A voltaic cell is constructed in which the anode is a Fe2+|Fe3+ half cell and the cathode is a Br-|Br2 half cell. The half-cell compartments are connected by a salt bridge.
(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
The anode reaction is:
(aq)(s)(l)(g) + (aq)(s)(l)(g) -----> (aq)(s)(l)(g) + (aq)(s)(l)(g)
The cathode reaction is:
(aq)(s)(l)(g) + (aq)(s)(l)(g) --------> (aq)(s)(l)(g) + (aq)(s)(l)(g)
The net cell reaction is:
(aq)(s)(l)(g) + (aq)(s)(l)(g) ------> (aq)(s)(l)(g) + (aq)(s)(l)(g)
In the external circuit, electrons migrate _____(from/to) the Fe2+|Fe3+ electrode _____(from/to) the Br-|Br2 electrode.
In the salt bridge, anions migrate _____(from/to) the Br-|Br2 compartment _____(from/to) the Fe2+|Fe3+ compartment.
#2
A voltaic cell is constructed in which the following cell reaction occurs. The half-cell compartments are connected by a salt bridge.
I2(s) + Pb(s) --->2I-(aq) + Pb2+(aq)
The anode reaction is:
(aq)(s)(l)(g) + (aq)(s)(l)(g) ------> (aq)(s)(l)(g) + (aq)(s)(l)(g)
The cathode reaction is:
(aq)(s)(l)(g) + (aq)(s)(l)(g) ------> (aq)(s)(l)(g) + (aq)(s)(l)(g)
In the external circuit, electrons migrate _____(from/to) the Pb|Pb2+ electrode _____(from/to) the I-|I2 electrode.
In the salt bridge, anions migrate _____(from/to) the Pb|Pb2+ compartment _____(from/to) the I-|I2 compartment.

An expression for the solubility product constant (Ksp) of manganese(II) hydroxide is Ksp = [[tex]Mn^{2+}[/tex]][[tex]OH^-[/tex]]. b) Molar solubility (S) of Mn(OH)[tex]_2[/tex] in water is  2.3 x [tex]10^{-21}[/tex]. c) The Ksp value is 2.3 x  [tex]10^{-21}[/tex].

a) The expression for the solubility product constant (Ksp) of manganese (II) hydroxide, Mn(OH)[tex]_2[/tex] is given below:

Ksp = [[tex]Mn^{2+}[/tex]][[tex]OH^-[/tex][tex]]^{2}[/tex]

Mn(OH)[tex]_2[/tex] ⇌ [tex]Mn^{2+}[/tex] + 2[tex]OH^-[/tex]

The equation shows that the stoichiometry of the reaction is one mole of Mn(OH)[tex]_2[/tex] dissociating to give one mole of [tex]Mn^{2+}[/tex] and two moles of [tex]OH^-[/tex].

b) Given the concentration of hydroxide ([tex]OH^-[/tex]) in a saturated solution of Mn(OH)[tex]_2[/tex], which is 7.5 x [tex]10^{-8}[/tex] M.

The molar solubility (S) of Mn(OH)[tex]_2[/tex] in water is determined by using the stoichiometry of the dissociation reaction and the equilibrium expression.

Using the stoichiometry of the dissociation reaction and the equilibrium expression;[[tex]Mn^{2+}[/tex]] = S[[tex]OH^-[/tex]] = 2S

Therefore, Ksp = [[tex]Mn^{2+}[/tex]][[tex]OH^-[/tex][tex]]^{2}[/tex]

= S × [tex](2S)^2[/tex]

= 4[tex]S^3[/tex]

= 4 (7.5 x[tex]10^{-8})^3[/tex]

= 2.3 x [tex]10^{-21}[/tex]

c) The Ksp value is 2.3 x  [tex]10^{-21}[/tex] and the molar solubility (S) is 6.6 x [tex]10^{-8}[/tex] M which are calculated above.

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Answer:

m = 204.16 grams

Explanation:

Given that,

No. of moles, n = 3.49

Molar mass of NaCl, M = 58.5 g/mol

We need to find the mass of 3.49 moles NaCl​. We know that,

[tex]n=\dfrac{m}{M}\\\\m=n\times M\\\\m=3.49 \times 58.5\\\\m=204.16\ g[/tex]

So, there are 204.16 grams in 3.49 moles NaCl​.

Answer:

24 hrs it takes for the earth to rotate

The pH after the addition of 30.0 mL of NaOH at 25 °C is 8.82.

To calculate the pH after the addition of NaOH, we need to determine the moles of CH3COOH and NaOH that react, and then use the stoichiometry to find the resulting concentration of CH3COOH and OH-.

Given:

Volume of CH3COOH = 30.0 mL = 0.0300 L

Concentration of CH3COOH = 0.50 M

Ka for CH3COOH = 1.8×10^(-5)

Volume of NaOH = 30.0 mL = 0.0300 L

Concentration of NaOH = 0.50 M

First, we calculate the moles of CH3COOH:

moles of CH3COOH = concentration × volume

moles of CH3COOH = 0.50 M × 0.0300 L

moles of CH3COOH = 0.015 mol

Since CH3COOH and NaOH react in a 1:1 ratio, the moles of NaOH are also 0.015 mol.

Next, we calculate the moles of OH- produced:

moles of OH- = moles of NaOH

moles of OH- = 0.015 mol

Now, we calculate the concentration of OH-:

concentration of OH- = moles of OH- / total volume

concentration of OH- = 0.015 mol / (0.0300 L + 0.0300 L)

concentration of OH- = 0.250 M

Using the equilibrium expression for the dissociation of water, we can calculate the concentration of H+ (or H3O+):

[H+][OH-] = Kw

[H+] = Kw / [OH-]

[H+] = 1.0 × 10^(-14) / 0.250 M

[H+] = 4.0 × 10^(-14) M

Finally, we calculate the pH:

pH = -log[H+]

pH = -log(4.0 × 10^(-14))

pH = 8.82

The pH after the addition of 30.0 mL of NaOH at 25 °C is 8.82.

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Answer:

c

Explanation:

The balanced equation is: NaHCO₃ + HCl → NaCl + H2₃.

We can see that for every mole of NaHCO₃, one mole of HCl is required to convert it to H2CO₃. Therefore, the maximum amount of strong acid that can be added to the buffer is 0.35 mol HCl.

A buffer is a solution that resists pH change upon the addition of small amounts of acid or base. It is a solution of a weak acid and its corresponding conjugate base or a weak base and its corresponding conjugate acid. A buffer system consists of a weak acid or a weak base, and its conjugate base or conjugate acid.

When an acid or base is added to the buffer solution, the equilibrium is shifted either to the left or right, thereby minimizing the change in pH. Therefore, to determine the maximum amount of strong acid that can be added to a buffer made by the mixing of 0.35 mol sodium hydrogen carbonate with 0.50 mol sodium carbonate, we first need to identify the components of the buffer.

Sodium hydrogen carbonate (NaHCO₃) is a weak acid and sodium carbonate (Na2CO₃) is its conjugate base. The maximum amount of strong acid that can be added to the buffer is the amount that will convert all of the buffer components to their conjugate acid.

In this case, we need to convert all of the NaHCO₃ to H2CO₃.

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I'm pretty sure ita A. acceleration

Answer:

it's a friction force

The correct answer is O(a) describes a transition-state complementarity model, whereas (b) describes a strict "lock and key" model.

The two reaction coordinate diagrams given represent two different models that describe the catalysis of enzyme-catalyzed reactions. Both models have a single intermediate which is the ES complex. In each case, the contribution of binding energy is given, which is 5 in (a) and by 7 in (b). The ES complex is given by 2 in (a) and 4 in (b). The activation energy for the uncatalyzed reaction is given by 5+6 in (a) and by 7+4 in (b). The models presented in the diagrams are the transition-state complementarity model (a) and the strict "lock and key" model

(b). The two models are different from each other. They differ based on the mechanism by which the enzyme catalyzes the reaction. The lock-and-key model assumes that the active site of the enzyme is rigid and complementary to the substrate that fits into it. The substrate is said to "fit" into the active site of the enzyme as a lock fits into a key. In the transition-state complementarity model, the enzyme's active site has flexibility, which allows it to interact with the substrate in such a way that it stabilizes the transition state, reducing the energy required to reach it. The activation energy for the catalyzed reaction is 5 in (a) and is 7 in (b).

Hence, the correct answer is O(a) describes a transition-state complementarity model, whereas (b) describes a strict "lock and key" model.

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Answer:

nitrogen

Explanation:

Answer:

B. Ca

Explanation:

Let's look at the electron configurations of all four elements (I am going to write it in noble gas configuration to make it simpler):

Mg electron configuration: [Ne]3s2

Ca electron configuration: [Ar]4s2

Ar electron configuration: [Ar]

K electron configuration: [Ar] 4s1

We notice that Ca has two electrons in the 4s sublevel, which satisfies what the question is asking for.

The answer is thus B. Ca.

Answer:

I would choose pear fruits are a good source of carbohydrates hopefully its right sorry if its not

Answer: Fruit

Fruits have a lot of carbohydrates in them :) Hope this helped!

Answer:

a, they are the building blocks of proteins.

Explanation:

hope this helps!

~mina

Answer:

A. They are the building blocks of proteins.

(Photo for proof at the bottom.)

Explanation:

Glycine, leucine, and lysine are all amino acids. There are 20 different kinds of amino acids, including the 3 mentioned here. Different amino acids bond together to form different structures, depending on how many there are. Generally, 50 or more amino acids bonded together makes a protein. But there are some exceptions. Amino acids (including glycine, leucine, and lysine) are the building blocks of proteins.

Here's a photo of Edge incase you're doubtful.

Please click the heart if this helped.

In the following reduction-oxidation reaction; ZnO(s) + C(s) → Zn(s) + CO(g), the species which functions as the oxidizing agent is option (B) C(s).

In the reaction ZnO(s) + C(s) → Zn(s) + CO(g), C(s) gains oxygen and goes from being a carbon atom to a carbon dioxide molecule which contains two oxygen atoms. Therefore, C(s) has been oxidized. Carbon is being oxidized in the above reaction because it has gained oxygen.

The term oxidation refers to the loss of electrons by an atom or molecule. Oxidation is a type of chemical reaction in which a substance loses electrons to another substance. The oxidation process usually occurs with a substance's interaction with oxygen.

Therefore, the correct answer is option B) C(s).

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Answer:

Corrosion is the process of deterioration of materials as a result of chemical, electrochemical or other reactions. Rusting is a part of corrosion and is a chemical process which results in the formation of red or orange coating on the surface of metals. ... Rust or rusting can affect only iron and its alloys.

Explanation:

Answer:

D

Explanation:

It states that D has oxygen, and is closer to the Sun.

Hope this helped, and please mark as Brianliest <3

Answer:

I am pretty sure it is B

Explanation:

The others would either burn u like a crispy chicken nugget or freeze you like a popsicle

A phase diagram is a graphical representation or map that shows the different phases (solid, liquid, and gas) of a substance as a function of pressure and temperature. It provides valuable information about the conditions under which a substance exists in each phase or undergoes phase transitions.

In a phase diagram, pressure is typically represented on the y-axis, and temperature is represented on the x-axis. The diagram is divided into regions corresponding to different phases, and boundaries or lines separating these regions represent phase transitions.

The phase diagram allows us to determine the conditions at which a substance can exist as a solid, liquid, or gas, and it provides insights into the effects of temperature and pressure on the phase behavior of the substance.

It helps in understanding processes such as melting, boiling, and sublimation and provides a useful tool in various fields including chemistry, physics, and materials science.

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Answer: it think the answer is c.) The cell gets help from white blood cells or b.) The cell gets help from its neighbor cells.

Explanation:

Answer:

the limiting reactant is aluminum

Explanation:

The amounts of isopropyl 99% alcohol and sterile water you need to use to make the 70% isopropyl solution using the alligation method are 707.1 mL of isopropyl 99% and 292.9 mL of water. The correct answer is option C.

Alligation method is a type of mathematical process used to calculate the quantities of two or more components of a mixture to obtain a mixture with a specific concentration or quality. This method uses the concept of averaging and the weighted average to get the desired results.

The allegation formula for isopropyl alcohol solution: Required % strength —— Higher % strength | DifferenceLower % strength | Difference70% ——- 99% | 29%70-29=41% | 29%

So, the required ratio of the alcohol to be mixed with water should be 41:29. Thus, the volumes of the 99% alcohol and water needed are (41/70) L and (29/70) L, respectively. Let's substitute the values and calculate:

41/70 × 1L = 0.586 L of 99% isopropyl alcohol

29/70 × 1L = 0.414 L of sterile water

Now we have the volumes of the 99% isopropyl alcohol and sterile water required. By adding these, the total volume of the solution is 1L, as given in the question. Therefore, the answer is 707.1 mL of 99% isopropyl alcohol and 292.9 mL of sterile water (707.1 mL + 292.9 mL = 1000 mL = 1 L).

Hence, option C) 707.1 mL of isopropyl 99% and 292.9 mL of water is the correct answer.

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Answer:

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

E

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E

E

E

BRAINLIEST PLEASE

In a 0.25 M aqueous solution of boric acid, the species with the highest concentration would be H_{3}BO_ 3}^{-} option B.

Boric acid (H_{3}BO_{3}) is a weak acid that can undergo multiple ionization steps. The given equilibrium constants (Ka) provide information about the ionization reactions. The first ionization reaction of boric acid is as follows:

[tex]H_{3}BO_{3}[/tex] ⇌ H^{+}+ H_{3}BO_ 3}^{-}

Ka1 = 7.3 x 10^-10

The second ionization reaction is:

[tex]H_{3}BO_ 3}^{-}[/tex] ⇌H^{+} + H_{3}BO_ 3}^{-2}

Ka2 = 1.8 x 10^-13

The third ionization reaction is:

H_{3}BO_ 3}^{-2} ⇌ H+ + [tex]BO_{3} ^{-3}[/tex]

Ka3 = 1.6 x [tex]10^{-14}[/tex]

To determine the species with the highest , we need to compare the equilibrium constants. The larger the Ka value, the more the reaction favors the dissociation of the species.

Based on the given Ka values, the second ionization reaction (H2BO3- ⇌ [tex]H^{+}[/tex] + H_{3}BO_ 3}^{-2}) has the highest Ka value (1.8 x 10^-13). This indicates that the concentration of H_{3}BO_ 3}^{-} in the solution would be higher compared to the other species. Therefore, option B, H_{3}BO_ 3}^{-}, would have the highest concentration in a 0.25 M aqueous solution of boric acid.

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The equation of the reaction is [tex]2PbS + 3O_{2} --- > 2Pb + 2SO_{3}[/tex] and the mass of lead produced is 2.28 g

The term equation is the means by which we could be able to represent what is going on in the reaction system on a piece of paper. Thus, when we write a reaction equation, it is a representation of the process that is going on in the system. By the use of the stoichiometry of the reaction we could obtain the parameters of the equation.

The balanced reaction equation is; [tex]2PbS + 3O_{2} --- > 2Pb + 2SO_{3}[/tex].

Number of moles of lead (II) sulfide = 2.54 grams/239 g/mol = 0.011 moles

Number of moles of oxygen =  1.88 g/32 g/mol = 0.059 moles

If 2 moles of  lead (II) sulfide reacts with 3 moles of oxygen

0.011 moles of  lead (II) sulfide reacts with  0.011 moles  *  3 moles /2 moles

= 0.0017 moles

Hence  lead (II) sulfide is the limiting reactant.

If 2 moles of lead (II) sulfide produces 2 moles of lead

Mass of the Lead produced =   0.011 moles * 207 g/mol = 2.28 g

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Answer:

Yes.

Explanation:

Yes, we have a problem with sending it to a landfill of copper oxide because it has harmful effect on the health of humans as well as more weight of the copper oxide. Copper oxide usually found in powder form which can easily be inhaled that causes death of the cell due to toxic effect on the mitochondria and lysosomes of the cell. It makes problem of health in carrying the copper oxide from the basement of the factory to the landfill region due to its power form so we can say that it can do problems  to human health while carrying from one place to another.

The element with the ground state electron configuration [Xe]4f¹⁴5d⁶6s² is actually Ir, which represents iridium. Option A is correct.

The electron configuration is a representation of how electrons are distributed among the energy levels, subshells, and orbitals within an atom. It follows a specific notation that describes the arrangement of electrons.

The electron configuration indicates that the electrons are arranged in the following manner;

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁶

The [Xe] notation represents the electron configuration of the noble gas xenon (Xe), which includes all the electrons up to the 5p level. After the noble gas core, we have 4f¹⁴ 5d⁶, which corresponds to the electron configuration of iridium (Ir).

Hence, A is the correct option.

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--The given question is incomplete, the complete question is

"Identify the element with the ground state electron configuration [Xe]4f¹⁴5d⁶6s². Options; A) Ir B) Ta C) Os D) Ru E) Au."--

a. Metals such as zinc (Zn) or aluminum (Al) could be reasonable guesses as they have higher standard reduction potentials than copper. b. Reduction half-reaction is Cu₂+(aq) + 2e⁻ → Cu(s), Oxidation half-reaction is Unknown metal (M)(s) → M₂+(aq) + 2e⁻.

a. To make a reasonable guess for the unknown metal in the electrochemical cell, we can consider the standard reduction potentials (E°) of different metals. The standard reduction potential is a measure of the tendency of a metal to undergo reduction (gain electrons) compared to a standard hydrogen electrode (SHE).

Since the free energy change under standard conditions is negative (-293.4 kJ/mol), it indicates a spontaneous redox reaction, where the copper solution undergoes reduction and the unknown metal solution undergoes oxidation.

A reasonable guess for the unknown metal would be a metal with a higher standard reduction potential than copper (Cu). This is because for the overall reaction to be spontaneous, the unknown metal should have a greater tendency to undergo oxidation (lose electrons) compared to copper. Based on this, metals such as zinc (Zn) or aluminum (Al) could be reasonable guesses as they have higher standard reduction potentials than copper.

b. The reduction and oxidation half-reactions can be written as follows

Reduction half-reaction

Cu₂+(aq) + 2e- → Cu(s)

Oxidation half-reaction

Unknown metal (M)(s) → M₂+(aq) + 2e⁻

In the reduction half-reaction, copper ions (Cu₂⁺) in solution are reduced and gain two electrons to form solid copper (Cu).

In the oxidation half-reaction, the unknown metal (M) in solid form is oxidized, losing two electrons to form metal ions (M₂⁺) in solution.

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In the Lewis structure of HCN (hydrogen cyanide), we can determine the number of bonding electrons and lone pair electrons by considering the valence electrons of each atom involved.

Hydrogen (H) is in Group 1 of the periodic table and has one valence electron. Carbon (C) is in Group 14 and has four valence electrons, and nitrogen (N) is in Group 15 with five valence electrons. Cyanide (CN) is a polyatomic ion, and since we have one carbon and one nitrogen, we have a total of nine valence electrons (four from carbon and five from nitrogen). In the Lewis structure, we first connect the atoms using a single bond between carbon and nitrogen, and carbon forms a single bond with hydrogen. This accounts for two of the nine valence electrons. To complete the octets around each atom, we distribute the remaining seven valence electrons as lone pairs around the nitrogen and carbon atoms. The lone pairs do not participate in bonding and are considered nonbonding or lone pair electrons. Therefore, in the Lewis structure of HCN, there are two bonding electrons and seven lone pair electrons (nonbonding electrons). The bonding electrons are involved in forming the bonds between the atoms, while the lone pair electrons are localized on the nitrogen and carbon atoms, fulfilling their octets.

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Answer:

85.5%

Explanation:

To get the experimental value, you need to convert 15.0 grams of Na2SO4 to grams of Na3PO4. You do this with stoichiometry.

Convert grams of Na2SO4 to moles with molar mass. Then convert to moles of Na3PO4 with the mole-to-mole ratio according to the balanced chemical equation. Then convert moles of Na3PO4 to grams with the molar mass.

15.0 g Na2SO4 x (1 mol/142.04 g) x (2 Na3PO4/3Na2SO4) x (163.94 g/1 mol) = 11.7 g Na3PO4

Percent Yield = (actual value/experimental value) x 100

Actual Value = 10.0 g

Experimental Value = 11.7 g

10.0g/11.7 g = 85.5%

The two emissions scenarios that most closely represent the current trend in CO2 emissions are the Representative Concentration Pathway (RCP) 4.5 and RCP 6.0 scenarios. RCP 4.5 assumes moderate emission reduction efforts, while RCP 6.0 represents a higher emission trajectory, reflecting the current trend where emissions reductions are not keeping pace with necessary targets.

The Representative Concentration Pathways (RCPs) are scenarios used to assess the potential impacts of greenhouse gas emissions on the climate system. RCP 4.5 assumes a moderate emission reduction pathway, with emissions peaking around 2040 and declining gradually. On the other hand, RCP 6.0 represents a higher emission trajectory, with emissions peaking later and declining more slowly compared to RCP 4.5. This scenario aligns with the current trend of rising CO2 emissions, indicating that global efforts to reduce emissions have not been sufficient. The current trend is closer to RCP 6.0, highlighting the challenges of achieving widespread emission reductions in various sectors of the global economy.

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The product of the reaction between [tex]CH_{3}[/tex]-CH=[tex]CH_{2}[/tex] and HBr is 2-bromo-3-methylbutane.

The reaction proceeds through the addition of a proton from HBr to the double bond, followed by the addition of a bromide ion.

The addition of the proton is stereospecific, and the bromide ion will add to the carbon atom that is least substituted by hydrogen. In this case, the carbon atom that is least substituted by hydrogen is the carbon atom that is attached to two hydrogen atoms.

Therefore, the bromide ion will add to the carbon atom that is attached to the double bond and the methyl group. The product of the reaction is 2-bromo-3-methylbutane.

here is the predominant product of the reaction of [tex]CH_{3}CHCH=CH-CH_{3}[/tex] with HBr:

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The complete question is:

Draw the predominant product(s) of the following reactions including stereochemistry when appropriate. CH CH CH -CEC-H HBr Consider EIZ stercochemistry of alkenes. Do not show stereochemistry in other cases If no reaction occurs_ raw the organic starting material. Draw one stnicture per sketcher Add additional sketchers using the drop down menu in the bottom right corner Separate multiple products using the sign from the drop-down menu: ChemDoodle Submit Answver Retry Entire Group more cirovp attempts remaining