I need help. Please help me on the question

Answer:

Explanation:

Ionic compounds are formed by the electrostatic attraction of cations and anions.

Cations, positive ions, are formed when atoms lose electrons, and anions, negative ions, are formed when atoms gain electrons.

When two different atoms have similar atraction for electrons (electronegativity) they will not donate to nor catch electrons from each other, so cations and anions will not be formed. Instead, the atoms would prefer to share electrons forming covalent bonds to complete their outermost shell (octet rule).

Then, in order to form ionic compounds the electronegativities have to substantially different. This situation does not happen between two nonmetal elements, which nitrogen and sulfur are. Then, you can predict safely that nitrogen and sulfur will not form an ionic compound.

Ionic compounds, then require the electronegativity difference that exist between some metals and nonmetals. Being magnesium an alkaline earth metal, its electronegativity is very low. On the other hand, fluorine the first element of the group 17, has the highest electronegativity of all the elements.Thus magnesium and fluorine will have enough electronegativity difference to justify the exchange of electrons, forming ions and, consequently, ionic compounds.

Magnesium and Fluorine are the elements most likely to form an ionic compound.

The pair of elements most likely to form an ionic compound is Magnesium and Fluorine.

Binary ionic compounds are composed of a metal (cation) and a nonmetal (anion). Magnesium is a metal and has a low ionization potential, meaning it easily loses electrons. Fluorine is a nonmetal and has a high electron affinity, meaning it readily gains electrons. These properties make it highly likely for magnesium and fluorine to form an ionic compound, such as magnesium fluoride (MgF2).

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Answer:

[tex]\boxed{\text{10 mL}}[/tex]

Explanation:

1. Write the balanced chemical equation.

[tex]\text{2NaOH} + \text{H$_{2}$SO$_{4}$} \longrightarrow\ \text{Na$_{2}$SO$_{4}$} + 2\text{H{$_{2}$O}}[/tex]

2. Calculate the moles of NaOH

[tex]\text{Moles of NaOH} =\text{60.0 mL NaOH} \times \dfrac{\text{0.5 mmol NaOH}}{\text{1 mL NaOH}} = \text{30 mmol NaOH}[/tex]

3. Calculate the moles of H₂SO₄.

[tex]\text{Moles of H$_{2}$SO$_{4}$}=\text{30 mmol NaOH} \times \dfrac{\text{1 mmol H$_{2}$SO$_{4}$} }{\text{2 mmol NaOH}} = \text{30 mmol H$_{2}$SO$_{4}$}[/tex]

4. Calculate the volume of H₂SO₄

[tex]c = \text{30 mmol H$_{2}$SO$_{4}$} \times \dfrac{\text{1 mL H$_{2}$SO$_{4}$}}{\text{3.0 mmol H$_{2}$SO$_{4}$}} = \text{10 mL H$_{2}$SO$_{4}$}[/tex]

The titration will require [tex]\textbf{10 mL}[/tex] H₂SO₄.

Answer:

[tex]\boxed{\text{2 mol/L}}[/tex]

Step-by-step explanation:

[tex]c = \dfrac{ \text{moles} }{ \text{litres}}\\\\c = \dfrac{n }{V }[/tex]

Data:

n = 5 mol

V = 2.5 L

Calculation:

[tex]c = \dfrac{ \text{5 mol} }{\text{2.5 L}} = \text{2 mol/L}[/tex]

The molar concentration of the solution is [tex]\boxed{\textbf{2 mol/L}}[/tex].

The answer is B.

ATP + Oxygen ---> ADP + P

The partial pressure of oxygen in a mixture with a total pressure of 800.0 mm Hg, where oxygen makes up 40% of the mixture, is 320.0 mm Hg.

To find the partial pressure of oxygen in a mixture of gases, you can use the formula P = (Patm) X (percent content in mixture). Given that the total pressure of the mixture is 800.0 mm Hg and oxygen makes up 40% of this mixture by volume, the partial pressure of oxygen can be calculated as follows:

PO2 = (800.0 mm Hg) X (0.40)

PO2 = 320.0 mm Hg

Therefore, the partial pressure of oxygen in this mixture is 320.0 mm Hg.

-59 kj/mol exothermic

The enthalpy change for the reaction of ethylene with hydrogen bromide to form ethyl bromide is -59 kJ/mol. This means that the reaction is exothermic, meaning that it releases heat.

The bond energies of the relevant bonds are as follows:

C-C bond energy: 347 kJ/mol

C-H bond energy: 413 kJ/mol

H-Br bond energy: 363 kJ/mol

C-Br bond energy: 276 kJ/mol

Reactants:

Ethylene: 2 C-H bonds + 1 C-C bond

Hydrogen bromide: 1 H-Br bond

Products:

Ethyl bromide: 3 C-H bonds + 1 C-C bond + 1 C-Br bond

Enthalpy change:

ΔH = Σ(bond energies broken in reactants) - Σ(bond energies formed in products)

ΔH = (2 × 413 kJ/mol + 1 × 347 kJ/mol) - (1 × 363 kJ/mol + 3 × 413 kJ/mol + 1 × 276 kJ/mol)

ΔH = -59 kJ/mol

Therefore, the enthalpy change for the reaction of ethylene with hydrogen bromide to form ethyl bromide is -59 kJ/mol. This means that the reaction is exothermic, meaning that it releases heat.

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Answer:

1.0 x 10⁻¹².

Explanation:

Cu₂S(s) ⇌ 2Cu⁺(aq) + S²⁻(aq),

The equilibrium constant (Keq) = [Cu⁺]²[S²⁻]/[Cu₂S].

[Cu⁺] = 1.0 × 10⁻⁵ M, [S²⁻] = 1.0 × 10⁻² M.

[Cu₂S] = 1.0, since the concentration of solid is always can be considered = 1.0.

∴ Keq = [Cu⁺][S²⁻]/[Cu₂S] = (1.0 × 10⁻⁵)²(1.0 × 10⁻²)/(1.0) = 1.0 x 10⁻¹².

Answer: The equilibrium constant for the above reaction is [tex]1.0\times 10^{-12}[/tex]

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{c}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The expression for [tex]K_{c}[/tex] is written as:

[tex]K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

Concentration of solid and liquid substances in a chemical reaction is taken to be 1.

For the given chemical reaction:

[tex]Cu_2S(s)\rightarrow 2Cu^+(aq.)+S^{2-}(aq.)[/tex]

The expression for [tex]K_{c}[/tex] is written follows:

[tex]K_c=[Cu^+]^2\times [S^{2-}][/tex]

We are given:

[tex][Cu^+]=1.0\times 10^{-5}M[/tex]

[tex][S^{2-}]=1.0\times 10^{-2}M[/tex]

Putting values in above expression, we get:

[tex]K_c=(1.0\times 10^{-5})^2\times (1.0\times 10^{-2})\\\\K_c=1.0\times 10^{-12}[/tex]

Hence, the equilibrium constant for the above reaction is [tex]1.0\times 10^{-12}[/tex]

Answer:

Heterogenous mixture is a mixture that has different properties throughout.

Explanation:

Mixture refers to substances that are not chemically mixed together, that is, they can easily be separated by physical methods. There are two major types of mixture; these are heterogeneous and homogeneous mixtures. Homogeneous mixtures refer to those mixtures that are uniform in composition. A good example of this is tap water. Heterogeneous mixture on the other hand refers to those mixtures, which are not uniform in composition. A good example of this is a mixture of water and sand.  

Explanation:

Hi!

Let's solve this!

Metamorphism is a process that involves rocks. This process occurs when they are subjected to a lot of pressure or temperatures.

This process also happens in the presence of fluids. There are different types of metamorphism such as dynamic, hydrothermal, shock, and others.

2.5 x 10^24 molecules of C4H10

Answer:

[tex]\boxed{\text{B. HC}}[/tex]

Explanation:

The blow-by gases that escape past the piston rings and get into the crankcase are mostly unburned fuel-air mixture.

The fuel is largely a mixture of hydrocarbons (HCs) .

The PCV system captures these gases and feeds them back to the cylinder for further combustion.

A. is wrong. Carbon dioxide exits through the exhaust.

C. is wrong. Most of the oxygen in the incoming air is used for combustion in the cylinders.

D is wrong. The NOx gases exit via the exhaust and are trapped by the catalytic converter.

Answer : The enthalpy change for this reaction is -193.8 kJ.

Solution :

The balanced chemical reaction is,

[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{SO_3}\times \Delta H_{SO_3})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{SO_2}\times \Delta H_{SO_2})][/tex]

where,

n = number of moles

[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

[tex]\Delta H=[(2\times -395.7)]-[(1\times 0)+(2\times -298.8)][/tex]

[tex]Delta H=-193.8kJ[/tex]

Therefore, the enthalpy change for this reaction is, -193.8 KJ

Final answer:

The heat of reaction for the formation of sulfur trioxide from sulfur dioxide and oxygen, using the provided enthalpies of formation, is calculated to be -193.8 kJ.

Explanation:

The student has asked what the heat of reaction is when sulfur dioxide reacts with oxygen to form sulfur trioxide. The given balanced chemical equation is 2SO2(g) + O2(g) → 2SO3(g). The standard enthalpy of formation (ΔHf0) for SO2 is −298.8 kJ/mol, and for SO3 it's −395.7 kJ/mol.

Calculating the Heat of Reaction

To find the heat of reaction, we use the following formula:

ΔH = [∑ (ΔHf0 products)] - [∑ (ΔHf0 reactants)]

ΔH = [2 mol × (−395.7 kJ/mol SO3)] - [2 mol × (−298.8 kJ/mol SO2) + 1 mol × (0 kJ/mol O2)]

ΔH = [−2 × 395.7 kJ] - [−2 × 298.8 kJ]

ΔH = (−791.4 kJ) - (−597.6 kJ)

ΔH = −193.8 kJ

Therefore, the heat of reaction for the formation of sulfur trioxide from sulfur dioxide and oxygen is −193.8 kJ.

Americium is an element named for a country in the Western Hemisphere

The element named after a location in the Western Hemisphere is Tennessine (Ts), named after Tennessee. Naming elements after locations or scientists is customary in chemistry, pending approval by IUPAC.

The element named after a country in the Western Hemisphere is Tennessine, with the symbol Ts. Element 117, Tennessine, was proposed by a team of scientists from Russia and the United States, named after the state of Tennessee. Naming elements after locations, particularly cities, regions, or countries is one of the traditions followed in the scientific community, as seen when a Russian research team named element 118 Oganesson, symbol Og, after the scientist Yuri Oganessian, who made significant contributions to the discovery of heavy elements.

Traditionally, it is typical for the discoverers of a new element to propose a name, which is then subject to approval by the International Union of Pure and Applied Chemistry (IUPAC) before it becomes official. Before getting their final names, elements may carry temporary systematic names based on Latin numbers corresponding to their atomic numbers.

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Answer:

2.38

Explanation:

119 divided by 50 =2.38

i hope you get it right :)

A neutral atom of tin has 50 protons, 50 electrons, and 69 neutrons. The number of protons and electrons is determined by the atomic number, while the number of neutrons is calculated by subtracting the atomic number from the mass number.

An atom of tin (Sn) is described by its atomic number and its mass number. The atomic number refers to the number of protons in the nucleus of the atom, and in a neutral atom, it also equals the number of electrons. Therefore, a neutral tin atom has 50 protons and 50 electrons.

The mass number, on the other hand, is the sum of protons and neutrons in the nucleus. Therefore, to find out the number of neutrons, you have to subtract the atomic number from the mass number: 119 - 50 = 69. So, a neutral tin atom has 69 neutrons.

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Answer:

Particle /                          mass number                atomic number

type of radiation

alpha / ⁴₂ He                              4                                     2

beta / e⁰                                     0                                   - 1

gamma / γ                                  0                                     0

neutron / n                                 1                                      0

Explanation:

These are the basic types of radiation: alpha (α), beta (β),  gamma (γ), and neutrons (n).

The radiation is emitted by unstable nuclei when undergo radiactive decay or by nuclei that are shooted by other particles.

Alpha radiation (⁴₂ He):

They are nuclei of helium-4 atoms: 2 protons and 2 neutrons.

Hence, the atomic number, which is the number of protons,  of these particles, is 2; and the mass number, which is the sum of protons and neutrons, is 2 + 2 = 4.

The symbol of this radiation is ⁴₂ He, where the superscript to the left of the chemical symbol is the mass number and the subscript to the left of the chemical symbol is the atomic number.

Beta (⁰₋ ₁ e)

These are electrons emiited from an unstable nucleus. The symbols used for this particle are either ⁰₋ ₁β or ⁰₋ ₁e.

The superscript 0 indicates that the relative mass of this particle is practically zero and the subscript -1 tells that the emission of electrons increases the atomic number of the nucleus that emits it.

Gamma (⁰₀γ)

The gamma rays are electromagnetic radiation of high (the highest) energy.

The both superscript and subscript are zero, meaning that this radiation does not change either the mass or atomic numbers of the nucleus.

Neutron (¹₀n):

Neutrons are also emiited from the nucleus and so they may be considered a radiation.

The atomic number of neutrons is 0 (since it does not have protons) and its mass number is 1.

Radioactive rays are divided into three, namely alpha rays, beta rays, and gamma rays.  

Check the picture below!

Types of Radioactive Rays

Based on the constituent particles, radioactive rays are divided into three, namely alpha rays, beta rays, and gamma rays.

Alfa ray (α ray)

Alpha rays are rays emitted by radioactive elements. This ray was discovered simultaneously with the discovery of the phenomenon of radioactivity, which is the decay of the nucleus which takes place spontaneously, uncontrolled, and produces radiation. Alpha rays consist of two protons and two neutrons. The following are the nature of alpha rays.

Beta Rays (β Rays)

Beta rays are high-energy electrons that originate from the nucleus. Here are some of the nature of beta rays.

Gamma Rays (Rays ˠ)

Gamma rays are radiation from electromagnetic waves that emanate from very high energy nuclei that have no mass or charge. Gamma rays also emit when a core emits alpha rays and beta rays. Gamma-ray decay does not cause changes in atomic number or atomic mass.

Gamma rays have the following natural properties:

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Detail

Class: High School

Subject: Chemistry

Keyword: Alpha, Beta, Gamma

Answer:

C. Pulverizing the zinc metal into a fine powder

Explanation:

Answer: C. Pulverizing the zinc metal into a fine powder

Explanation:

Answer:

B. 92.4 g

Explanation:

the balanced equation for the reaction is as follows

CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O

molar ratio of CaCO₃ to CaCl₂ is 1:1

number of CaCO₃ moles reacted - 104 g / 100 g/mol = 1.04 mol

therefore number of CaCl₂ moles reacted - 1.04 mol

mass of CaCl₂ expected to be formed = 1.04 mol x 111 g/mol = 115.44 g

percentage yield = actual yield / theoretical yield x 100 %

theoretical yield = 115.44 g

percentage yield = 80.15 %

substituting these values in the above equation

80.15 % = actual yield / 115.44 g x 100 %

actual yield = 92.5 g

therefore answer is B. 92.4 g

Answer:

Endothermic proceses

Exothermic proceses

Explanation:

By definition an exothermic processes are those in which the system releases energy to the enviroment, while endothermic processes are those in which the system absorbs energy from the environment.

That change in energy is generally measured as heat.

Then an exothermic process releases heat and and endothermic process absorbs heat.

From that, as a first consequence, in an exothermic process the final substances (products in the case of a chemical reaction) ends with lower energy and lower temperature than the initial substances (reactants in the case of a chemical reaction).

The heat content of the substances is usually measured as enthalpy, then, for a chemical reaction you can write th is equation for the enthalpy change:

Thus, for an exothermic reaction ΔH products  < ΔH reactants ⇒ ΔH < 0, whilced for an endothermic reaction ΔH products > ΔH reactants ⇒ ΔH > 0.

The difference in energy between the products and the reactants is a result of the chemical potential energy of the substances, which is the energy stored in the chemical bonds. Then, in an endothermic reaction more energy is needed to break the chemical bonds of the reactants than what is released from the formation of the chemical bonds of the products. Of course the opposite is true: in an exothermic reaction, energy is released because it is required less energy for the formation of the new bonds of the products than what is needed to break the bonds of the reactants.

Answer:

where's the question... ?

4 carbon chain is butane.  With a triple bond it is butane.  Since the triple bond is on the first carbon it can be called 1-butyne, or but-1-yne.

Answer:

It is A. Because charged particles of solar wind ignite different gases in Earth's atmosphere.

Explanation:

Since the solar wind from the sun is too radioactive for humans (they would die), once the charged particles hit the earth's atmosphere it shows its color. Every element has its own color and once it hits the atmosphere it really starts to show.

Because charged particles of solar wind ignite different gases in Earth's atmosphere is the correct statement.

The northern lights, also known as the auroras, are a natural phenomenon that occurs when charged particles from the Sun's solar wind interact with the Earth's magnetosphere and atmosphere. The solar wind consists of charged particles, mainly electrons and protons, that are ejected from the Sun's outer atmosphere.

As these charged particles enter the Earth's magnetosphere, they follow the planet's magnetic field lines and are funneled toward the polar regions. When they reach the Earth's atmosphere, they collide with atoms and molecules, particularly those of oxygen and nitrogen.

These collisions excite the atoms and molecules, causing them to release energy in the form of light. The specific colors observed in the auroras depend on the type of gas present in the atmosphere and the altitude at which the collisions occur. Oxygen atoms emit green and red light, while nitrogen atoms produce blue and purple light.

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Answer:

54.0 g.

Explanation:

2H₂ + O₂ → 2H₂O.

When 3.00 moles of hydrogen molecules and 1.50 moles of oxygen molecules react, they form 3.00 moles of water according to the balanced reaction.

∴ The no. of grams in 3.0 moles of water = no. of moles x molar mass = (3.0 mol)(18.0 g/mol) = 54.0 g.

Answer:

A) 54.0 g

Explanation:

Hello! The answer to your question is 54.0 grams.

2H2 + O2 --> 2H2O

Because of that, we know our result will possess 3 significant figures.

The only sensible answer here (without the calculations) would be 54.0, due to the decimal placement. Hope this helps you!

Answer:

1.99 grams of methane

Explanation:

A combustion reaction always produces the product carbon dioxide and water. So the chemical equation for this reaction would be:

CH₄ + O₂ → CO₂ + H₂O

To answer this question, you need to first determine which is the excess reactant. First step is to balance the equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Next we get the number of moles of each reactant we actually have:

8.00 g of O₂ = ? moles of O₂

You can get the number of moles, by first computing how many grams of the molecule is present in 1 mole.

O = 15.999g/mole. Since there are 2 oxygens in one mole of O₂, all you need to do is add up the atomic mass of 2 oxygens. You will get:

O₂= 31.998 g/mole

You can use this then to determine how many moles of O₂ there are in 8.00g.

[tex]8.00g\times\dfrac{1mole}{31.998g}=0.250moles[/tex]

So there are 0.250 moles of O₂ in 8.00 g of O₂.

We do the same for CH₄

4.00g of CH₄=? moles of CH₄

            C           H₄

CH₄= 12.011 + 1.008(4) = 16.043 g/mole

[tex]4.00g\times\dfrac{1mole}{16.043g}=0.249moles[/tex]

So let's sum up our new given. We now have:

0.250 moles of O₂

0.249 moles of CH₄

Next we look at the molar ratio of reactants to produce products:

CH₄ + 2O₂ → CO₂ + 2H₂O

According to this equation we can assume the following:

We need 1 mole of CH₄ for every 2 moles of O₂ in this reaction. Using what we have, we will see how much of reactant of the other reactant we need to use up the other.

[tex]0.25molesofO_{2}\times\dfrac{1moleofCH_{4}}{2molesofO_{2}}=0.125molesofCH_{4}\\\\0.249molesofCH_{4}\times\dfrac{2molesofO_{2}}{1moleofCH_{4}}=0.498molesofO_{2}[/tex]

Compare the results with what we have:

What we have                   What we need

0.250 moles of O₂   <     0.498 moles of O₂

0.249 moles of CH₄ >     0.125 moles of CH₄

This means that since we have less O₂ that what we need to use up CH₄, then O₂ is the limiting reactant and CH₄ is the excess.

To compute how much we have in excess, we use the number of moles produced when we use up limiting reactant which we did earlier and convert it into grams to determine how much in grams we used up.

Earlier we solved that we need 0.125 moles of CH₄ to use up all the O₂. Now convert that value into grams:

[tex]0.125molesofCH_{4}\times\dfrac{16.043gofCH_{4}}{1moleofCH_{4}}=2.005g of CH_{4}[/tex]

This means that 2.005g of CH₄ will be used up.

Subtract that from the CH₄ we already have:

4.00 g - 2.005 g =1.99 g of CH₄

Answer:

[tex]\boxed{\text{KNO}_{3}}[/tex]

Explanation:

The reaction is

KOH(aq) + HNO₃(aq) ⟶ KNO₃(aq) + H₂O(ℓ)

If you evaporate the water, the solid substance is the compound, potassium nitrate.

[tex]\boxed{\textbf{KNO}_{3}}[/tex]

KNO₃(aq) ⟶ KNO₃(s)

Specific heat capacity of a substance is the amount of heat required to raise the temperature by one degree Celsius of one gram of a substance. Therefore, the correct option is option A.

Enthalpy term is basically used in thermodynamics to show the overall energy that a matter have. Mathematically, Enthalpy is directly proportional to specific heat capacity of a substances.

Mathematically,

q = n ×ΔH

where

q = amount of heat

n = no of moles

ΔH = enthalpy

n = w / M.M

w = given mass

M.M = molar mass

A) C₆H₆

melting point =5.5° C

boiling point =80.1 ° C

Heat of fusion = 9.92 kJ/mol

Heat of vaporization =30.8kJ/mol

B) CH₃CH₂CH₂CH₂CH₂COH₆ is solid

C) NaCl is solid

D) NH₃

melting point = -77.73 °C

boiling point =  -33.34 °C

Therefore, the correct option is option A.

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Answer:

B. The number of complete electron shells stays the same.

Explanation:

The periods on a periodic table consists of those elements that are arranged in the horizontal row in the periodic table.

Elements in the same period have the same number of electron shells and this corresponds to the period. This implies that elements of period 2 have two electron shells, those of period 3 have three shells and so on.

The number of Valence electrons of the elements on the same period increases progressively by one across the period from left to right.

A group is the vertical column to which elements are arranged on the periodic table. Elements in the same group have the same number of Valence electrons and the number of shell increases down a group.

When we go down a period, the number of shell stays the same.

The answer is D. 6 you multiply 2 and 3

Answer:

The correct answer is D

Explanation:

because you would take 2 and 3 and multiply them and get 6 or

3×2=6

Answer:

(a) The change in cell voltage is 0.05V

(b) The change in cell voltage is 0.03V

Explanation:

Redox reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases.

Further explanation:

The image taken in context is attached below.

The standard reduction potentials for iron and silver are:

[tex]E^o_{(Fe^{2+}/Fe)}=-0.44V\\E^o_{(Ag^{+}/Ag)}=+0.80V[/tex]

In the given cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction.

So, silver will undergo reduction reaction will get reduced. Iron will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode): [tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]

Reduction half reaction (cathode): [tex]Ag^{+}+e^-\rightarrow Ag[/tex]

Thus, the anode and cathode will be [tex]E^o_{(Fe^{2+}/Fe)}[/tex]

and [tex]E^o_{(Ag^{+}/Ag)}[/tex] respectively.

The overall cell reaction will be,

[tex]2Ag^{+}+Fe\rightarrow Fe^{2+}+2Ag[/tex]

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

[tex]E^o=E^o_{(Ag^{+}/Ag)}-E^o_{(Fe^{2+}/Fe)}[/tex]

[tex]E^o=(+0.80V)-(-0.44V)=1.24V[/tex]

Now we have to calculate the cell potential.

Using Nernst equation :

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]}{[Ag^{+}]^2}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 2

[tex]E_{cell}[/tex] = emf of the cell = ?

Now put all the given values in the above equation, we get:

[tex]E_{cell}=1.24-\frac{0.0592}{2}\log \frac{(1M)}{(1M)^2}[/tex]

[tex]E_{cell}=1.24V[/tex]

Thus, the emf of cell potential is 1.24 V

Part (a):

The ion concentrations in the cathode half-cell [tex](Ag^+/Ag)[/tex] are increased by a factor of 10 from 1 M to 10 M.

The emf of the cell potential will be,

Using Nernst equation :

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]}{[Ag^{+}]^2}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 2

[tex]E_{cell}[/tex] = emf of the cell = ?

Now put all the given values in the above equation, we get:

[tex]E_{cell}=1.24-\frac{0.0592}{2}\log \frac{(1M)}{(10M)^2}[/tex]

[tex]E_{cell}=1.29V[/tex]

The change in cell voltage will be,

[tex]E_{cell}=1.29V-1.24V=0.05V[/tex]

Thus, the change in cell voltage is 0.05V

Part (b):

The ion concentrations in the anode half-cell [tex](Fe^{2+}/Fe)[/tex] are increased by a factor of 10 from 1 M to 10 M.

The emf of the cell potential will be,

Using Nernst equation :

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]}{[Ag^{+}]^2}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 2

[tex]E_{cell}[/tex] = emf of the cell = ?

Now put all the given values in the above equation, we get:

[tex]E_{cell}=1.24-\frac{0.0592}{2}\log \frac{(10M)}{(1M)^2}[/tex]

[tex]E_{cell}=1.27V[/tex]

The change in cell voltage will be,

[tex]E_{cell}=1.27V-1.24V=0.03V[/tex]

Thus, the change in cell voltage is 0.03V

Learn more:

Spontaneity of reaction; https://brainly.com/question/13151873 (answer by Kobenhavn)

Standard reduction potential;  https://brainly.com/question/8739272 (answer by RomeliaThurston)

Keywords:

Nernst equation, standard reduction potential, spontaneity of the reaction.

The change in the cell voltage when the ion concentrations in the ANODE half-cell are increased by a factor of 10 is 0.030  V

What is the change in the cell voltage when the ion concentrations in the ANODE half-cell are increased by a factor of 10?

The Nernst equation

[tex]E=E^o - \frac{RT}{nF} lnQ[/tex]

The given electrochemical cell has both aqueous species Fe 2 +  and  Ag +  at  1  M  concentration

If the anode concentration is increased by a factor of  10 , the cell potential will change by the correction term:  

[tex]\Delta E = -\frac{RT}{nF} lnQ[/tex]

Now we determine the overall reaction

[tex]Fe(s) -> Fe^{2+}(aq)+2e [anode]\\2*(Ag^+(aq)+e -> Ag(s)) [cathode]\\Fe(s)+2Ag^+(aq)->Fe^{2+}(aq)+2Ag(s)  [overall][/tex]

n=2 electrons were transferred

[tex]Q=\frac{[Fe^{2+}]}{[Ag^+]^2}[/tex]

The change in cell potential is

[tex]\Delta E = - \frac{RT}{nF} ln \frac{[Fe^{2+}]}{[Ag^+]^2}[/tex]

Note that [tex][Fe^{2+}]=10M[/tex] and assume that T=298.15 K

[tex]\Delta E = - \frac{8.314J/molK*298.15K}{2*96485C/mol} ln \frac{10M}{1M^2}\\\Delta E = -0.030 V[/tex]

Therefore the cell potential will decrease by 0.030 V

Grade:        9

Subject:  chemistry

Chapter:   the ion concentrations

Keywords:  the cell voltage, the ion concentrations, the cathode half-cell,   the anode, factor

Answer:

Immoral means not moral and connotes evil or licentious behavior. Amoral, nonmoral, and unmoral, virtually synonymous although the first is by far the most common form, mean utterly lacking in morals (either good or bad), neither moral nor immoral.

so I would say no

Hope This Helps!  Have A Nice Day!!

Question 1:

1. One form of niobium oxide crystallizes in the unit cell shown above.

A. How many niobium atoms per unit cell? 3

B. How many oxygen atoms per unit cell? 3

C. What is the formula of niobium oxide? Nb2O5

Answer: A) 3 B) 3 C) Nb205

Question 2:

2. Quartz (left) and glass (right) are both forms of silicon dioxide. A piece of quartz breaks into a collection of smaller regular crystals with smooth faces. A piece of glass breaks into irregular shards. Use molecular structures to explain why the two solids break so differently. Select all that apply.

1. the bonding and geometry in quartz shows regular repeating patterns *

2.  the bonding and geometry in quartz shows irregular repeating patterns

3. the bonding and geometry in glass shows irregular repeating patterns *

4. the atomic geometries of the two solids can explain why they break so differently  *

5. the atomic geometries of the two solids cannot explain why they break so differently

6. the bonding and geometry in glass shows regular repeating patterns

Answer: 1,3,4

3. Solid silver adopts a face-centered cubic lattice. The metallic radius of a silver atom is 144 pm.

a. How many silver atoms occupy one unit cell of solid silver? 4

b. Calculate the length (in pm) of one side of the unit cell. 407

c. Calculate the volume (in m3) of one the unit cell.  6.74 x10^-29

d. What percentage (expressed to four significant figures) of the volume is empty? 25.95

Answer: a)4 b)407  c)6.74 x10^-29   d)25.95

4. Why hexane is insoluble in water? Select all that apply.

1. Water is polar solvent. *

2. Polar solvents dissolve both polar and non-polar compounds.

3. Polar solvents dissolve only polar compounds. *

4. Solubility does not depends on the polarity of the solvents and reagents.

5. Hexane is non-polar compound. *

Answer: 1,3,5

<3 goodluck

Final answer:

The empirical formula for the niobium oxide with oxide ions at the middle of each edge and niobium atoms at the center of each face is Nb2O, obtained by counting the ions per unit cell and adjusting for their location in the structure.

Explanation:

To determine the empirical formula of the niobium oxide, first, we need to consider how many ions are associated with each other in the cubic unit cell structure. A cubic unit cell with oxide ions at the middle of each edge and niobium atoms at the center of each face can be described as follows: each edge has 1/4 of an oxide ion since it is shared by 4 cubes, resulting in 1 oxide ion per unit cell (there are 12 edges, thus 12 x 1/4 = 3, but each oxygen is set twice, thus 3 / 2 = 1.5 but oxygen only gives 1 charge so we only take half). Each face-centered niobium contributes 1/2 of an atom to the unit cell since each face is shared between two unit cells, leading to 3 niobium atoms per unit cell (6 faces x 1/2 = 3). Therefore, for every 1 niobium atom, there is 0.5 of an oxide ion, and by multiplying both the niobium and oxide amounts by 2 to get whole numbers, we obtain the empirical formula Nb2O.

The coordination number of an ion in a crystalline structure is the number of counterions that surround it. In the niobium oxide unit cell described, the oxide ions are at the midpoint of each edge, and niobium atoms are at the center of each face. Without additional information, however, it is not possible to determine the exact coordination number, as this would depend on the 3D arrangement of the atoms within the cell.