a) The electronic configuration of arsenic is1s2 2s2 2p6 3s2 3p6 3.pdf
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a) The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [Ground state configuration] The number of electron in each subshell is 2 8 18 5 (K,L,M,N) b) one permissible quantum number would be 4,0,0,-1/2 c) paramagnetic element are those element that have unpaired electron in the outer most shell and diamagnetic are the one that have paired electron. Now in Arsenic, in the 4p there are 3 electron which are distributed among the three sub-orbitals of P (Px, Py and Pz) so they have 3 unpaired electron in 4p. Hence it is paramagnetic. d) Now the compound Na3As, AsCl3 and AsF3, if we look at it, in all the compound As combine with 3 atoms of another element either by covalent or ionic bond. This is because As requires 3 electron to attend the noble gas configuration or the octet state If we consider Na3As, Na will loses 1 e forming Na+ Na======>Na+ + 1e , Since there are 3 Na, so there will be 3 electron. These 3 e are taken up by As. So the configuration becomes 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6. Hence it has attain octet state. Therefore the ground state is consistent. Solution a) The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [Ground state configuration] The number of electron in each subshell is 2 8 18 5 (K,L,M,N) b) one permissible quantum number would be 4,0,0,-1/2 c) paramagnetic element are those element that have unpaired electron in the outer most shell and diamagnetic are the one that have paired electron. Now in Arsenic, in the 4p there are 3 electron which are distributed among the three sub-orbitals of P (Px, Py and Pz) so they have 3 unpaired electron in 4p. Hence it is paramagnetic. d) Now the compound Na3As, AsCl3 and AsF3, if we look at it, in all the compound As combine with 3 atoms of another element either by covalent or ionic bond. This is because As requires 3 electron to attend the noble gas configuration or the octet state If we consider Na3As, Na will loses 1 e forming Na+ Na======>Na+ + 1e , Since there are 3 Na, so there will be 3 electron. These 3 e are taken up by As. So the configuration becomes 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6. Hence it has attain octet state. Therefore the ground state is consistent..
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![a) The electronic configuration of arsenic is
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [Ground state configuration]
The number of electron in each subshell is 2 8 18 5 (K,L,M,N)
b) one permissible quantum number would be 4,0,0,-1/2
c) paramagnetic element are those element that have unpaired electron in the outer most shell
and diamagnetic are the one that have paired electron. Now in Arsenic, in the 4p there are 3
electron which are distributed among the three sub-orbitals of P (Px, Py and Pz) so they have 3
unpaired electron in 4p. Hence it is paramagnetic.
d) Now the compound Na3As, AsCl3 and AsF3, if we look at it, in all the compound As
combine with 3 atoms of another element either by covalent or ionic bond. This is because As
requires 3 electron to attend the noble gas configuration or the octet state
If we consider Na3As, Na will loses 1 e forming Na+
Na======>Na+ + 1e , Since there are 3 Na, so there will be 3 electron. These 3 e are taken up
by As.
So the configuration becomes 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6. Hence it has attain octet state.
Therefore the ground state is consistent.
Solution
a) The electronic configuration of arsenic is
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [Ground state configuration]
The number of electron in each subshell is 2 8 18 5 (K,L,M,N)
b) one permissible quantum number would be 4,0,0,-1/2
c) paramagnetic element are those element that have unpaired electron in the outer most shell
and diamagnetic are the one that have paired electron. Now in Arsenic, in the 4p there are 3
electron which are distributed among the three sub-orbitals of P (Px, Py and Pz) so they have 3
unpaired electron in 4p. Hence it is paramagnetic.
d) Now the compound Na3As, AsCl3 and AsF3, if we look at it, in all the compound As
combine with 3 atoms of another element either by covalent or ionic bond. This is because As](https://image.slidesharecdn.com/atheelectronicconfigurationofarsenicis1s22s22p63s23p63-230412105232-73c39900/85/a-The-electronic-configuration-of-arsenic-is1s2-2s2-2p6-3s2-3p6-3-pdf-1-320.jpg)
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The atom with less electronegative is central atom and the atoms with more electronegtive are terminal atoms.
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Essay answers are limited to about 500 words (3800 characters maximum, including spaces).
3800 Character(s) remaining
Solution
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S ( Z = 16 ) 1st excited state electronic configuration is,
1s2 s2 2p6 3s2 3p3 3d1
S in its first exicted state undergoes sp2 hybirdisation due ot which it has 1 paired sp2 hybrid orbital and 2 unpaired sp2 hybrid orbitals. And also S has one p and one d orbital with unpaired electrons.
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The reamining unpaired pure p-orbitals of S overlaps with unpaired p - orbital of each oxygen perpendicular to the internuclear axis to form one pi bonds with each oxygen.
Still S ia remained with one lone pair of electons.
Due to this SO 2 is in bent shape and has the bond angle < 120 0
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Unit three notes s05
This document provides a brief history of the atomic theory, from early philosophers' ideas to the modern quantum model. It describes Democritus' idea of indivisible atoms, Dalton's model of atoms as solid spheres, Thomson's discovery of electrons, Rutherford's gold foil experiment showing the nucleus, Bohr's planetary model, and the modern electron cloud/quantum model with protons and neutrons in the nucleus and electrons in clouds outside. It also discusses atomic structure, isotopes, ions, atomic mass, and calculating atomic properties using atomic number and mass number.
4 The Atom & Electronic Configuration
The document summarizes the Bohr model of the atom and electronic configuration. It explains that electrons can only occupy certain energy levels or orbits around the nucleus, and that electrons in orbits further from the nucleus require more energy to remove. It also discusses how the ionization energy increases with each subsequent electron removed from atoms and ions. Finally, it introduces how elements are written with their full electron configuration including shells and subshells.
Non metals generally want to gain electrons and a.pdf
Non metals generally want to gain electrons and a negative charge and metals
generally want to donate electrons and gain a positive charge. In this case, you need to look at
these elements on the periodic table and look at their common oxidation states (usually below the
element on the table). As will probably be As3- Se2- Br- K+ Ca2+ Ga3+ These explanations
might be beyond the chemistry level you\'re in right now, but I\'m not sure so I\'ll give brief
explanations anyway. Arsenic can be tricky because it has multiple oxidation states. It would
want to gain 3 electrons to fill its 4p orbital which in its base state has only 3/6 electrons.
Gallium is tricky as well. It is willing to give up 3 electrons. This can be explained by filling
orbital shells too. It only has 1 4p electron, but if it gives up 3 then the 3d shell can give up some
electrons and put 1 electron in all of its 4p and 3d orbitals, which is more stable than having just
1/6 electrons in its 4p orbital. Now, drawing the dots on the element should be a little easier.
Look at their normal amount of valence electrons and then add electrons for a negative charge
and subtract electrons for a positive charge.
Solution
Non metals generally want to gain electrons and a negative charge and metals
generally want to donate electrons and gain a positive charge. In this case, you need to look at
these elements on the periodic table and look at their common oxidation states (usually below the
element on the table). As will probably be As3- Se2- Br- K+ Ca2+ Ga3+ These explanations
might be beyond the chemistry level you\'re in right now, but I\'m not sure so I\'ll give brief
explanations anyway. Arsenic can be tricky because it has multiple oxidation states. It would
want to gain 3 electrons to fill its 4p orbital which in its base state has only 3/6 electrons.
Gallium is tricky as well. It is willing to give up 3 electrons. This can be explained by filling
orbital shells too. It only has 1 4p electron, but if it gives up 3 then the 3d shell can give up some
electrons and put 1 electron in all of its 4p and 3d orbitals, which is more stable than having just
1/6 electrons in its 4p orbital. Now, drawing the dots on the element should be a little easier.
Look at their normal amount of valence electrons and then add electrons for a negative charge
and subtract electrons for a positive charge..
Ch. 7 electron configuration
- Electron configuration describes the arrangement of electrons in an atom's orbitals and energy levels according to 4 quantum numbers.
- The Pauli Exclusion Principle states that no two electrons can have the same set of 4 quantum numbers.
- Elements are arranged on the periodic table based on their electron configurations, with groups reflecting which orbitals are being filled.
- A noble gas configuration can be used to abbreviate long electron configurations by writing the nearest noble gas followed by the remaining electrons.
1. struktur atom, sistem periodik unsur, ikatan kimia
1) The document discusses the structure of atoms, including subatomic particles like protons, neutrons, and electrons. It describes the historical models of the atom from Democritus to the modern quantum mechanical model.
2) Notations for elements are introduced, including atomic number, mass number, and electron configuration. Various terms like isotopes, isobars, and isotones are defined.
3) Electron configuration is explained according to both the Aufbau principle and quantum numbers like principal, azimuthal, magnetic, and spin quantum numbers. Rules for writing electron configurations are provided.
Quantum numbers
Erwin Schrodinger developed wave equations to describe the probabilities of electron locations in atoms, since electrons cannot be precisely located. He used sets of integers called quantum numbers to describe possible electron locations, including: 1) The principal quantum number representing the electron's energy level, 2) The azimuthal quantum number representing the subshell (s, p, d, f), and 3) The magnetic quantum number representing the number of orbitals in the subshell. The spin quantum number represents the electron's spin as either +1/2 or -1/2. Quantum numbers are used to describe electron configurations of elements.
Valence Bond Theory
Valence Bond Theory (VBT) explains the nature of metal-ligand bonding in complex compounds. VBT assumes that the central metal atom forms hybrid orbitals with its outer shell orbitals that overlap with ligand orbitals to form sigma bonds. There are two types of hybridization that result in octahedral geometry: d2sp3 for "inner-orbital" complexes with paired electrons, and sp3d2 for "outer-orbital" complexes with unpaired electrons. Examples that form via d2sp3 hybridization, making them inner-orbital complexes, include ferricyanide ([Fe(CN)6]3-), ferrocyanide ([Fe(CN)6]4-
Quantum Numbers
This document discusses quantum numbers and their role in describing the size, shape, and orientation of atomic orbitals. It explains that there are four quantum numbers - principal, angular, magnetic, and spin. The principal quantum number determines the electron shell or energy level, while the angular and magnetic quantum numbers further specify the subshell and orbital within that subshell. The spin quantum number refers to the spin of the electron. Factors that influence ionization energy such as atomic radius, nuclear charge, and electron shielding are also summarized.
Electron configuration cheat sheet
This document provides a summary of key concepts for electron configuration in high school chemistry, including:
1) Electrons fill subshells according to the aufbau principle to achieve lowest energy, with Hund's rule specifying that electrons occupy each orbital singly before pairing up.
2) The four subshells are s, p, d, and f, with set numbers of orbitals and maximum electrons in each. Valence electrons are in the outermost shell.
3) Electron configuration can be written using boxes and arrows, spectroscopic notation, or noble gas notation, with examples provided.
ATOMIC AND MOLECULAR STRUCTURE
This document provides a 3-paragraph summary of the contents of an inorganic pharmaceutical chemistry course titled "Atomic and Molecular Structure/Complexation". The course will cover 9 topics including atomic and molecular structure, electrolytes, essential and trace ions, non-essential ions, gastrointestinal agents, antacids, protective adsorbents, radiopharmaceutical preparations, and radio opaque and contrast media. It will examine the fundamental atomic structure including protons, neutrons, electrons, isotopes and quantum mechanics. The course will also cover electronic configurations, orbitals, quantum numbers, ionization, periodic trends, and transition metal ions. The instructor is Hayder R. Fadhil and the course is listed as the 1st semester, 3
Chapter 5 Electrons in Atoms
1) The document discusses the quantum mechanical model of the atom, including atomic orbitals and electron configurations.
2) It describes how electrons can occupy different atomic orbitals based on their energy levels and how the electron configuration notation is used to show the filling of these orbitals.
3) It also mentions exceptions to the expected electron configuration filling order that result in more stable half-filled or filled orbitals.
QUESTION 4 Manganese has an elec tron configuration 1s 2s 2p 3s 3p 4s.docx
QUESTION 4 Manganese has an elec tron configuration 1s 2s 2p 3s 3p 4s 3d3. Which electrons are considered valence electrons? 4s 3d O 3s23p 63d 3d 4s QUESTION 5 lling diagrams below correctly represents the electron configuration for the valence electrons of oxygen, O? 2s 2p 2s 2p 3s 3p 2s 2p 3s 3p
Solution
4)
valence shell is outermost shell
Here outermiost shell is n=4
SO, 4s2 is valance shell
Answer: 4s2
5)
Electronic configuration is 1s2 2s2 2p4
Noble gas having lesser electron than O and closest to it is Helium (He)
Number of electron in He = 2
electronic configuration of He is: 1s2
So, electronic configuration of O in noble gas form is: [He] 2s2 2p4
So, it should have 2 electrons in 2s and 4 electrons in 2p
Remember pairing of electrons do not start until all orbitals are singly occupied
So, option 4 is wrong
Answer: 1st option
.
Molecular orbital theory(mot) of SF6/CO2/I3-/B2H6
1) Molecular orbital theory views a molecule as delocalized molecular orbitals formed from linear combinations of atomic orbitals. Bonding molecular orbitals are lower in energy due to constructive interference, while antibonding orbitals are higher in energy due to destructive interference.
2) The document provides examples of applying molecular orbital theory to SF6, CO2, B2H6, and I3- molecules. It describes the atomic orbitals and molecular orbitals formed, including bonding, antibonding, and non-bonding orbitals, and explains how the molecular orbitals rationalize the electronic structures and bonding patterns in these molecules.
Unit 1.pdf
1. The document provides information on molecular structure and bonding theories including atomic and molecular orbitals, linear combination of atomic orbitals, molecular orbital diagrams of diatomic molecules like N2, O2 and F2, π molecular orbitals of butadiene and benzene, and crystal field theory.
2. Key concepts covered include how molecular orbitals are formed from the overlap and combination of atomic orbitals, bonding and anti-bonding molecular orbitals, and how molecular orbital diagrams can be used to explain bonding properties.
3. Crystal field theory is introduced as explaining the color, magnetic properties and other characteristics of crystalline substances based on interactions between the d-orbitals of metal ions and ligand ions or molecules
2016 Topic 2: Electron Configuration
This document discusses electron configuration and how it is described using quantum numbers. It explains the main concepts like energy shells (n), subshells (s, p, d, f), orbitals and how electrons fill these according to the Aufbau principle and Hund's rule. Examples are provided to show the electron configurations of elements like hydrogen, helium, lithium and how the configurations change as the atomic number increases. Practice problems are included at the end to determine configurations of additional elements.
Atomic structure
This document discusses atomic structure and properties. It begins by defining the three main subatomic particles - protons, neutrons, and electrons - and their relative masses and charges. It then explains atomic mass units and discusses quantum numbers like principal, azimuthal, magnetic, and spin quantum numbers. The rest of the document covers electron configurations, orbital diagrams, atomic orbitals and subshells, ionization energy, electronegativity, and how these properties relate to an element's position on the periodic table.
Similar to a) The electronic configuration of arsenic is1s2 2s2 2p6 3s2 3p6 3.pdf (20)
Based on differences in electronegativity- how would you characterize.docx
Based on differences in electronegativity- how would you characterize.docx
Non metals generally want to gain electrons and a.pdf
Non metals generally want to gain electrons and a.pdf
1. struktur atom, sistem periodik unsur, ikatan kimia
1. struktur atom, sistem periodik unsur, ikatan kimia
QUESTION 4 Manganese has an elec tron configuration 1s 2s 2p 3s 3p 4s.docx
QUESTION 4 Manganese has an elec tron configuration 1s 2s 2p 3s 3p 4s.docx
More from dilipanushkagallery
1) a)Solution 1) a).pdf
Este documento no contiene información sustancial para resumir en 3 oraciones o menos. Parece estar incompleto o vacío.
We need the averages obtained.SolutionWe need the averages obt.pdf
The document requests that averages be obtained but provides no other context or details. It consists of a single sentence repeated twice stating that averages are needed but does not specify what is being averaged or for what purpose.
Please rateAnswer The accuracy was poor, but the precision was go.pdf
Please rate
Answer: The accuracy was poor, but the precision was good. All of the answers were very
different from the true value, but were very similar to each other.
Solution
Please rate
Answer: The accuracy was poor, but the precision was good. All of the answers were very
different from the true value, but were very similar to each other..
resultreceptoreffectorPATELLAR REFLEXa stretch reflexGolgi.pdf
result
receptor
effector
PATELLAR REFLEX
a stretch reflex
Golgi tendon organ in quadriceps femoris
quadriceps femoris & hamstrings
Hitting the patellar tendon with a reflex sledge just underneath the patella extends the muscle
axle in the quadriceps muscle. This creates a flag which heads out back to the spinal line and
neural connections (without interneurons) at the level of L4 in the spinal line, totally autonomous
of higher focuses. From that point, an alpha engine neuron leads an efferent motivation back to
the quadriceps femoris muscle, activating withdrawal. This withdrawal, composed with the
unwinding of the opposing flexor hamstring muscle causes the leg to kick. This is a reflex of
proprioception which keeps up stance and adjust, permitting to keep one\'s adjust with little
exertion or cognizant thought.
The patellar reflex is a clinical and exemplary case of the monosynaptic reflex curve. There is no
interneuron in the pathway prompting to withdrawal of the quadriceps muscle. Rather the bipolar
tactile neuron neurotransmitters specifically on an engine neuron in the spinal string. Be that as it
may, there is an inhibitory interneuron used to unwind the hostile hamstring muscle (Reciprocal
innervation).
This trial of an essential programmed reflex might be impacted by the patient intentionally
hindering or overstating the reaction; the specialist may utilize the Jendrassik move as a
diversion or redirection so as to guarantee a more legitimate reflex test.
ACHILLES REFLEX
stretch reflex
muscle spindle
gastrocnemius
The lower leg jolt reflex, otherwise called the Achilles reflex, happens when the Achilles
ligament is tapped while the foot is dorsi-flexed. A positive result would be the twitching of the
foot towards its plantar surface. Being a profound ligament reflex, it is monosynaptic. It is
likewise an extend reflex. These are monosynaptic spinal segmental reflexes. When they are in
place, uprightness of the accompanying is affirmed: cutaneous innervation, engine supply, and
cortical contribution to the relating spinal fragment.
Lower leg of the patient is casual. It is useful to bolster the bundle of the foot at any rate to some
degree to put nearly strain in the Achilles ligament, however don\'t totally dorsiflex the lower
leg. A little strike is given on the Achilles ligament utilizing an elastic mallet to evoke the
reaction. In the event that you are not ready to evoke a reaction, a Jendrassik move can be
attempted by having the patient container their fingers on every hand and attempt to pull the
hands separated. A positive reaction is set apart by an energetic plantarflexion of the foot. The
reaction is additionally evaluated into Grade 1-4 as indicated by the reflex reviewing framework
plantar reflex
reflex elicited
downward response of the hallux
upward response (extension) of the hallux
The plantar reflex is a reflex inspired when the sole of the foot is invigorated with a limit
instrument. The reflex can take one of two.
Please find the answers and explanations belowPart 1Critical l.pdf
Please find the answers and explanations below:
Part 1:
Critical level of soil/plant nutrient is a term used to define the quantized nature of plant yield/soil
yeild upto a percentage of maximum possible yield. Technically, the critical levels of plant
growth or soil nutrient is defined as the quantitative amount of nutrient/growth of the soil/plant
which results in atleast 90% of the maximum possible yield for that soil or plant. This term is
routinely used to define and quantize the nature and constituents of the soil.
Part 2:
Every nutrient shows some specific features of deficiency in plants which can be utilized to
diagnose the nutritional status of a plant. According to the information, a plant shows deficiency
symptoms on the bottomleaves but not on top leaves. This can be a possible sign of magnesium
deficieny in plants. It has been proven scientifically that magnesium deficiency worns out the
bottom leaves only while the top leaves remain normal. On the other hand, a plant with zinc
deficiency strictly shows worning out of the top leaves but the bottom leaves remain normal.
Thus, the information suggests clear signs of magnesium deficiency in this plant.
Part 3:
Addition of element \"X\" might not exert beneficial effects on improving the plant growth due
to below mentioned reasons:
1. Nutritional status: The element \"X\" might be a non-essential element which is required in
exceedingly low amounts or altogether not required essentially for plant growth improvement.
For example, increasing the calcium content of soil has no beneficial effect on plant growth
because calcium is not an essential element for plant growht.
2. Bio-availability: At times, some nutrients are though present in the soil but are not bio-
available for the plant in the complex form. For example, nitrogen is present in the environment
(soil and air) but not absorbed by the plant until is is fixed into soluble form by nirogen fixing
bacteria/plants. Thus, the element \"X\" might not be bio-available for the plant in this form and
hence increasing its concentration might not have any beneficial effect on plant growth.
Part 4:
According to the information, the plant shows potassium deficiency whereas the soil has more
than sufficient amount of potassium content. A possible reason behind this observation could be
imbalance in sodium and potassium concentrations in the soil. It is well known that a gradient of
sodium and potassium ions is essentially required to maintain the optimum concentration of
these ions in the plant and to perform physiological functions such as opening and closing of
stomata. However, an inappropriate increase in the ionic concentration of either of these ions can
lead to deterioration of this ionic balance thus resulting in nutrient deficiency. Thus, although a
sufficient amount of potassium is present, the plant shows symptoms of potassium deficiency
probably due to imabalance in sodium/potassium balance.
Part 5:
The nutrients which are req.
Psychoanalysis is a method of treating mental disorders, shaped by p.pdf
Psychoanalysis is a method of treating mental disorders, shaped by psychoanalytic theory, which
emphasizes unconscious mental processes. Example: The violent behavior of criminals is often
traced to the violent ways they themselves were treated during their formative years.
Solution
Psychoanalysis is a method of treating mental disorders, shaped by psychoanalytic theory, which
emphasizes unconscious mental processes. Example: The violent behavior of criminals is often
traced to the violent ways they themselves were treated during their formative years..
pH+pOH=14 pH=14-pOH=14-11.97=2.03SolutionpH+pOH=14 pH=14-p.pdf
pH+pOH=14
pH=14-pOH=14-11.97=2.03
Solution
pH+pOH=14
pH=14-pOH=14-11.97=2.03.
No solutionSolutionNo solution.pdf
The document discusses whether a solution exists or not. It first states that there is no solution, then that there is a solution, and finally again that there is no solution.
The concentration of the reactants decreases .pdf
The concentration of the reactants decreases in the solution. The document states that the concentration of the reactants decreases twice, indicating that over time the amount of reactants is reduced in the solution.
let E0 be given, a belongs to X= d(f(x),f(a)) = Ld(x,a)=.pdf
let E>0 be given, a belongs to X
=>
d(f(x),f(a)) <= Ld(x,a)
=>
|f(x)-f(a)| <= L|x-a|
take d = E/L
for |x-a|
|f(x)-f(a)| < E for |x-a| < d
=>
f is continuous at a
thus proved
Solution
let E>0 be given, a belongs to X
=>
d(f(x),f(a)) <= Ld(x,a)
=>
|f(x)-f(a)| <= L|x-a|
take d = E/L
for |x-a|
|f(x)-f(a)| < E for |x-a| < d
=>
f is continuous at a
thus proved.
import java.util.ArrayList; import java.util.Iterator; .pdf
import java.util.ArrayList;
//import java.util.Iterator;
/**
* A PokemonSpecies entry in the Pokedex. Maintains the number of candies associated
* with the Pokemon species as well as the Trainer\'s inventory of Pokemon of this
* species.
*/
public class PokemonSpecies {
private int pokedexNumber;
private String speciesName;
private int candies;
/**
* Maintains the list of Pokemon of this species in the Trainer\'s inventory.
*/
private ArrayList caughtPokemon;
/**
* Constructor suitable for a newly encountered Pokemon species during the course of the
* game and for loading species data from a save file.
*
*/
public PokemonSpecies(int pokedexNumber, String speciesName, int candies) {
this.pokedexNumber = pokedexNumber;
this.speciesName = speciesName;
this.candies = candies;
// construct caughtPokemon
caughtPokemon = new ArrayList();
}
/**
* Getter methods
*/
public Integer getPokedexNumber() {
return pokedexNumber;
}
public String getSpeciesName() {
return speciesName;
}
public int getCandies() {
return candies;
}
/**
* Add a newly caught Pokemon to the player\'s inventory and
* increase the number of candies for this PokemonSpecies
*
* @param pokemon the newly caught Pokemon
*/
public void addNewPokemon(Pokemon pokemon) {
caughtPokemon.add(pokemon);
this.addNewPokemonCandies();
}
/**
* Helper function to load Pokemon from a save file into the player\'s inventory for this
* Pokemon Species
*/
public void loadPokemon(Pokemon pokemon) {
caughtPokemon.add(pokemon);
}
/**
* Find a Pokemon of the given combatPower in the player\'s inventory for this species.
*
*/
public Pokemon findPokemon(int cp) throws PokedexException {
for(int i = 0; i < caughtPokemon.size(); i++){
Pokemon mon = caughtPokemon.get(i);
if(mon.getCombatPower() == cp){
return mon;
}
}
throw new PokedexException(Config.POKEMON_NOT_FOUND);
}
/**
* Transfer a Pokemon with the provided combatPower from the player\'s inventory
* to the Professor. This removes the Pokemon from the player\'s inventory and
* also increases the number of candies the player has associated with this
* PokemonSpecies.
*/
public Pokemon transferPokemon(int cp) throws PokedexException {
for(int i = 0; i < caughtPokemon.size(); i++){
if(caughtPokemon.get(i).getCombatPower() == cp){
Pokemon mon = caughtPokemon.remove(i);
this.addTransferCandies();
return mon;
}
}
throw new PokedexException(Config.POKEMON_NOT_FOUND);
}
/**
* Check if the player has any Pokemon of this species
*/
public boolean isEmpty() {
return caughtPokemon.isEmpty();
}
/**
* Increment candies when a new pokemon is caught
*/
private void addNewPokemonCandies() {
this.candies += PokemonGO.NEW_POKEMON_CANDIES;
}
/**
* Increment candies when a pokemon is transferred to the professor
*/
private void addTransferCandies() {
this.candies += PokemonGO.TRANSFER_POKEMON_CANDIES;
}
/**
* Prepare a listing of all the Pokemon of this species that are currently in the
* player\'s inventory.
*
*/
public String caughtPokemonToString() {
String pokemonStr = \"\";
.
Insufficient data. Figure missing.SolutionInsufficient data. F.pdf
The document could not be summarized as it contained insufficient data and was missing a figure to provide context. No meaningful information was presented in the short text to create an informative summary.
Hershay and Chase designed an experiment to determine which molecule.pdf
Hershay and Chase designed an experiment to determine which molecule acted as the genetic
material in ohages.They use the technique know as Radioactive labelling.Here , They used
certain radioactive isotopes.
Steps used by there experiments:
1.Developed two batches of phages .
2.One Batch-Growth in presence of radioactive phosphorous.Phosphorous is present in DNA but
not in bacteri or phage.
3.Second Batch-Grown in presence of radioactive sulphur.Sulphur is found in Proteins but not in
DNA.
4.Separetly infected bacteria and measured radioactivity.
5. Result-RadioLabelled DNA infected bacteria by phage become radioactive
Hence concluded that during phage infection DNA which are injected are dominant. So, DNA
must be the genetic material that works on the bacteria.(reprogram)
Solution
Hershay and Chase designed an experiment to determine which molecule acted as the genetic
material in ohages.They use the technique know as Radioactive labelling.Here , They used
certain radioactive isotopes.
Steps used by there experiments:
1.Developed two batches of phages .
2.One Batch-Growth in presence of radioactive phosphorous.Phosphorous is present in DNA but
not in bacteri or phage.
3.Second Batch-Grown in presence of radioactive sulphur.Sulphur is found in Proteins but not in
DNA.
4.Separetly infected bacteria and measured radioactivity.
5. Result-RadioLabelled DNA infected bacteria by phage become radioactive
Hence concluded that during phage infection DNA which are injected are dominant. So, DNA
must be the genetic material that works on the bacteria.(reprogram).
Humans being a diploid organism inherit two copies of each gene, fro.pdf
Humans being a diploid organism inherit two copies of each gene, from each parent except for
the males who possess only one X chromosome from the mother, meanwhile lacks several
compensatory genes present on Y chromosome to be inherited from the father.
In this context, brothers and sisters even when they share the same genotype, differ considerably
due to the variation occurring the genetic makeup occurring through mutation, recombination
and migration of genes. It also happens that alleles are segregated at different loci, which play a
key role in the distinction of siblings.
Solution
Humans being a diploid organism inherit two copies of each gene, from each parent except for
the males who possess only one X chromosome from the mother, meanwhile lacks several
compensatory genes present on Y chromosome to be inherited from the father.
In this context, brothers and sisters even when they share the same genotype, differ considerably
due to the variation occurring the genetic makeup occurring through mutation, recombination
and migration of genes. It also happens that alleles are segregated at different loci, which play a
key role in the distinction of siblings..
Given, pH of buffer = 8.86 [H+] = 1.380 10-9 M .pdf
Given, pH of buffer = 8.86
[H+] = 1.380 * 10-9 M
and [OH-] = (1.0*10-14 ) / [H+]
= 7.2*10-6 M
B (aq) + H2O (l) <-----> BH+ (aq) + OH- (aq)
I(M) 0.42 0.28 0
C(M) -7.2*10-6 +7.2*10-6 +7.2*10-6
-------------------------------------------------------
E(M) 0.42 0.28 7.2*10-6
Kb = [BH+][OH-] / [B]
= 0.28 * 7.2*10-6 / 0.42
= 4.8*10-6
When 0.0020 mol HCl is added to 0.25 L of this solution, concentration is 0.0020 mol / 0.25 L
= 8.0*10-3 M
B + HCl ------> BH+ + Cl-
I(M) 0.42 8*10-3 0.28
C(M) -8*10-3 - 8*10-3 +8*10-3
------------------------------------
E(M) 0.412 0 0.288
Kb = [BH+][OH-] / [B]
4.8*10-6 = 0.288 * [OH-] / 0.412
[OH-] = 6.9*10-6 M
Then, [H+] = 1.4*10-9 M
so pH = 8.85
I(M) 0.42 8*10-3 0.28
C(M) -8*10-3 - 8*10-3 +8*10-3
------------------------------------
E(M) 0.412 0 0.288
Kb = [BH+][OH-] / [B]
4.8*10-6 = 0.288 * [OH-] / 0.412
[OH-] = 6.9*10-6 M
Then, [H+] = 1.4*10-9 M
so pH = 8.85
4.8*10-6 = 0.288 * [OH-] / 0.412
[OH-] = 6.9*10-6 M
Then, [H+] = 1.4*10-9 M
so pH = 8.85
Solution
Given, pH of buffer = 8.86
[H+] = 1.380 * 10-9 M
and [OH-] = (1.0*10-14 ) / [H+]
= 7.2*10-6 M
B (aq) + H2O (l) <-----> BH+ (aq) + OH- (aq)
I(M) 0.42 0.28 0
C(M) -7.2*10-6 +7.2*10-6 +7.2*10-6
-------------------------------------------------------
E(M) 0.42 0.28 7.2*10-6
Kb = [BH+][OH-] / [B]
= 0.28 * 7.2*10-6 / 0.42
= 4.8*10-6
When 0.0020 mol HCl is added to 0.25 L of this solution, concentration is 0.0020 mol / 0.25 L
= 8.0*10-3 M
B + HCl ------> BH+ + Cl-
I(M) 0.42 8*10-3 0.28
C(M) -8*10-3 - 8*10-3 +8*10-3
------------------------------------
E(M) 0.412 0 0.288
Kb = [BH+][OH-] / [B]
4.8*10-6 = 0.288 * [OH-] / 0.412
[OH-] = 6.9*10-6 M
Then, [H+] = 1.4*10-9 M
so pH = 8.85
I(M) 0.42 8*10-3 0.28
C(M) -8*10-3 - 8*10-3 +8*10-3
------------------------------------
E(M) 0.412 0 0.288
Kb = [BH+][OH-] / [B]
4.8*10-6 = 0.288 * [OH-] / 0.412
[OH-] = 6.9*10-6 M
Then, [H+] = 1.4*10-9 M
so pH = 8.85
4.8*10-6 = 0.288 * [OH-] / 0.412
[OH-] = 6.9*10-6 M
Then, [H+] = 1.4*10-9 M
so pH = 8.85.
exists. ThenSolutionexists. Then.pdf
This short document does not contain any substantive information that can be summarized in 3 sentences or less. It appears to be an incomplete fragment without enough context to extract a meaningful high-level summary.
D. A mount point is the connection point to the folder on a logical .pdf
D. A mount point is the connection point to the folder on a logical drive containing the required
data.
Solution
D. A mount point is the connection point to the folder on a logical drive containing the required
data..
Both Sfe = n and Sfe = SfoSolutionBoth Sfe = n and Sfe = Sfo.pdf
This short document states that the surface area of an end face (Sfe) of a solid is equal to both the number of squares (n) in the end face as well as the surface area of the opposite face (Sfo).
Answer7).DNA is the long chains of nucleotides. Chromosomes and.pdf
Answer:
7).
DNA is the long chains of nucleotides. Chromosomes and histone proteins together are called as
chromosomes. A segment of chromosomal DNA is gene.
8)
Animal cell mitosis
Plant cell mitosis
The animal cell shape becomes round before the cell division strats
No change in the cell shape
Centrosome is essential
Absent
Mitotic apparatus contains asters. So the spindle is amphiastral
Mitotic apparatus is without asters. So the spindle is anastral
Spindle fibers are degenerated during cytokinesis
Spindle fobers are used in the cell wall formation
Cleavage of cell centripetally during cytokinesis
Cleavage of cell centrifually during cytokinesis
A furrow is formed between two daughter cells
Cell plate is formed that proceeds to the edges
9. Chromosomes mover toward poles: Anaphase
10. DNA replicates: Interphase
11. Chromosomes line up at the equator: Metaphase
12. Nuclear membrane breaks down and chromatids form: Prophase
13. Cytoplasm begins to divide: Telophase
Animal cell mitosis
Plant cell mitosis
The animal cell shape becomes round before the cell division strats
No change in the cell shape
Centrosome is essential
Absent
Mitotic apparatus contains asters. So the spindle is amphiastral
Mitotic apparatus is without asters. So the spindle is anastral
Spindle fibers are degenerated during cytokinesis
Spindle fobers are used in the cell wall formation
Cleavage of cell centripetally during cytokinesis
Cleavage of cell centrifually during cytokinesis
A furrow is formed between two daughter cells
Cell plate is formed that proceeds to the edges
Solution
Answer:
7).
DNA is the long chains of nucleotides. Chromosomes and histone proteins together are called as
chromosomes. A segment of chromosomal DNA is gene.
8)
Animal cell mitosis
Plant cell mitosis
The animal cell shape becomes round before the cell division strats
No change in the cell shape
Centrosome is essential
Absent
Mitotic apparatus contains asters. So the spindle is amphiastral
Mitotic apparatus is without asters. So the spindle is anastral
Spindle fibers are degenerated during cytokinesis
Spindle fobers are used in the cell wall formation
Cleavage of cell centripetally during cytokinesis
Cleavage of cell centrifually during cytokinesis
A furrow is formed between two daughter cells
Cell plate is formed that proceeds to the edges
9. Chromosomes mover toward poles: Anaphase
10. DNA replicates: Interphase
11. Chromosomes line up at the equator: Metaphase
12. Nuclear membrane breaks down and chromatids form: Prophase
13. Cytoplasm begins to divide: Telophase
Animal cell mitosis
Plant cell mitosis
The animal cell shape becomes round before the cell division strats
No change in the cell shape
Centrosome is essential
Absent
Mitotic apparatus contains asters. So the spindle is amphiastral
Mitotic apparatus is without asters. So the spindle is anastral
Spindle fibers are degenerated during cytokinesis
Spindle fobers are used in the cell wall formation
Cleavage of cell centripetally during cyto.
Any function consisting of one or more terms of variables with whole.pdf
Any function consisting of one or more terms of variables with whole number exponents is a
polynomial function. If it is in one variable (usually x), it is in the following form:
Solution
Any function consisting of one or more terms of variables with whole number exponents is a
polynomial function. If it is in one variable (usually x), it is in the following form:.
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- 1. a) The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [Ground state configuration] The number of electron in each subshell is 2 8 18 5 (K,L,M,N) b) one permissible quantum number would be 4,0,0,-1/2 c) paramagnetic element are those element that have unpaired electron in the outer most shell and diamagnetic are the one that have paired electron. Now in Arsenic, in the 4p there are 3 electron which are distributed among the three sub-orbitals of P (Px, Py and Pz) so they have 3 unpaired electron in 4p. Hence it is paramagnetic. d) Now the compound Na3As, AsCl3 and AsF3, if we look at it, in all the compound As combine with 3 atoms of another element either by covalent or ionic bond. This is because As requires 3 electron to attend the noble gas configuration or the octet state If we consider Na3As, Na will loses 1 e forming Na+ Na======>Na+ + 1e , Since there are 3 Na, so there will be 3 electron. These 3 e are taken up by As. So the configuration becomes 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6. Hence it has attain octet state. Therefore the ground state is consistent. Solution a) The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p3 [Ground state configuration] The number of electron in each subshell is 2 8 18 5 (K,L,M,N) b) one permissible quantum number would be 4,0,0,-1/2 c) paramagnetic element are those element that have unpaired electron in the outer most shell and diamagnetic are the one that have paired electron. Now in Arsenic, in the 4p there are 3 electron which are distributed among the three sub-orbitals of P (Px, Py and Pz) so they have 3 unpaired electron in 4p. Hence it is paramagnetic. d) Now the compound Na3As, AsCl3 and AsF3, if we look at it, in all the compound As combine with 3 atoms of another element either by covalent or ionic bond. This is because As
- 2. requires 3 electron to attend the noble gas configuration or the octet state If we consider Na3As, Na will loses 1 e forming Na+ Na======>Na+ + 1e , Since there are 3 Na, so there will be 3 electron. These 3 e are taken up by As. So the configuration becomes 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6. Hence it has attain octet state. Therefore the ground state is consistent.