What is the change in oxidation number of sulfur when the dithionite anion, s2o42−, is converted to so2 in an oxidation-reduction reaction? does each sulfur atom gain or lose electrons in the reaction? is sulfur oxidized or reduced in the reaction?

The theoretical yield of iron (II) sulfide would be 155.92 g

It is the total stoichiometric product from a reaction.

From the equation of the reaction:

2Fe(s) + 3S(s) → Fe2S3(s)

The mole ratio of Fe to Fe2S3 is 2:1

Mole of 83.77 g Fe = 83.77/55.85

                               = 1.4999 moles

Equivalent mole of Fe2S3 = 1.4999/2

                                            = 0.75 moles

Mass of 0.75 mole Fe2S3 = 0.75 x 207.9

                                             = 155.92 g

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Answer: Option (c) is the correct answer.

Explanation:

A molecule which has hydrogen bonding will have the highest boiling point. So, out of the given options only [tex]NH_{3}[/tex] will have hydrogen bonding.

Whereas in [tex]CHCl_{3}[/tex] there will be dipole-dipole interactions and no hydrogen bonding within the molecule.

In [tex]OF_{2}[/tex] and [tex]C_{6}H_{6}[/tex], there will be dipole-dipole interaction in both the molecules.

Thus, we can conclude that [tex]NH_{3}[/tex] will have the highest boiling point.

Seafloor spreading provides evidence for D. formation of sedimentary rocks.

In conclusion, when seafloor spreading occurs, there are sediments that overtime accumulates and forms sedimentary rock.

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Examples of matter include;

Keywords: Matter, states of matter, particles, atoms, molecules, elements, compounds  

Level: High school  

Subject: Chemistry  

Topic: Matter and particles of matter  

Sub-topic: Classification of matter

[tex]{{\mathbf{K}}^{\mathbf{+}}}[/tex] will be dragged towards the side of [tex]{\mathbf{O}}{{\mathbf{H}}^{\mathbf{-}}}[/tex] while [tex]{\mathbf{B}}{{\mathbf{r}}^{\mathbf{-}}}[/tex] will be dragged towards the side of [tex]{{\mathbf{H}}^{\mathbf{+}}}[/tex]. (Refer to the attached image)

Further explanation:

Electronegativity:

It is defined as the tendency of an atom to attract the shared electrons in the bond towards itself is known as electronegativity. The more electronegative atom will more attract the bonding electrons towards itself than the less electronegative one. Therefore the electrons will spend more time with the more electronegative atom than an electropositive atom. The electronegative atom will acquire the partial negative charge, and the electropositive atom will acquire a partial positive charge.

The polarity of the bond can be estimated by the electronegativity difference [tex]\left({\Delta {\text{EN}}}\right)[/tex]. [tex]\Delta{\text{EN}}[/tex]is the electronegativity difference between the two atoms that are bonded to each other. The formula to calculate [tex]\Delta{\text{EN}}[/tex]in XY bond is as follows:

[tex]{\mathbf{\Delta EN=}}\left({{\mathbf{electronegativity of Y}}} \right){\mathbf{-}}\left({{\mathbf{electronegativity of X}}}\right)[/tex]

Here, X is the electropositive atom and Y is the electronegative atom.

Higher the electronegativity difference between the two atoms, more will be the polarity of the bond and vice-versa.

Water is a polar molecule. Hydrogen atom acquires partial positive charge while oxygen atom is partially negatively charged. So the end of the water molecule with hydrogen is positively charged, and that with oxygen is negatively charged.

In the case of a potassium-bromine bond, bromine is more electronegative than potassium. So the electrons will be more attracted towards bromine due to which it develops a partial negative charge. Potassium, being less electronegative than bromine, in turn, acquires a partial positive charge. This separation of charge results in the formation of a polar bond between potassium and bromine.

So [tex]{{\text{K}}^+}[/tex] will be dragged towards the side of [tex]{\text{O}}{{\text{H}}^-}[/tex] while [tex]{\text{B}}{{\text{r}}^-}[/tex] will be dragged towards the side of [tex]{{\text{H}}^+}[/tex].

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Covalent bonding and molecular structure

Keywords: electronegativity difference, electropositive, electronegative, electrons, polar, KBr, K, Br, K+, Br-, H+, OH-, potassium, bromine, hydrogen, oxygen.

the answer is A: decrease

Answer: 2.4 grams

Explanation:

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]  

For [tex]O_2[/tex]

Given mass = 19 g

Molar mass of  [tex]O_2[/tex] = 32 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of}O_2=\frac{19}{32}=0.60moles[/tex]

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

1 mole of [tex]O_2[/tex] reacts with 2 moles of [tex]H_2[/tex]

0.60 moles of [tex]O_2[/tex] wil react with =[tex]\frac{2}{1}\times 0.60=1.20[/tex] moles of [tex]H_2[/tex]

mass of [tex]H_2=moles\times {\text {molar mass}}=1.20moles\times 2g/mol=2.4g[/tex]

Thus 2.4 grams of hydrogen will react completely with 19 g of oxygen.

Answer:

Strengthens the evidence and support for a scientific theory.

Explanation:

Hello,

Scientific method provides a compelling tool scientists use to both develop and demonstrate new theories. As it involves both the observation and experimentation towards a specific subject of matter, it turns out convenient to consider alternative explanations substantiating such subject of matter in light of obtaining a more precise explanation for it. In such a way, this investigation of alternative explanations strengthens the evidence and support for a scientific theory.

Best regards.

Explanation:

Acetals are geminal diethers derivatives of aldehyde formed by the addition to equivalent molecules of an alcohol and removal of water.

When ethanol is added to the ethanal in acidic medium:

Ethanal + Ethanol → Hemiacetal

Hemiacetal + Ethanol → Acetal

Acetal produced when ethanol is added to ethanal are given in the image attached.

The concentration of drug after 4.5 hoursis [tex]\boxed{{\text{0}}{\text{.25 mg/100 mL}}}[/tex].

Further Explanation:

Radioactive decayis responsible to stabilizeunstable atomic nucleusaccompanied by the release of energy.

Half-life is the duration after which half of the original sample has been decayed and half is left behind. It is represented by [tex]{t_{{\text{1/2}}}}[/tex].

The relation between rate constant and half-life period for first-order reaction is as follows:

[tex]k = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}[/tex]                                                                                                    …… (1)

Here,

[tex]{t_{{\text{1/2}}}}[/tex] is half-life period.

k is rate constant.

Substitute 90 min for [tex]{t_{{\text{1/2}}}}[/tex] in equation (1).

[tex]\begin{aligned}k &= \frac{{0.693}}{{90{\text{ min}}}} \\&= 7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}} \\\end{aligned}[/tex]  

Radioactive decay formula is as follows:

[tex]{\text{A}} = {{\text{A}}_0}{e^{ - kt}}[/tex]                                        …… (2)

Here

A is the concentration of sample after time t.

t is the time taken.

[tex]{{\text{A}}_0}[/tex] is the initial concentration of sample.

k is the rate constant.

The time for the process is to be converted into min. The conversion factor for this is,

[tex]1{\text{ hr}} = 60\min[/tex]  

Therefore time taken can be calculated as follows:

[tex]\begin{aligned}t&= \left( {4.5{\text{ hr}}} \right)\left( {\frac{{60{\text{ min}}}}{{1{\text{ hr}}}}} \right) \\&= 270{\text{ min}} \\\end{aligned}[/tex]  

Substitute 2 mg/100 mL for [tex]{{\text{A}}_0}[/tex] , [tex]7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}[/tex] for k and 270 min for t in equation (2).

[tex]\begin{aligned}{\text{A}} &= \left( {\frac{{2{\text{ mg}}}}{{100{\text{ mL}}}}} \right){e^{ - \left( {7.7 \times {{10}^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}} \right)\left( {270{\text{ min}}} \right)}} \\&= \frac{{0.25{\text{ mg}}}}{{100{\text{ mL}}}} \\\end{aligned}[/tex]  

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Radioactivity

Keywords: half-life, 0.25 mg/100 mL, 2 mg/100 mL, A, k, t, rate constant, 4.5 h, 270 min, radioactive decay formula.

Answer:

30

Explanation:

To find out the neutron number of an element we must first know what the mass value of that element is and the value of the atom number of that element. To find out these values, just look for the element in the periodic table. Once discovered simply subtract the values using the formula:

No. of neutrons = mass - atom number.

In the case of Cr, the mass is 54, while the atom number is 24. So we can find the number of neutrons.

Cr Neutrons = 54-24 = 30

Answer:

The rate will be doubled by a Factor of 18

Explanation:

Good luck to everyone doing this!! you got this

The rate of the reaction will increase by a factor of 18 if the concentration of C2O42- is tripled and the concentration of HgCl2 is doubled.

The question asks how the rate of the reaction 2 HgCl2(aq) + C2O42-(aq) → 2 Cl-(aq) + 2 CO2 (g) + Hg2Cl2 (s) will change if the concentration of C2O42- is tripled and the concentration of HgCl2 is doubled. Given the rate law rate = k[HgCl2][C2O42-]2, we can predict the effect on the rate. If the rate law is followed, tripling [C2O42-] will increase the rate by a factor of 32 or 9 because the concentration of C2O42- is squared in the rate law. Doubling [HgCl2] will double the rate of reaction. Therefore, tripling [C2O42-] and doubling [HgCl2] together will increase the reaction rate by a factor of 2 * 9 = 18.

For a first-order reaction, the rate constant can be determined using the concentration of the reactant at a given time. In this case, the rate constant is 0.25 s^-1.

A first-order reaction is one in which the rate of reaction is directly proportional to the concentration of the reactant. The rate law expression for a first-order reaction is given by rate = k[A], where [A] is the concentration of the reactant and k is the rate constant.

In this case, the concentration of the reactant decreases to 25% of its initial value in 3.0 minutes. We can use this information to determine the rate constant (k).

25% of the initial concentration corresponds to 0.25 times the initial concentration, so the concentration at that time is 0.25[A]. We can substitute this value into the rate law expression and solve for k:

0.25[A] = k[A]

0.25 = k

Therefore, the value for the rate constant (k) for the reaction is 0.25 s-1.

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To find the rate constant (k) for a first-order reaction, we can use the half-life formula. The given information states that it takes 3.0 minutes for the concentration of the reactant to decrease to 25% of its initial value. By substituting these values into the equation, we can find the rate constant.

To determine the rate constant (k) for the first-order reaction, we can use the formula for the half-life of a first-order reaction. The half-life is the amount of time it takes for the concentration of the reactant to decrease to half of its initial value. In this case, it takes 3.0 minutes for the concentration to decrease to 25% of its initial value, which is equivalent to one half-life.

The formula for the first-order half-life is: t1/2 = ln(2)/k
Since the concentration decreases to 25% of its initial value after one half-life, we can use this information to solve for k:

25% = (1/2) * 100% = e-kt1/2
ln(1/2) = -k * t1/2
k = -ln(1/2) / t1/2

Substituting the given values, we have:
k = -ln(1/2) / 3.0 minutes

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The benzhydrol radical is more stable than the hydroxyl isopropyl radical because it can delocalize its unpaired electron through resonance across the aromatic ring, while the hydroxyl isopropyl radical lacks such extensive resonance stabilization.

When considering the stability of hydroxyl isopropyl and benzhydrol radicals, it's important to look at resonance stabilization. Radicals that can delocalize their unpaired electron through resonance are generally more stable. In the case of a benzhydrol radical, also known as a benzylic radical, there is significant stabilization due to the possibility of multiple resonance structures. This is because the unpaired electron in the benzhydrol radical can be delocalized through the aromatic ring's system of conjugated pi bonds.

On the other hand, the hydroxyl isopropyl radical, which is an alkyl radical, lacks the extended conjugated system that allows for such delocalization. While it may have some hyperconjugative stabilization, it does not benefit from the extensive resonance stabilization that the benzhydrol radical enjoys. As such, the benzhydrol radical is more stable than the hydroxyl isopropyl radical due to the resonance effects and the subsequent delocalization of the unpaired electron across the aromatic ring.

Answer:

c

Explanation:

The value of Kc for the second reaction is mathematically given as

Kc' = 3.769  

What isthe value of Kc for the second reaction?

Question Parameters:

kc = 7.04 × 10-2 for the reaction

2 hbr(g) ⇌h2(g) + br2(g)

1/2 h2(g) + 1/2 br2 ⇌hbr(g)

Generally, the equation for the reaction  is mathematically given as

2HBr(g) ⇄ H2(g) + Br2(g)

Therefore

Kc = [H2] [Br2] / [HBr]^2

7.04X10^-2 = [H2][Br] / [HBr]^2

Upon final reaction

Kc' = [HBr] / [H2]^1/2*[Br2]^1/2

Hence

[tex]\sqrt{(1/7.04X10^-2)}= [HBr] / [H2]^1/2*[Br]^1/2}\\\\Kc' = \sqrt{(1/7.04X10^-2)[/tex]

Kc' = 3.769  

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The pH of a buffer containing NH3 and NH4Cl is determined by the equilibrium between NH4+ and NH3 in water, with the reaction NH4+ (aq) + H2O(l) → H3O+ (aq) + NH3(aq). The reaction demonstrates the action of the ammonium ion as a weak acid, and its Ka is calculated using the Ka = Kw/Kb relationship. The chloride ion does not undergo significant hydrolysis, so it does not affect the pH of the buffer.

The chemical reaction responsible for the pH of a buffer containing NH3 (ammonia) and NH4Cl (ammonium chloride) involves the equilibrium between NH4+ and NH3 in water:

NH4+ (aq) + H2O(l) ⇒ H3O+ (aq) + NH3(aq)

This represents the dissociation of the ammonium ion, which is a weak acid, in water to produce hydronium ions (H3O+) and ammonia. Since ammonia is a weak base, the corresponding acid dissociation constant (Ka) can be calculated using the relation Ka = Kw/Kb, where Kw is the ion product of water and Kb is the base dissociation constant of ammonia.

The chloride ion, being the conjugate base of the strong acid hydrochloric acid (HCl), does not undergo significant hydrolysis in water:

Cl-(aq) + H2O(l) → HCl(aq) + OH−(aq)

Since HCl is a strong acid, the equilibrium constant (Ka) for its conjugate base, Cl-, is essentially zero, which means Cl- does not affect the buffer solution's pH appreciably.

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Answer:

               Option-2 (0.8830 mol C₃H₈) is the correct answer.

Solution:

                 In statement we are given with the amount of propane gas and are asked to find out the moles for given mass.

As we know mass is related to moles as follow,

                                      Moles = Mass / M.mass ----- (1)

Data Given:

                Moles = ??

                Mass = 38.95 g

                M.mass = (C)3 + (H)8 = (12)3 + (1)8 = 36 + 8 = 44 g/mol

Putting values in equation 1,

                                    Moles = 38.95 g / 44 g/mol

                                    Moles = 0.8854 Moles

Answer: Value of A is 2 and value of B is 1.

Explanation:

Lithium is the third element of the periodic table. It belongs to the Period 2 and Group 1 and is considered as an alkali metals.

To write the electronic configuration, we need to find the number of electrons this element contain, which is equal to the atomic number.

Number of electrons = Atomic number = 3

So, the electronic configuration will be: [tex]1s^22s^1[/tex]

The s-orbital in total can contain only 2 electrons.

Hence, the value of A is 2 and value of B is 1.

The periodic table showcases the table of the chemical elements listed in order of atomic number, usually in rows(period) from the left to the right, in such a way that elements with similar atomic structures as well as similar chemical properties can be located in vertical columns (groups).

Lithium is the third chemical element on the periodic table. It can be located on group 1 element also known as the alkali metals. It is soft in nature and has a whitish color.

The electronic configuration explains how electrons are distributed in their atomic orbitals.

The electronic configuration for Lithium can be expressed as:

Li: 1s² 2s¹

Therefore, from the given information, we can conclude that the value for the electronic configuration of A and B  in the Lithium sample given is 2 and 1 respectively

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The basic building block of chemistry is known as the atom. The smallest particle of an element that still retains all the properties of the element is known as the atom. The correct option is C.

The atom can be considered as the basic building blocks of matter which possess the properties of the chemical element. An atom don't exist independently, instead they form ions and molecules which in turn combine in large numbers to form matter.

An atom is an indivisible particle and it contains the sub-atomic particles like protons, electrons and neutrons. The positively charged particles are called protons, the negatively charged particles are called electrons. The neutrons are chargeless particles.

All atoms of the same element are identical but different elements have different types of atoms. The chemical reactions occur when the atoms are rearranged.

Thus the correct option is C.

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You couldn't substitute 3M H2SO4 for concentrated HNO3 in Part 1 of the experiment due to the significant differences in their reactivity and properties. HNO3 is a powerful oxidizing agent and provides specific reactions, while H2SO4 is a strong acid but not as strong an oxidizing agent.

The reason you couldn't substitute 3M H2SO4 for concentrated HNO3 in Part 1 of the experiment is due to the difference in their reactivity and properties. Nitric Acid (HNO3) is a powerful oxidizing agent. This means it has the ability to oxidize other substances, bringing about specific chemical reactions. On the other hand, Sulfuric Acid (H2SO4), while still a strong acid, is not as strong an oxidizing agent and would not yield the same results if used as a substitute for HNO3 in this experiment.

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Final answer:

To produce a single molecule of water (H₂O), exactly two hydrogen atoms are required. This 2:1 ratio of hydrogen to oxygen atoms is consistent no matter how many molecules of water are being made.

Explanation:

To produce a single molecule of water, which has the chemical formula H₂O, you need two hydrogen atoms. The subscript '2' in H₂ indicates that there are two hydrogen atoms for every oxygen atom in a water molecule. Therefore, for each molecule of water you create, you must have this 2:1 ratio of hydrogen to oxygen atoms.

For example, if you want to create two water molecules, you would need a total of 4 hydrogen atoms (2 molecules × 2 atoms/molecule). Similarly, to produce five water molecules, you would require 10 hydrogen atoms in all (5 molecules × 2 atoms/molecule). No matter the number of water molecules you are looking to produce, you will always need twice as many hydrogen atoms as you have water molecules because of this consistent ratio.

Final answer:

C6H5CH2CH2Br is not formed during the radical bromination of C6H5CH2CH3 because the intermediate benzyl radical, formed at the carbon adjacent to the aromatic ring, is much more stable than the primary radical needed for the other product. Selectivity is due to bromine's preference for stable radicals, supported by Hammond's postulate.

Explanation:

The reason why C6H5CH2CH2Br is not formed during the radical bromination of C6H5CH2CH3 involves the relative stability of the radical intermediate. Radical bromination tends to occur at the position that forms the most stable radical, which for a benzyl compound is the carbon atom directly adjacent to the aromatic ring. The radical formed at this position, a benzyl radical, is highly stabilized by resonance. In contrast, the radical that would be required to form C6H5CH2CH2Br is a primary radical, which is less stable and thus less likely to form. This selectivity is due to the fact that bromine radicals are relatively selective and prefer to abstract hydrogen atoms from positions that lead to more stable radical intermediates. Moreover, Hammond's postulate suggests that since the radical formation with bromine is endothermic, the transition state will more closely resemble the stable radical intermediate, leading to more selective radical formation.