1. When an element has a positive charge :

Given the following partial (valence-level) electron configurations,<br> (a) identify each element,

Kamila [148]

The arrangement of electrons within an atom is called the electronic configuration and the electrons are filled up according to the energy of the levels as: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f.

<h3>What is the difference between electron configuration and valence electron configuration?</h3>

Electron configuration is the number of electrons that are present in the atom and is repressed through the different sub-shells. So for example, the electron configuration for oxygen is 1s^2 2s^2 2p^4. While the valence electrons only refer to the electrons in the outermost shell of the atom.

<h3>What is a partial orbital diagram?</h3>

A partial orbital diagram shows only the highest energy sublevels being filled. Al (Z = 13) 1s22s22p63s23p1. A condensed electron configuration has the element symbol of the previous noble gas in square brackets. Al has the condensed configuration [Ne]3s23p1.

Learn more about electronic configuration here:

<h3>brainly.com/question/26084288</h3><h3 /><h3>#SPJ4</h3>

4 0

1 year ago

Please help i will give 20 points

NNADVOKAT [17]

Answer:

It's A., thats conduction and its a heat transfer

Explanation:

6 0

2 years ago

Use the following half-reactions to construct a voltaic cell:

velikii [3]

<u>Answer:</u> The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

<u>Explanation:</u>

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

4 0

2 years ago

Please help

Kay [80]

Answer:

Can you post the pic to it so I can give you more info?

Explanation:

5 0

3 years ago

Which balanced equation represents a fusion reaction?

lyudmila [28]

<span> </span><span>Fusion reaction is a type of nuclear reaction where two or more nuclei combine or collide to form an element with a higher atomic number. This happens when the collision is in a very high speed. In this process, some of the matter of the fusing nuclei is converted to energy.</span>

3 0

3 years ago