A potential difference Δ V exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 1.00 × 10 − 20 J of work is required to eject a positive potassium ion (K + ) from the interior of the cell, what is the magnitude of the potential difference (in millivolts) between the inner and outer surfaces of the cell?

The magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0625 V

Explanation:

Given that,

Work done

We need to calculate the magnitude of the potential difference between the inner and outer surfaces of the cell

Using formula of wok done

Where,

V = volt

W = work done

q = charge

Put the value into the formula

Hence, The magnitude of the potential difference between the inner and outer surfaces of the cell is 0.0625 V

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.

  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