula:

Electricity (Q) = Power (P) × Time (t)

In this case, the power is given as 10 kW (kilowatts) and the time is not provided. Therefore, we cannot determine the amount of electricity passed without knowing the time.

Without the time value, we cannot calculate the amount of zinc obtained during the electrolysis of ZnSO4.