A potential difference Δ V ΔV exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 2.20 × 10 − 20 J 2.20×10−20 J of work is required to eject a positive potassium ion (K + ) (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?

Explanation:

Given

Work required to eject a positive Potassium ion

Charge on the Potassium ion

During ejection Potential difference between the inner and outer membrane exert a force opposite to the direction of motion so some amount of work is required to over come this force

Work done=Potential difference\times charge

i guess that visual aids present persuasive examples to what you are presenting, some factors would be theme (does it fit with the color, information, what you are presenting), size (can you see it, is it a big enough span of data/information), and overall look of the visual aid.

hope this : )

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.