A clock battery wears out after moving 10,400 C of charge through the clock at a rate of 0.530 mA. (a) How long (in s) did the clock run? s (b) How many electrons per second flowed?

Explanation:

Given that:

Charge Q = 10400 C

current I = 0.530 mA

From our previous knowledge, we know that charge Q can be expressed as:

Q = I×t

where

I = current and t = time

∴

replacing our values

10400 = (0.530×10⁻³) t

t = 10400/(0.530×10⁻³)

t = 19622641.51 seconds

However, the required flow of electrons per seconds can be calculated by using the formula:

number of electrons/second = current (I)/ charge on electron

= (0.530×10⁻³)/(1.6×10⁻¹⁹)

= 3.3125×10¹⁵ electrons

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz

when a device is used in a circuit in which the voltage is 81 v, the current flowing through the device is 3 a. what is the resistance of the device?

a. 27 ω

b. 40.5 a

c. 243 ω

d. 243 a

the correct answer is a. 27