Adiverging lens has a focal length of 23.9 cm. an object 2.1 cm in height is placed 100 cm in front of the lens. locate the position of the image. answer in units of cm. 007 (part 2 of 3) 10.0 points what is the magnification? 008 (part 3 of 3) 10.0 points find the height of the image. answer in units of cm.

Explanation:

Given that,

Focal length of the diverging lens, f = -23.9 cm

Height of the object, h = 2.1 cm

Object distance, u = -100 cm

(a) Let v is the image distance. It can be calculated using lens formula as :

v = -19.28 cm

So, the image is located in from of the lens at a distance of 19.28 cm.

(b) The magnification of lens is given by :

m = 0.192

(c) Let h' is the height of the image. It can be calculated as :

h' = 0.403 cm

Hence, this is the required solution.

  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

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