1.) The modern atomic theory has been updated over the years as new observations of the atom have been made. What do these changes say about the strength of the modern atomic theory?

A. The changes make the theory weaker because they demonstrate that the scientific community can't agree on the properties of atoms.
B. The changes make the theory weaker because they demonstrate that the original atomic theory was flawed, so all resulting theories are also flawed.
C. The changes make the theory stronger because it has been tested and edited multiple times, making it more durable.
D. The changes make the theory stronger because each time the theory is changed, it becomes more popular.


2.) In Rutherford's famous experiment, he shot heavy, fast moving, positive alpha particles at a thin gold foil. Based on Thomson's plum pudding model of the atom, what did Rutherford expect to happen, and why?

A. All the alpha particles would be deflected by the foil because of the even distribution of mass and charge throughout the atom.
B. All the alpha particles would pass straight through the foil because of the even distribution of mass and charge throughout the atom.
C. Most of the alpha particles would become embedded in the foil because they are attracted to the negative electrons in the atom.
D. Some of the alpha particles would be deflected and some would pass through, because of the negative charge of the electron cloud.

Answer:

B. >>> Image 3  

Step-by-step explanation:

Ionic equation :

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ca²⁺(aq) + 6Cl⁻(aq) ⟶ 6Na⁺(aq) + 6Cl⁻(aq) + Ca₃(PO₄)₂(s)

=====

Net ionic equation:

Cancel all ions that appear on both sides of the reaction arrow (underlined).

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ca²⁺(aq) + 6Cl⁻(aq) ⟶ 6Na⁺(aq) + 6Cl⁻(aq) + Ca₃(PO₄)₂(s)

3Ca²⁺(aq) + 2PO₄³⁻(aq )⟶ Ca₃(PO₄)₂(s)

A. >>> Image 2 is wrong, because it is the molecular equation

C. >>> Image 4 is wrong, because it is not balanced.

D. >>> Image 5 is wrong, because the phosphate ion behaves as a unit. It does not break up into phosphorus and oxide ions.

Answer:

B is correct

Explanation:

Edge

3MgCl2 + 2Na3P => Mg3P2 + 6NaCl

I believe it is D. because stoichiometry allows people to determine how much reactant is consumed in a chemical reaction.

Final answer:

The answer to the chemical reaction question is D) limiting reactant. It is the substance that is completely used up first in a reaction, determining the maximum amount of product that can be produced.

Explanation:

In a chemical reaction, the limiting reactant is consumed completely during the reaction. The correct answer to the question is D) limiting reactant. A limiting reactant (or limiting reagent) is the substance that is entirely used up first in a chemical reaction, and it limits the amount of product that can be formed. Once the limiting reactant is used up, the reaction cannot proceed further, even if other reactants are still present in excess.

The excess reagent, on the other hand, is the reactant that remains after the reaction has gone to completion because it was present in a larger quantity than necessary according to the stoichiometric ratio defined by the balanced chemical equation.

To determine the limiting reactant in a chemical reaction, it is essential to compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation. The reactant in the smallest stoichiometric amount is the limiting reactant.

  The arrangement of valence electrons in metals can be best described by the

A sea of electrons.

The heat absorbed is calculated with the help of the formula,

Q = mCΔT

Q is heat absorbed, m is mass given as 56 gms, C is the specific heat of iron, that is, 0.449 J/g°C, and delta T is the change in temperature, which is 28 - 7 = 21°C

Thus, Q = 56 × 0.449 × 21 = 528 J

The wavelength (λ), speed (c) and the frequency (ν) of a wave are related as:

ν = c/λ  -----(1)

The speed of the waves is a constant in vacuum i.e. it remains at 3*10⁸ m/s. therefore, it will not change with changes in wavelength or frequency.

Let us consider:

ν1 = initial frequency and λ1 = initial wavelength

ν2 = final frequency and λ2 = final wavelength = 2(λ1)

based on equation (1) we have:

ν1/ν2 = λ2/λ1 = 2(λ1)/λ1 = 2

therefore: ν2 = 1/2(ν1)

The frequency will get reduced by one-half as the wavelength doubles.

Answer: If the number and kinds of atoms on both sides of the chemical equation are same, the chemical equation is called as balanced chemical equation.

If the number and kinds of atoms on both sides of the chemical equation are different, the chemical equation is called as skeletal chemical equation.

[tex]CH_4+O_2\rightarrow CO_2+H_2O[/tex]

As the number of atoms on both sides of the chemical equation are different, the above equation represents a skeletal chemical equation.

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

As the chemical equations must follow the law of conservation of mass, the number of atoms on both sides of the above chemical equation must be same so as the mass is same on both sides of the equation. So the equation was balanced.

Answer:

In the reactant side of the chemical equation, there are 4 carbon atoms, 12 hydrogen atoms, and 14 oxygen atoms. In the product side of the chemical equation, there are 4 carbon atoms, 12 hydrogen atoms, and 14 oxygen atoms. The number and kinds of atoms on both sides of the chemical equation are the same . So, this chemical equation is balanced

Explanation:

The reactants have 4 carbon atoms, 14 oxygen atoms, and 12 hydrogen atoms. The product has the same numbers and kinds of atoms. So, the equation is balanced.

  The molarity  of a saline solution  that contain 0.900 g NaCl (58.44 g/mol) dissolved in 100.0 ml solution is  0.15 M

  calculation

 Molarity = moles /volume  in liters

volume in  liters = 100.0 ml / 1000 = 0.1 L

 moles =  mass÷ molar mass

=  0.900 g ÷58.44 g/mol =0.015 moles

molarity  is therefore =   0.015 moles /  0.1 L =0.15 M

Answer:

46. mmol  

Step-by-step explanation:

The equation for the equilibrium is:

HA + H₂O ⇌ A⁻ + H₃O⁺

The solution is a buffer, so we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻]/[HA])

   5.40 = 4.74 + log([A⁻]/(10.)     Subtract 4.74 from each side

   0.66 = log([A⁻]/(10.)                 Take the antilog of each dide

[A⁻]/10. = 10 ^0.66

[A⁻]/10. = 4.57                              Multiply each side by 10.

    [A⁻] = 46. mmol

You will add 46. mmol of sodium acetate.

The amount of acetate needed to add to the solution is ; 46 mmol

Given data:

pH of buffer solution = 5.4

volume of  Solution ( H A ) = 10 mmol

pKa of acetic acid = 4.74

Given that the solution is a buffer solution we will apply  Henderson-Hasselbalch equation:

pH = [tex]pKa + log ( [A^-] / [HA] )[/tex]  -------- ( 1 )

Insert values into equation 1

[tex]5.40 = 4.74 + log([A^-]/(10 )[/tex]

0.66 = [tex]log([A^-]/(10.)[/tex] -------- ( 2 )  ( after subtracting pKa value from both side )

∴  [tex][A^-]/10. = 10^{0.66}[/tex]  ----- ( 3 )  ( antilog )

Multiply both sides of equation 3 by  10

[tex][A^-] = 46 mmol[/tex].

Hence we can conclude that the amount of acetate needed to add to the solution is  46 mmol

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Answer: 2 g

Explanation: [tex]2Pb(NO_3)_2(s)\rightarrow 2PbO(s)+4NO_2(g)+O_2(g)[/tex]

As can be seen from the balanced chemical equation, 2 moles of lead nitrate produce 4 moles of nitrogen dioxide.

[tex]2\times 331.2g=662.4g[/tex] of lead nitrate produces [tex]4\times 46=184g[/tex] of nitrogen dioxide.

184 g of nitrogen dioxide will be produced by 662.4 g of lead nitrate

So 11.5 g of nitrogen dioxide will be produced by=[tex]\frac{662.4}{184}\times {11.5}=41.4 g[/tex] of lead nitrate

As can be seen from the balanced chemical equation, 2 moles of lead nitrate produce 1 mole of oxygen.

[tex]2\times 331.2g=662.4g[/tex] of lead nitrate produces 32 g of oxygen.

41.4 g of lead nitrate produces =[tex]\frac{32}{662.4}\times {41.4}=2g[/tex] of oxygen.

The accepted name for N2O5 is dinitrogen pentoxide.

The accepted name for the compound N2O5 is dinitrogen pentoxide. It is formed by the combination of two nitrogen atoms and five oxygen atoms. The prefix 'di' indicates that there are two nitrogen atoms, and the prefix 'pento' indicates that there are five oxygen atoms.

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Equation 1, because Zn being more reactive, replaces Ag from AgNO3

Zn being highly reactive can replace Ag from AgNO3 and produce Ag metal.

Equation 1: 2AgNO3(aq) + Zn(s) → 2Ag(s) + Zn(NO3)2

This equation is the right one to extract silver metal from silvernitrate solution as its single displacement reaction.

Answer:

Equation 1, because Zn being more reactive, replaces Ag from AgNO3

Explanation:

The preferential activity of one metal to displace the other in a chemical reaction can be deduced based on the reactivity or activity series.

The elements at the top of the series are highly reactive and can displace metals that fall below them. Both Zn and Mg appear above Ag in the reactivity  series and can displace silver from silver nitrate.

1) AgNO3 + Zn → ZnNO3 + Ag

Metallic silver can be extracted through the above single displacement reaction.

2) 2AgNO3 + MgCl2 → 2AgCl + Mg(NO3)2

This is double displacement reaction which instead forms a precipitate of AgCl

Empirical Formula is Fe1S1O4

(the numbers are suppose to be subscript)

Answer: [tex]FeSO_4[/tex]

Solution : Given,

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of Fe = 36.76 g

Mass of S = 21.11 g

Mass of O = 42.13 g

Step 1 : convert given masses into moles.

Moles of Fe =[tex] \frac{\text{ given mass of Fe}}{\text{ molar mass of Fe}}= \frac{36.76g}{56g/mole}=0.65moles[/tex]

Moles of S = \frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{21.11g}{32g/mole}=0.65moles

[/tex]

Moles of O = \frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{42.13g}{16g/mole}=2.63[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Fe = [tex]\frac{0.65}{0.65}=1[/tex]

For S = [tex]\frac{0.65}{0.65}=1[/tex]

For O =[tex]\frac{2.63}{0.65}=4[/tex]

The ratio of Fe : S : O= 1 : 1 : 4

Hence the empirical formula is [tex]Fe_{1}S_{1}O_4[/tex]

Answer:

It is a double replacement reaction, and all four compounds are different.

Step-by-step explanation:

The equation for the reaction is

[tex]\underbrace{\hbox{ ionic compound + ionic compound}}_{\hbox{reactants}} \longrightarrow  \underbrace{\hbox{ ionic compound + ionic compound}}_{\hbox{products}}[/tex]

This must be a double replacement reaction, in which the metal cations trade partners and all four compounds are different.

(A) and (C) are wrong, because single replacements involve elements in reactions of the type

element + compound ⟶ element + compound

(D) is wrong because, if each compound retained the same set of ions, there would be no reaction.  

Answer:

It is a double replacement reaction, and all four compounds are different.

Explanation:

Just took the FLVS 4.03 Single and Double Replacement Reactions quiz. I got a 100.

Answer:

Without enough oxygen.  

Step-by-step explanation:

In the presence of excess oxygen, the combustion of fuel produces carbon dioxide. For example, propane burns according to the equation

C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

In the presence of a limited amount of oxygen, carbon monoxide is also formed.

C₃H₈ + ⁷/₂O₂ ⟶ 3CO + 4H₂O

The process is called incomplete combustion.

(a) is wrong. Sulfur impurities cause the formation of sulfur dioxide.

(c) is wrong. High temperatures usually lead to more complete combustion.

(d) is wrong. Lead compounds inhibit knocking in gasoline engines, but they do not affect the formation of carbon monoxide.

Properties of matter can be broadly classified into two categories:

Physical properties which usually involve a change in the state of matter and Chemical properties which involve a change in the chemical composition of matter.

Now, physical properties can be further classified as:

Extensive: these depend on the amount of the substance, eg: mass, volume  

Intensive: these do not depend on the amount of the substance eg: density, color, melting point, boiling point

Here we are given a 5.0 g and 1 cm3 silver cube :

Therefore:

Extensive properties are-

1) Mass of silver = 5.0 g

2) Volume of silver = 1 cm3

Intensive properties are:

1) Density of silver = mass/volume = 5.0 g/ 1 cm3 = 5.0 g/cm3

2) Melting point of silver = 962 C

3) Color = white/gray

In a titration, a molar mass lower than the actual one could be obtained by spilling part of the acid before titration or recording a dark red color instead of a light pink as the endpoint. These errors would lower the calculated molar mass because either too little acid or too much NaOH is assumed for the reaction.

In a titration experiment with a carboxylic acid and a standardized NaOH solution, there are factors that could lead to the measurement of a molar mass that is smaller than its actual value.

Error I: If some of the carboxylic acid is spilled when being transferred into the titration flask, this will decrease the amount of acid available for titration. This leads to an underestimation of the amount of NaOH needed for neutralization and therefore a smaller calculation for molar mass.

Error II: If the endpoint of the titration is recorded when the color is dark red rather than light pink, this means that the end-point has been overshot – too much NaOH has been added. This excess amount would lead to the calculation of a smaller molar mass since an increased volume of NaOH is wrongly assumed to have neutralized the acid.

Titration analysis therefore requires real accuracy both in process and observation to avoid such errors. It is important to determine and record the right endpoint based on a distinct, recognized color change with the suitable indicator, in this case, a change from colorless to light pink with phenolphthalein.

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The correct answer is that both errors I and II will lead to a molar mass that is smaller than the actual molar mass.

I. Some of the acid is spilled when being transferred into the titration flask.

When some of the acid is spilled, the mass of the acid that is actually titrated is less than the mass that was intended to be titrated.

Since molar mass is calculated by dividing the mass of the substance by the number of moles of that substance, a smaller mass will result in a smaller calculated molar mass if the number of moles is not adjusted accordingly.

The molar mass (M) is given by:

[tex]\[ M = \frac{m}{n} \][/tex]

where m is the mass of the acid and n is the number of moles of the acid.

If the mass m is smaller due to spillage, and n remains the same (because the titre does not change), then the calculated molar mass M will be smaller.

II. The endpoint is recorded when the solution is dark red in color rather than light pink.

Phenolphthalein is a common acid-base indicator that is colorless in acidic solutions and pink in basic solutions.

The endpoint of the titration is reached when the solution changes from colorless to light pink, indicating that the acid has been neutralized by the NaOH.

If the endpoint is recorded when the solution is dark red, it means that the titration has continued past the true endpoint, adding excess NaOH.

This excess NaOH will not react with the acid (since it's already been neutralized), but it will be included in the calculation of the moles of NaOH used.

The number of moles of acid [tex](\( n_{acid} \))[/tex] is calculated using the moles of NaOH [tex](\( n_{NaOH} \))[/tex] used to reach the endpoint:

[tex]\[ n_{acid} = n_{NaOH} \][/tex]

If [tex]\( n_{NaOH} \)[/tex] is overestimated due to continuing the titration past the endpoint, then [tex]\( n_{acid} \)[/tex] will also be overestimated. Since the mass of the acid m remains unchanged, the calculated molar mass M will be smaller:

[tex]\[ M = \frac{m}{n_{acid}} \][/tex]

In conclusion, both spilling some of the acid and recording the endpoint too late will result in a calculated molar mass that is smaller than the actual molar mass of the carboxylic acid.

The grams  of   carbon monoxide  is 148. 68 g

calculation

Fe₂O₃(s)  + 3CO (g) →   3 CO₂(g)  + 2Fe (s)

 Step 1: find  the moles  of  CO

   moles = mass÷ molar mass

 from periodic table the  molar mass of Fe =56 g/mol

 moles =  198.5 g÷ 56 g/mol =3.54  moles

Step 2 : use the mole ratio to determine the  moles of  Co

 from given equation  Co: Fe is 3: 2

therefore the  moles of CO = 3.54 moles  x 3/2= 5.31 moles

Step 3 Find mass of Co

mass  =  moles × molar mass

from periodic table the  molar mass of CO = 12+ 16 = 28 g/mol

= 5.31  moles × 28 g/mol = 148.68  g

Answer:

-12 258 kJ  

Step-by-step explanation:

Reaction 1: C₃H₈(g)+5O2(g) ⟶ 3CO₂(g)+ 4H₂O(g);          ΔH₁  = - 2 043 kJ  

Reaction 2: 6C₃H₈(g)+30O2(g) ⟶ 18CO₂(g)+ 24H₂O(g); ΔH₂ = -12 258 kJ

Reaction 2 is Reaction 1 multiplied by 6.

ΔH is an extensive property, so you must also multiply ΔH by 6.

Final answer:

The enthalpy change for Reaction 2 is calculated by multiplying the given enthalpy change for Reaction 1 by the factor of six. Since enthalpy is an extensive property, Reaction 2's enthalpy change is –12258 kJ.

Explanation:

ΔH, or change in enthalpy, is an extensive property, which means it is proportional to the amounts of reactants and products in a reaction. In the student's question, we need to calculate the enthalpy change for Reaction 2 based on the given enthalpy change for Reaction 1. We are given that the enthalpy change for Reaction 1, which is the combustion of propane [tex](C_3H_8)[/tex], is [tex]\Delta H_1 = -2043 kJ[/tex]:

[tex]C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g), \Delta 2 \Delta H_1 = -2043 kJ[/tex]

For Reaction 2, the equation is the same but scaled up by a factor of six:

[tex]6C_3H_8(g) + 30O_2(g) \rightarrow 18CO_2(g) + 24H_2O(g), \Delta 2 \Delta H_2 = ?[/tex]

To find ΔH2, we multiply ΔH1 by the scaling factor, which is six, because enthalpy is an extensive property:

[tex]\Delta H_2 = 6 \times \Delta H_1 = 6 \times (-2043 kJ) = -12258 kJ[/tex]

Therefore, the enthalpy change for Reaction 2 is ΔH2 = –12258 kJ.

Answer: D

Electrons have negative charges. Atoms are originally neutral. They turn positive and form cations when they lose electrons. They turn negative and form anions when they gain electrons.

The superscript next to an ion shows its charge. By convention, the size of the charge is written in front of the sign of the charge. For example, "[tex]1-[/tex]" indicates a negative charge of one. "[tex]2+[/tex]" indicates a positive charge of two.

The choices include only elements from the s block and the p block. They are main group elements. Charges of their ions can be predicted.

Also by convention, cations shall go in front of anions in a chemical formula. Their should balance the charge on each other. For example, [tex](+2) + 2 \times (-1) = 0[/tex]. Two ions of charge -1 would combine with one ion of charge +2 to form a neutral compound.

Choice A:

[tex]\text{Cl}^{-}[/tex] is an anion, not a cation.

Choice B:

Magnesium Mg is in IUPAC group 2. It has two valence electrons. It tends to lose both of them to form an [tex]\text{Mg}^{2+}[/tex] ion with a charge of +2. The [tex]\text{Mg}^{+}[/tex] ion with a charge of +1 might be too unstable to become part of an ionic compound.

Choice C:

[tex]\text{Na}^{+}[/tex] is a cation and has a positive charge. By convention, it should go in front of [tex]\text{Cl}^{-}[/tex], which is an anion and has a negative charge. The formula should be NaCl.

Choice D:

Calcium Ca is also in IUPAC group 2. Similar to Mg, it has two valence electrons and tends to form a [tex]\text{Ca}^{2+}[/tex] ion. Each [tex]\text{Ca}^{2+}[/tex] ion shall combine with two  [tex]\text{Cl}^{-}[/tex] ions for their charges to balance. Hence the formula CaCl₂.

Option D is the correct choice. The cation is Ca+2, the anion is Cl-, and the compound they form is CaCl2, called Calcium chloride. The chemical formula is written such that the charges balance out.

The correct choice that identifies the cation, the anion, and the formula for the compound they create correctly is Option D. The cation in this case is Ca+2 and the anion is Cl–. When these two combine, they form the compound CaCl2, also known as Calcium chloride. In a compound, the cation is always written first followed by the anion. The charges must balance out, hence for Ca+2 (Calcium) and Cl- (Chloride) to balance, two chloride ions are needed for every one calcium ion, hence the formula CaCl2.

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Answer: [tex]2H_2+O_2\rightarrow 2H_2O[/tex]

The molar relationship says that every 1 mole of an element or compound weighs equal to its molecular weight.

As 1 mole of molecular hydrogen [tex]H_2[/tex] weighs 2 g.

2 moles of molecular hydrogen will weigh[tex]=\frac{2}{1}\times {2}=4g[/tex]

1 mole of molecular oxygen [tex]O_2[/tex] weighs 32 g

1 mole of water [tex] H_2O[/tex] weighs = 18g

2 moles of water [tex] H_2O[/tex] weigh=[tex]\frac{18}{1}\times {2}=36g[/tex]

As can be seen from the balanced chemical equation above, 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Thus 4g of [tex]H_2[/tex] combines with 32 g of [tex]O_2[/tex] to give 36 moles of [tex] H_2O[/tex].

The study of chemicals and bonds is called chemistry. There are different types of elements, and these are metals and nonmetals.

The correct answer is 36 mole

According to the question, the 2 moles of water react with the mass of water:-

[tex]\frac{18}{1}*2 = 36g[/tex].

Hence, 36 g is used to react with hydrogen and oxygen.

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Answer: D. They are made up of hard spheres that are in random motion.

Explanation:

A gas is a state of aggregation of matter in which, under certain conditions of temperature and pressure, its molecules interact weakly with each other, without forming molecular bonds, adopting the shape and volume of the container that contains them and tending to separate everything possible because of its high concentration of kinetic energy.

The molecules of a gas are practically free and have the ability to be distributed throughout the space in which they are contained because the gravitational forces and attraction between them are practically negligible compared to the speed at which they move. .

Therefore, gas molecules do not travel specific trajectories or vibrate in a stationary position, instead they move quickly and randomly through the entire space of the container that contains them.

The true statement about gases is that they are made up of hard spheres that are in random motion, which is option D.

Among the given options, statement D is true. Gases are composed of particles, such as atoms or molecules, that are in constant random motion. These particles move in straight lines until they collide with other particles or the walls of the container, where they undergo elastic collisions. The random motion of gas particles is responsible for their ability to fill the entire volume of the container and to exert pressure on the container walls.

Hence, the correct option is option d.

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the answer is D 0.0556 mol

divide the given number of atoms 3.35x10^22 by 6.02x10^23 Avogadro's number(the number of atoms in one mol of anything) to get the total number of Moles. It's merely a fraction of a whole. 3.35/60.2 if you cancel out the tens.   1.0 mol mercury(the molar mass is the mass number on a periodic table) It couldn't possibly be so few atoms is to be gold or atoms of mercury as there are 6.02x10^23 atoms in a 0.0556 mol.  

have a nice day hope this helps !!!!!!!

The electropositive elements or metals elements tend to lose electrons and form positive ions, while the  electronegative elements or non-metals tend to gain electrons and form negative ions.

Metals are electropositive elements with extra electrons in their atom. Hence, they easily loss electron to achieve octet. Metals donate electrons to electron deficient non-metals forming ionic compounds.

Non -metal are located in right side of the periodic table. Except group 18 elements all the non-metals are electron deficient. Some of them such as oxygen, nitrogen, fluorine are highly electronegative and easily gain electrons from metal atoms forming ionic compounds or share electrons from other non-metals forming covalent compounds.

Therefore,  metallic elements tend to lose electrons and form positive ions, while the  electronegative elements or non-metals tend to gain electrons and form negative ions.

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Metals lose electrons and form positive ions, while non-metals gain electrons and form negative ions due to their atomic structure.

The metals elements tend to lose electrons and form positive ions, while the non-metals elements tend to gain electrons and form negative ions. This is due to the atomic structure of these elements. Metals have fewer electrons in their outer shell, and thus they can easily lose them to become stable, which leads to the formation of positively charged ions. On the other hand, non-metals have more electrons in their outer shell and thus they need more electrons to become stable. As a result, they tend to receive electrons, forming negatively charged ions.

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Final answer:

The VSEPR theory provides information about molecular shape and bonding by predicting electron pairs' arrangement around central atoms in a molecule. So the correct option is  B.

Explanation:

The theory that provides information about both molecular shape and molecular bonding is the valence-shell electron-pair repulsion (VSEPR) theory. The VSEPR theory predicts the arrangement of electron pairs around each central atom and usually the correct arrangement of atoms in a molecule. This theory considers electron-pair repulsions to determine the geometry of the molecule, which is essential for understanding molecular bonding and shape.

To reconcile the geometries of s, p, and d atomic orbitals with molecular shapes, we introduce the concept of hybrid orbitals. Hybridization is an extension of VSEPR theory and valence bond theory, providing a more complete understanding of how atomic orbitals combine to form the electronic structure of molecules.

 N and H covalently bond to form the correct formula NH3.

A. N3H

B. NH3

C. NH2

D. NH

Answer:

B. Excited state

Explanation:

Energy levels higher than the ground state are called the excited states. This concept is based on the premise that electrons can move round the nucleus in certain permissibe orbits or energy levels.

The ground state is the lowest energy state available to the electron. This is usually the most stable state.

The excited state is any level higher than the ground state. An electron  in an energy level has a definite amount of energy associated with it at that level.