Using data in Appendix E in the textbook,
calculate the standard emf for each of the following reactions:
1.H2(g)+F2(g)→2H+(aq)+2F−(aq).
2.Cu2+(aq)+Ca(s)→Cu(s)+Ca2+(aq).
3.2ClO−3(aq)+10Br−(aq)+12H+(aq)→Cl2(g)+5Br2(l)+6H2O(l).

The half-reaction Br[tex]_2[/tex](l) + 2[tex]e^-[/tex] → 2[tex]Br^-[/tex](aq) is a reduction half-reaction, while the half-reaction Zn(s) → [tex]Zn^{2+}[/tex](aq) + 2[tex]e^-[/tex] is an oxidation half-reaction.

The given half-reactions are:

Br[tex]_2[/tex](l) + 2[tex]e^-[/tex] → 2[tex]Br^-[/tex](aq)

Zn(s) → [tex]Zn^{2+}[/tex](aq) + 2[tex]e^-[/tex]

The oxidation and reduction reactions are defined as follows:

Oxidation reaction: A half-reaction that includes the loss of electrons is referred to as an oxidation reaction. The oxidation number of the species involved in the reaction is increased in this process.

Reduction reaction: A half-reaction that involves gaining electrons is referred to as a reduction reaction. The oxidation number of the species involved in the reaction is decreased in this process.

Now let us identify which half-reaction is oxidation and which is reduction:

Br[tex]_2[/tex](l) + 2[tex]e^-[/tex] → 2[tex]Br^-[/tex](aq) (reduction reaction)

Zn(s) → [tex]Zn^{2+}[/tex](aq) + 2[tex]e^-[/tex](oxidation reaction)

Thus, the half-reaction Br[tex]_2[/tex](l) + 2[tex]e^-[/tex] → 2[tex]Br^-[/tex](aq) is an example of a reduction half-reaction, while the half-reaction Zn(s) → [tex]Zn^{2+}[/tex](aq) + 2[tex]e^-[/tex] is an example of an oxidation half-reaction.

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The most electrically conductive solution among the three given solutions is hydrobromic acid, which ionizes completely in the solution and produces a high amount of ions. Therefore, option (b) hydrobromic acid is the correct answer to this question.

Acids produce ions in solution, which leads to the solution being more conductive. The more ions an acid produces, the higher the electrical conductivity of the solution. Hence, to determine the electrical conductivity of the acid, we need to know the number of ions generated by the acid in the solution. Here, three acids: hydrobromic acid, acetic acid, and chloric acid have been taken at the same concentration. So, let's check the number of ions produced by each of the acids: Hydrobromic acid: It is a strong acid that dissociates completely in the aqueous solution.

The dissociation reaction is: HBr → H+ + Br−Since it ionizes completely in the solution, the electrical conductivity of the solution would be high. Acetic acid: It is a weak acid that dissociates partially in the aqueous solution. The dissociation reaction is: CH3COOH ↔ H+ + CH3COO−Since it does not ionize completely in the solution, the electrical conductivity of the solution would be low compared to hydrobromic acid. Chloric acid: It is a strong acid that dissociates completely in the aqueous solution. The dissociation reaction is: HClO3 → H+ + ClO3−Since it ionizes completely in the solution, the electrical conductivity of the solution would be high. So, the most electrically conductive solution among the three given solutions is hydrobromic acid, which ionizes completely in the solution and produces a high amount of ions. Therefore, option (b) hydrobromic acid is the correct answer to this question.

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Answer:

B. One element changes into another.

Explanation:

We can see from the equation that uranium is changed to thorium and helium. Transmutation is the process by which one element is changed to another through radioactive decay, nuclear bombardment, or other similar processes.

Answer:b

Explanation:

The bromine atom in BrF3 is sp3d hybridized, in BrO2- it is sp3 hybridized, and in BrF5 it is sp3d2 hybridized.

The hybridization of bromine in each of the following compounds is as follows:

a) BrF3:

The central atom, bromine (Br), in BrF3 undergoes sp3d hybridization. It forms three sigma bonds with three fluorine atoms and has two lone pairs of electrons. The three sigma bonds are formed by overlapping of the sp3d hybrid orbitals of bromine with the 2p orbitals of the fluorine atoms.

b) BrO2-:

The central atom, bromine (Br), in BrO2- undergoes sp3 hybridization. It forms two sigma bonds with two oxygen atoms and has two lone pairs of electrons. The two sigma bonds are formed by overlapping of the sp3 hybrid orbitals of bromine with the 2p orbitals of the oxygen atoms.

c) BrF5:

The central atom, bromine (Br), in BrF5 undergoes sp3d2 hybridization. It forms five sigma bonds with five fluorine atoms and has one lone pair of electrons. The five sigma bonds are formed by overlapping of the sp3d2 hybrid orbitals of bromine with the 2p orbitals of the fluorine atoms.

Hybridization is a concept used to describe the mixing of atomic orbitals to form new hybrid orbitals in a molecule. It helps to explain the molecular geometry and bonding in the molecule. The hybridization of an atom depends on the number of sigma bonds it forms and the number of lone pairs of electrons it possesses.

In summary, the bromine atom in BrF3 is sp3d hybridized, in BrO2- it is sp3 hybridized, and in BrF5 it is sp3d2 hybridized. The hybridization of bromine determines the shape and geometry of the molecules and provides insights into the nature of their chemical bonding.

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Answer:

Newtons second law says that acceleration (gaining speed) happens when a force acts on a mass (object)

Explanation:

For example riding your bicycle, your bicycle is the mass and your leg muscles pushing on the pedals of your bicycle is the force.

The minimum hydroxide concentration required for the precipitation of [tex]Fe(OH)_2[/tex] to begin is [tex]1.691* 10^-^8 M.[/tex]

The balanced chemical equation for the precipitation reaction is:

[tex]Fe(OH)_2[/tex](s) ⇌ [tex]Fe^2^+[/tex](aq) + [tex]2OH^-[/tex](aq)

Ksp = [tex][Fe^2^+][OH^-]^2[/tex]

Ksp [tex]Fe(OH)_2[/tex] = [tex]4.87*10^-^1^7[/tex]

Ksp = [tex][Fe^2^+][OH^-]^2[/tex]

[tex]4.87*10^-^1^7[/tex] = (0.085)(2x)²

[tex]4.87*10^-^1^7[/tex] = 0.17x²

x² =[tex]4.87*10^-^1^7[/tex] / 0.17

x² = [tex]2.8647* 10^-^1^6[/tex]

x = [tex]1.691*10^-^8[/tex] M

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To calculate the binding energy per nucleon for each isotope of silver, we need to determine the total binding energy of each isotope and divide it by the number of nucleons (protons and neutrons) in each isotope.

The binding energy per nucleon represents the average energy required to remove a nucleon from the nucleus. It is a measure of the stability of the nucleus. To calculate the binding energy per nucleon, we need to determine the total binding energy and the number of nucleons for each isotope. First, we calculate the total binding energy for each isotope using the mass-energy equivalence equation E = mc², where E is the energy, m is the mass, and c is the speed of light.

For the first isotope, 10747Ag, with an atomic mass of 106.905095 g/mol, we multiply the atomic mass by the Avogadro's number to obtain the mass of one mole of atoms. Then we divide the mass by the number of atoms in one mole to get the mass of one atom. The difference between the mass of one atom and the mass of the individual nucleons (protons and neutrons) gives the mass defect. Multiplying the mass defect by c² gives the binding energy of one atom. Finally, dividing the binding energy by the number of nucleons in the isotope gives the binding energy per nucleon.

We repeat the same calculations for the second isotope, 10947Ag, with an atomic mass of 108.904754 g/mol. The resulting values represent the binding energy per nucleon for each isotope of silver.

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The reaction of (CH₃)₃CO- with iodomethane or 1-chlorobutane will increase the reaction rate due to an increased frequency of collisions and an increased concentration of electrophiles.

The reaction between (CH₃)₃CO- (tert-butoxide ion) and iodomethane or 1-chlorobutane is an example of an SN₂ nucleophilic substitution reaction. In SN₂ reactions, the rate-determining step involves the simultaneous attack of the nucleophile (tert-butoxide ion) on the electrophile (iodomethane or 1-chlorobutane) with the inversion of configuration.

When the concentration of iodomethane is increased in the reaction mixture, it will lead to an increase in the reaction rate. This increase occurs due to the collision theory and the effect of concentration on the reaction kinetics.

According to the collision theory, for a reaction to occur, the reactant molecules must collide with sufficient energy and proper orientation. By increasing the concentration of iodomethane, there are more iodomethane molecules available in the reaction mixture. Consequently, there will be an increased frequency of collisions between the nucleophile (tert-butoxide ion) and the electrophile (iodomethane), increasing the likelihood of successful collisions leading to a reaction.

Additionally, the rate of an SN₂ reaction is dependent on the concentration of both the nucleophile and the electrophile. In this case, as the concentration of iodomethane increases, the concentration of the electrophile increases. This increase in electrophile concentration will result in a higher reaction rate, as more electrophiles are available for reaction with the nucleophile.

Therefore, increasing the concentration of iodomethane or 1-chlorobutane in the reaction mixture will increase the reaction rate due to an increased frequency of collisions and an increased concentration of electrophiles.

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After considering the given data we conclude that the value of the equilibrium constant, Kp, for the reaction [tex]N_2(g) + 2NO_2(g)--- > 2NO_2(g)[/tex] at 298 K is [tex]4.8*10^{-17}[/tex].

To evaluate the value of the equilibrium constant, Kp, for the reaction [tex]N_2(g) + 2NO_2(g)--- > 2NO_2(g)[/tex] at 298 K, we can apply the following steps:
Write the balanced chemical equation for the reaction.
[tex]N_2(g) + 2NO_2(g) - - - > 2NO_2(g)[/tex]
Write the expression for Kp for each of the two reactions given.
[tex]Kp_1 = [NO]^2/[N_2][O_2][/tex]
[tex]Kp_2 = [NO_2]^2/[NO]^2[O_2][/tex]
Here,[tex][NO], [N_2], [O_2], and [NO_2][/tex]are the partial pressures of the respective species at equilibrium.
Stage the given values of [tex]Kp_1[/tex] and [tex]Kp_2[/tex] into the expression for Kp for the desired reaction.
[tex]Kp = Kp_2/Kp_1 = ([NO_2]^2/[NO]^2[O_2])/([NO]^2/[N_2][O_2]) = ([NO_2]^2[N_2])/[NO]^4[/tex]
Restructure the expression for Kp to solve for the unknown partial pressure of [tex]NO_2[/tex].
[tex][NO_2]^2 = Kp[NO]^4/[N_2][/tex]
Stage the given values of [tex]Kp_1[/tex] and [tex]Kp_2[/tex], and the given partial pressures of N2 and O2, into the expression for [NO].
[tex][NO] = \sqrt(Kp_1[N_2][O_2])[/tex]
Stage the calculated value of [NO] into the expression for[tex][NO_2][/tex] to obtain the equilibrium partial pressure of [tex]NO_2[/tex].
[tex][NO_2]^2 = Kp[NO]^4/[N_2][/tex]
[tex][NO_2]^2 = (2.4*10^{12} )(4.0*10^{-31} )([N_2][O_2])^2/[N_2][/tex]
[tex][NO_2]^2 = 9.6*10^{-19} [N_2][O_2]^2[/tex]
[tex][NO_2] = \sqrt(9.6*10^{-19} [N_2][O_2]^2)[/tex]
[tex][NO_2] = 3.1*10^{-10} atm[/tex]
Stage the evaluated values of [NO] and[tex][NO_2][/tex] into the expression for Kp to obtain the equilibrium constant.
[tex]Kp = ([NO_2]^2[N_2])/[NO]^4[/tex]
[tex]Kp = (9.6*10^{-19} )(3.1*10^{-10} )^2/[/\sqrt(4.0*10^{-31*1} atm)]^4[/tex]
[tex]Kp = 4.8*10^{-17}[/tex]
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The mass of the HCl that is required for the reaction is 2.92 g

The mass of the hydrogen gas produced is 0.08 g

Number of moles of the iron = 2.3 g/56 g/mol

= 0.04 moles

We have that 1 mole of the iron reacts with 2 moles of HCl

0.04 moles of iron would react with 0.04 * 2/1

= 0.08 moles

Mass of the HCl required = 0.08 moles * 36.5 g/mol

= 2.92 g

We have that the mole ratio of the iron and the hydrogen is 1:1 kit then follows that mass of the hydrogen produced would now be;

0.04 moles * 2 g/mol

= 0.08 g

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Answer:

It's 0.00015

Explanation:

got it right on edge 2021

Answer:

left is chemical because its chemicals, and right is physical because the ice cream is melting and thats physical

Explanation:

Barium nitrate and sodium chloride do not react with each other because both of them are soluble in water. The chemical equation for the reaction between barium nitrate and sodium chloride is given below.

Ba(NO₃)₂ + 2NaCl → BaCl₂ + 2NaNO₃

The reaction between barium nitrate and sodium chloride is a double displacement reaction, where barium cation is exchanged with sodium cation and nitrate anion is exchanged with chloride anion. But the reaction does not occur due to the solubility of barium nitrate and sodium chloride in water.The solution of barium nitrate and sodium chloride will remain clear and colorless with no precipitation forming. In fact, it is a method of testing the presence of sulfate ions in a solution. A small amount of barium nitrate is added to the solution to form barium sulfate. Since barium sulfate is insoluble, it forms a white precipitate.

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Answer:

The answer to this question is HCl(aq) + H2O(l) = H3O+(aq) + Cl-(aq). HCl is considered a strong acid meaning it is 100% ionized in solution. The equilibrium constant is therefore very large.

You divide by the mass of the solution.

Percent by mass is the mass of the solute divided by the mass of the solution, multiplied by 100.

Example:

You dissolve a 4.00 g sugar cube in 350 mL of tea at 80 °C. The density of water at 80 °C is 0.975 g/mL. What is the percent by mass of sugar in the solution?

Solution

Step 1 — Determine mass of the solute.

The solute is the sugar cube. Mass of solute = 4.00 g.

Step 2 — Determine mass of solvent.

The solvent is the 80 °C water. Use the density of the water to find the mass.

mass of solvent = 350 mL ×

0.975

g

1

mL

= 341 g

Step 3— Determine the total mass of the solution

mass of solution = mass of solute + mass of solvent = 4.00 g + 341 g = 345 g

Step 4 — Determine percent composition by mass of the sugar solution.

% by mass =

mass of solute

mass of solution

× 100 % =

4.00

g

345

g

× 100 % = 1.159 %

The percent composition by mass of the sugar solution is 1.159 %.

The new concentration of the solution is 0.30 M.

To calculate the new concentration of the solution, we need to consider the dilution process.

Given information:

Initial volume of the HCl solution = 20 mL

Initial concentration of the HCl solution = 0.60 M

Final volume of the diluted solution = 40 mL

The number of moles of HCl in the initial solution can be calculated using the formula:

Moles = Concentration × Volume (in liters)

Initial moles of HCl = 0.60 M × (20 mL / 1000 mL/L) = 0.012 moles

During dilution, the moles of solute remain constant. Therefore, the moles of HCl in the diluted solution will also be 0.012 moles.

The final concentration of the solution can be calculated using the formula:

Final Concentration = Moles / Volume (in liters)

Final concentration = 0.012 moles / (40 mL / 1000 mL/L) = 0.30 M

Therefore, the new concentration of the solution after dilution is 0.30 M.

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thats kinda hot not even gon'

Answer:

4- radioactive isotopes

Explanation:

I don't remember exactly but this question was on the regents

Uranium-238  is a non- fissile element. It is a radioactive isotope that can be best used to determine geological formations. Thus, option 4 is correct.

Radioactive isotopes are elements that have an unstable atomic nucleus and can undergo radioactive decay to produce new particles and energy. They have the same atomic number as that of their parent species.

Isobars are a substance that has atomic mass and do not include uranium. Nuclear fusion and fission are the processes of nuclear energy that combine or splits the unstable nucleus to form a new particle.

Therefore, option 4. Uranium-238 is a radioactive isotope.

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Answer:

same I was looking for a question like this I was asking question and someone kept commenting this bitly thing i dont kno I though it was just weird and frustrating lol

Answer:

Q = 114349.5 J

Explanation:

Hello there!

In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:

[tex]Q_1=45g*4.18\frac{J}{g\°C}*(100\°C-40\°C)=11286J\\\\Q_2=45g* 2260 \frac{J}{g} =101700J\\\\Q_3=45*2.02\frac{J}{g\°C}*(115\°C-100\°C)=1363.5J[/tex]

Thus, the total energy turns out to be:

[tex]Q_T=11286J+101700J+1363.5J\\\\Q_T=114349.5J[/tex]

Best regards!

Answer:

1. single replacement

Explanation:

Single replacement can be represented by AB+C ⇄B+AC, which matches the picture

Double replacement and double displacement are the same kind of reaction and can be represented by AB+CD ⇄BD+AC

Decomposition reaction can be represented by AB ⇄A+B

Answer:

Radiometric dating

Explanation:

Radiometric dating would be the most effective way since it is a technique that can establish the age of objects older than a few thousand years.

The different parts of the chemical equation below are identified as follows:

1. A: Solid

2. B: Liquid

3. C: Gas

4. D: Aqueous

5. (s): Solid

6. (l): Liquid

7. (g): Gas

A chemical equation is a symbolic representation of a chemical reaction where reactants are written on the left side and products on the right side. These reactants and products are identified with their respective states such as solid, liquid, gas, and aqueous.

The symbols for the physical states of the reactants and products are included in parentheses, alongside their respective chemical formulas.

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Answer:

what type of question?

Explanation:

whats your favorite color???

Answer:

Becxause there is no whee5r else for the flame to go out fromk

Explanation:

NaCl

Na = 1

Cl=1

Answer:

19. because fire needs oxygen or air to keep the fire burning.

20.A molecule of sodium chloride, NaCl, consists of one atom each of sodium and chlorine. Hence, each molecule of NaCl has 2 atoms total. Similarly, we can calculate the total number of atoms in 1 mole of sodium chloride. Since 1 molecule contains 1 atom each of Na and Cl, 1 mole of NaCl will contain 1 mole each of Na and Cl.

Explanation:

Hope this helps!

The enthalpy change of the reaction is -456.8 kJ.

The balanced chemical equation of the reaction is as follows:

Ca(s) + 1/2O2(g) + CO2(g) → CaCO3(s)

Given reactions:

Ca(s) + 1/2O2(g) → CaO(s) ΔH = -635.1 kJ/mol

CaCO3(s) → CaO(s) + CO2(g) ΔH = 178.3 kJ/mol

The target reaction is a combination of the two given reactions, hence the enthalpy change of the target reaction is the sum of the enthalpies of the two given reactions. It means that:

ΔHrxn = ΔH1 + ΔH2

We have:ΔH1 = -635.1 kJ/mol

ΔH2 = 178.3 kJ/mol

Therefore,ΔHrxn = ΔH1 + ΔH2= (-635.1 kJ/mol) + (178.3 kJ/mol)= -456.8 kJ/mol

Therefore, the enthalpy change of the reaction is -456.8 kJ.

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In the given reaction, 4 moles of KO2 react to produce 3 moles of O2. Therefore, if 10.5 moles of carbon dioxide (CO2) are used at STP, an equal amount of oxygen (O2) is produced. Hence, 10.5 moles of O2 will be generated.

According to the balanced chemical equation, the reaction 4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g) shows that for every 4 moles of KO2, 3 moles of O2 are produced. Therefore, the molar ratio between KO2 and O2 is 4:3.

Since 10.5 moles of CO2 are used, we can use the stoichiometry of the reaction to determine the amount of O2 produced. As the reaction equation does not involve CO2, the amount of CO2 used does not directly affect the production of O2. Thus, 10.5 moles of CO2 will yield an equal amount of O2, resulting in 10.5 moles of O2 being produced at STP.

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