Answer:
1.24 x 10 to the 5 ev = 124,000 ev its B
Explanation:
E = hc/lambda = 1.24 ev-micrometer/1.0x10 to the -5 micrometers = 1.24 x 10 to the 5 ev = 124,000 ev
h = Planck's constant = 6.626 × 10 to the -34 joule·s
c = speed of light = 2.998 × 10 to the 8 m/s
lambda is the given wavelength
E is the desired photon energy
Answer:
a) Batteries and fuel cells are examples of galvanic cell
b) Ag-cathode and Zn-anode
c) Cell notation: Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)
Explanation:
a) A galvanic cell is an electrochemical cell in which chemical energy is converted to electrical energy. The chemical reaction which drives a galvanic cell is a redox reaction i.e. a reduction-oxidation process.
A typical galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs. Batteries and fuel cells are examples of galvanic cells.
b) The nature of the electrode that will serve as an anode or cathode depends on the value of the standard reduction potential (E⁰) of that electrode. The electrode with a higher or more positive the value of E⁰ serves as the cathode and the other will function as an anode.
In the given case, the E⁰ values from the standard reduction potential table are:
E⁰(Zn/Zn2+) = -0.763 V
E°(Ag/Ag+)=+0.799 V
Therefore, Ag will be the cathode and Zn will be the anode
c) In the standard cell notation, the anode half cell is written on the left followed by the salt bridge '||' and finally the cathode half cell to the right.
Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)
From the given problem, a limit on the depression of a building is placed at 20 centimeters. To solve how many floors can be safely added, a quantity of how many cm will a building sink for each floor that is added is needed. Unfortunately, it is not found anywhere in the problem. However, we can provide a formula to solve for the depression. This is as follows:
Building depression < 20 cm
Building depression = (cm depression per floor) * (no. of floors)
Answer:
The size of the image is 1.04 m.
Explanation:
Given that,
Height of object = 2.40 m
Distance of object = 2.60 m
Radius of curvature =4.00 m
Focal length 
We need to calculate the image distance
Using mirror formula
We need to calculate the height of the image
Using formula of magnification
Put the value into the formula
Hence, The size of the image is 1.04 m
