b. testing the hypothesis
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
<h3>
In reference to the diagram attached </h3><h3>
Answer:</h3>
one battery, two switches, and three light bulbs
Explanation:
- A circuit is an arrangement that shows the flow of electrons from a current source such as a battery.
- The diagram shows a circuit that contains one battery, two switches, and three light bulbs.
- The switches serves to close the circuit when necessary.
- The battery supplies current to the circuit.
- The light bulbs are used for lighting.
Answer: 1.99 x 10²² molecules H2
Explanation:First we will solve for the moles of H2 using Ideal gas law PV= nRT then derive for moles ( n ).
At STP, pressure is equal to 1 atm and Temperature is 273 K.
Convert volume in mL to L:
750 mL x 1 L / 1000 mL
= 0.75 mL
n = PV/ RT
= 1 atm ( 0.75 L ) / 0.0821 L.atm/ mole.K ( 273 K)
= 3.3x10-² moles H2
Convert moles of H2 to atoms using Avogadro's Number.
3.3x10-² moles H2/ 6.022x10²³ atoms H2 / 1 mole H2
= 1.99x10²² atoms H2
