An electrical shock happens when electric ________enters the body at one point and leaves through another.

392.1388 g/mol is the molar mass of the given compound. The molar mass of a material is a bulk attribute rather than a molecular one.

The ratio among the mass with the quantity of a substance (measured in mole) of any sample of a chemical compound is known as the molar mass (M) in chemistry. The compound's molar mass represents an average over numerous samples, which frequently have different masses because of isotopes. A terrestrial average as well as a function of the relative distribution of the isotopes of the component atoms on Earth, the molar mass is most frequently calculated using the standard atomic weights.

FeSO[tex]_4[/tex](NH[tex]_4[/tex])[tex]_2[/tex](SO[tex]_4[/tex])[tex]_2[/tex].6H[tex]_2[/tex]O molar mass=

55.845 + 32.065 + 15.9994×4. + (14.0067 + 1.00794×4)×2 + 32.065 + 15.9994×4 + 6×(1.00794×2 + 15.9994)

=392.1388 g/mol

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To increase the amount of iron produced in the equation Fe3O4(s) + 4H2(g) → energy, you need to increase the amount of hydrogen gas (H2) in the reaction.

To increase the amount of iron produced in the equation Fe3O4(s) + 4H2(g) → energy, you would need to increase the amount of hydrogen gas (H2) in the reaction.

According to the equation, for every 4 moles of hydrogen gas, you can produce 1 mole of iron (Fe3O4). So, if you increase the amount of hydrogen gas, you will increase the amount of iron produced in the reaction.

For example, if you double the amount of hydrogen gas from 4 moles to 8 moles, you will also double the amount of iron produced from 1 mole to 2 moles.

Answer:

CaS

Explanation:

Calcium is a metal from Group 2, so it has 2 valence electrons. Sulfur is a non-metal from Group 16, so it has 6 valence electrons. Metals and non-metals form ionic bonds, in which metals lose electrons and non-metals gain electrons. In both cases, they follow the octet rule, trying to have the electron configuration of the closest noble gas, which has its valence shell complete with 8 electrons.

Calcium has 20 electrons and loses 2 electrons to have the electron configuration of Ar and form Ca²⁺.

S has 16 electrons and gains 2 electrons to have the electron configuration of Ar and form S²⁻.

In the formula unit, one atom of Ca²⁺ bonds to one atom of S²⁻ to maintain electroneutrality and form CaS.

A compound is a pure substance because its molecule cannot be broken down into simpler particles by physical means.

Answer:

A compound is a pure substance because its molecule cannot be broken down into simpler particles by physical means.

Explanation:

Answer:

Difference between 133g and 29g

Explanation:

The equation for the reaction is:

C₄H₈O₂ + C₂H₅OH = C₆H₁₂O₂ + H₂O

Now you see that the number of the moles of butanoic acid and etyl butyrate is equal in

the reaction. That means;

number of moles of C₄H₈O₂ = number of moles of C₆H₁₂O₂

mass of C₄H₈O₂/ Molar mass of C₄H₈O₂ = mass of C₆H₁₂O₂/ molar mass of C₆H₁₂O₂

mass of C₆H₁₂O₂ = molar mass of C₆H₁₂O₂ x mass of C₄H₈O₂/ Molar mass of C₄H₈O₂

Now, assuming 100% yield, the mass of ethyl butyrate produced is: 

= 7.45/88.11 x 116.16

=9.82g

Thus, the theoretical yield of ethyl butyrate is 9.82g.

Given 7.45 g of butanoic acid, a perfect 100% yield would produce approximately 9.82 g of ethyl butyrate. This is calculated based on the mole to mole correspondence between butanoic acid and ethyl butyrate and the molar masses of the two substances.

To solve this question, we first need to determine the molar mass of butanoic acid (C4H8O2), which is approximately 88.11 g/mol. Given that we have 7.45 g of butanoic acid, we can calculate the number of moles of butanoic acid to be 7.45 g / 88.11 g/mol ≈ 0.0845 mol.

The reaction of butanoic acid with ethanol produces ethyl butyrate in a 1:1 ratio. So, the same number of moles of ethyl butyrate (0.0845 mol) would be produced in an ideal case.

The molar mass of ethyl butyrate (C6H12O2) is around 116.16 g/mol. Thus, the grams of ethyl butyrate synthesized from the reaction would be 0.0845 mol x 116.16 g/mol ≈ 9.817 g.

Assuming a 100% yield, we would have 9.817 g of ethyl butyrate. However, to express the answer with three significant figures, we round it to 9.82 g.

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The correct answer is...

D. Only the hydrogen needs to be balanced. There are equal numbers of aluminum and chlorine.

To balance this chemical equation, we have to make sure that equal number of elements are located on the left side and the right side by placing coefficients:

CCl4  +  O2   <-- --> COCl2  +  Cl2

Balancing this equation:

2CCl4(g)+ O2(g) ⇌ 2COCl2(g) + 2Cl2(g)

Answer:

2,1,2,2

To balance the equation ccl4(g)+o2(g)⇌cocl2(g) + cl2(g), the coefficients should be: 1,2,1,1.

To balance the equation ccl4(g)+o2(g)⇌cocl2(g) + cl2(g), we need to ensure that the number of each atom is the same on both sides of the equation.

Starting with the carbon atoms, we have 1 on the left side and 1 on the right side. So, the coefficient for CCl4 remains 1.

Next, we have 2 chlorine atoms on the left side and 2 on the right side. So, the coefficient for Cl2 remains 1.

Finally, we have 2 oxygen atoms on the right side and 0 on the left side. To balance the oxygen atoms, we need a coefficient of 2 for O2.

Therefore, the balanced equation is: CCl4(g) + 2O2(g) ⇌ COCl2(g) + Cl2(g).

The number of protons and electrons of an atom is equal and the amount can be obtained by looking at the atomic number of that atom or element.

The atomic numbers are:

Gold = 79

Thallium = 81

So we are looking for the element with an atomic number of 80, that is:

Mercury

Answer:

Mercury (Hg)

The liquid metal that fits the description is mercury (Hg), which has more protons than gold (Au) and is liquid at room temperature. Mercury is unique due to its electronic configuration and is thermally conductive but a poor conductor of heat compared to other metals.

The liquid metal in question that has more protons than gold but fewer electrons than thallium is mercury (Hg). Gold (Au) has an atomic number of 79, meaning it has 79 protons. Thallium (Tl) has an atomic number of 81, meaning it has 81 electrons when neutral. Mercury has an atomic number of 80, placing it right between gold and thallium in terms of protons, and since it is a metal and liquid at room temperature, it fits the description given.

Mercury is a metal that is thermally conductive but compared to other metals like copper or silver, it is a poor conductor of heat. However, it is still a fair conductor of electricity. The electronic configuration of mercury is unique and makes it resistant to losing electrons, which contributes to its liquid state at room temperature and its conduction properties.

Treating the solution with ammonium vanadomolybdate is necessary to form a stable vanadium complex, enabling accurate quantification. This reagent helps to ensure proper reaction conditions for measuring vanadium concentration.

In the context of vanadium quantification, treating the solution with ammonium vanadomolybdate is essential to ensure the formation of a stable and detectable colored complex.

This complex, typically red-brown with the general formula (VO)₂(SO₄)₃, is formed in the presence of H₂O₂ and H₂SO4, and its intensity depends on the amount of vanadium present.

By optimizing the reaction conditions, such as the concentration of H₂O₂ and H₂SO4, and using ammonium vanadomolybdate as a reagent, we can maximize the absorbance at 450 nm for accurate quantitative analysis.

The addition of ammonium vanadomolybdate helps to maintain the proper chemical environment, preventing the interference of other ions and ensuring that the measurement of vanadium concentration is precise.

This reagent assists in forming the colorimetric complex necessary for detection and quantification.

Answer: The correct option is A.

Explanation:  Law of conservation of mass states that the total mass in a chemical reaction remains conserved that is total mass on the reactant side will always be equal to the product side.

We are given 4 chemical reactions, the total mass during a chemical reaction will be same in the case of reaction A because total number of atoms on the reactant side is equal to the total number of atoms on the product side.

[tex]Mg(s)+Cl_2(g)\rightarrow MgCl_2(s)[/tex]

Mass of Magnesium = 24 g/mol

Mass of Chlorine = 35.5 g/mol

[tex]\text{Mass on the reactant side}=24+(2\times 35.5)=95g/mol[/tex]

[tex]\text{Mass on the product side}=24+(2\times 35.5)=95g/mol[/tex]

From the above, it is visible that the Total mass during this chemical reaction is same.

All the other reactions are not balanced, therefore the total mass will not be the conserved.

To calculate the vapor pressure of a benzene solution with camphor, use Raoult's Law considering the negligibility of camphor's vapor pressure and determine the mole fraction of benzene.

Calculating Vapor Pressure of a Solution

To calculate the vapor pressure of a solution containing 28.2 g of camphor (C10H16O) dissolved in 94.8 g of benzene, we can use Raoult's Law. This law states that the partial vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. The formula is P1 = X1 * P1o, where P1 is the vapor pressure of the solvent in the solution, X1 is the mole fraction of the solvent, and P1o is the vapor pressure of the pure solvent.

First, we need to calculate the mole fraction of benzene in the solution. To do that, we must find the moles of both camphor and benzene using their respective molar masses (152.23 g/mol for camphor and 78.11 g/mol for benzene). After calculating the moles, we can find the mole fraction of benzene (X1) by dividing the moles of benzene by the total moles of both compounds. We can then apply the equation above to find the new vapor pressure of the benzene in the solution (vapor pressure of a solution).

Note that this calculation assumes camphor has negligible vapor pressure, which is a reasonable assumption given it is a low-volatility solid. The answer will be in mmHg because that is the unit given for the vapor pressure of pure benzene.

Electrochemistry concepts inform us that it takes three moles of electrons to generate one mole of chromium. The amount of charge transferred, coupled with the current allows us to calculate the required time which is 115134041.88 seconds to produce 4.94 mg of Chromium at a current of 0.234 A.

The student's question deals with the concepts in electrochemistry, specifically involving the reduction of chromium(III) to chromium(0). The key here is that it takes three moles of electrons to produce one mole of chromium. Therefore, we first need to find out how many moles of electrons are needed to produce 4.94 mg of Chromium, which is 0.0944 moles.

In electrochemistry, the total amount of charge (Q) passed is given by the product of the number of moles of electron (n), the charge per mole of electron (F), and the current (I). F, according to Faraday's constant, is 96485 C mol e. Therefore, Q = n x F x I = 0.0944 moles x 3 mol e per mol Cr x 96485 C mol e = 26942034.48 Coulombs. Now, to find out how much time this takes, we divide Q by the current, using the formula:

time = Q / I = 26942034.48 Coulombs / 0.234 A = 115134041.88 seconds

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To produce 4.94 mg of chromium metal using a current of 0.234 A, approximately 117.5 seconds are required. This is calculated by finding the moles of chromium, then determining the charge needed, and finally using the current to find the time. The process involves basic stoichiometry and electrochemistry principles.

To solve this, we need to follow several steps:

First, determine the moles of chromium to be produced. The molar mass of chromium (Cr) is approximately 52 g/mol:

4.94 mg = 0.00494 g

Moles of Cr

= 0.00494 g / 52 g/mol

= 9.5 x 10⁻⁵ mol

Next, calculate the total charge (Q) required to produce the given moles of chromium. The reduction of Cr(III) to Cr(0) uses 3 moles of electrons per mole of Cr:

Total charge (Q) = 9.5 x 10⁻⁵ mol Cr x 3 mol e- / 1 mol Cr x 96485 C/mol e-

Q = 27.5 C

Finally, use the current (I) to find the time (t):

= t = Q / I

= 27.5 C / 0.234 A

= 117.5 seconds

To produce 4.94 mg of chromium metal using a current of 0.234 A, approximately 117.5 seconds are required.

When only rough approximations are needed, we use a conical flask.

When we want to measure liquid volumes to an accuracy of within about 1%, we use the graduated cylinders.  They are for also used general purpose, but not for sensitive quantitative analysis.

When extreme precise or accurate measurements are needed for dilution or preparation of liquid samples, we use the volumetric flask to contain it.

If we are preparing to perform titration, we will have to use a pipette to transfer very accurate amounts of sample. 

Both the white solids, KCl and [tex]{\mathbf{PbC}}{{\mathbf{l}}_{\mathbf{2}}}[/tex] can be distinguished by dissolving them in water.

Further Explanation:

Solubility rules:

1. The common compounds of group 1A are soluble.

2. All the common compounds of ammonium ion and all acetates, chlorides, nitrates, bromides, iodides, and perchlorates are soluble in nature. Only the chlorides, bromides, and iodides of [tex]{\text{A}}{{\text{g}}^+}[/tex], [tex]{\text{P}}{{\text{b}}^{2+}}[/tex], [tex]{\text{C}}{{\text{u}}^+}[/tex] and [tex]{\text{Hg}}_2^{2+}[/tex] are not soluble.

3. All common fluorides, except for [tex]{\text{Pb}}{{\text{F}}_{\text{2}}}[/tex] and group 2A fluorides, are soluble. Moreover, sulfates except [tex]{\text{CaS}}{{\text{O}}_{\text{4}}}[/tex], [tex]{\text{SrS}}{{\text{O}}_{\text{4}}}[/tex], [tex]{\text{BaS}}{{\text{O}}_{\text{4}}}[/tex], [tex]{\text{A}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] and [tex]{\text{PbS}}{{\text{O}}_{\text{4}}}[/tex] are soluble.

4. All common metal hydroxides except [tex]{\text{Ca}}{\left({{\text{OH}}}\right)_{\text{2}}}[/tex], [tex]{\text{Sr}}{\left({{\text{OH}}}\right)_{\text{2}}}[/tex], [tex]{\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}[/tex] and hydroxides of group 1A and that of transition metals are insoluble in nature.

5. All carbonates and phosphates, except those formed by group 1A and ammonium ion, are insoluble.

6. All sulfides, except those formed by group 1A, 2A, and ammonium ion are insoluble.

7. Salts that contain [tex]{\text{C}}{{\text{l}}^-}[/tex], [tex]{\text{B}}{{\text{r}}^-}[/tex] or [tex]{{\text{I}}^-}[/tex] are usually soluble except for the halide salts of [tex]{\text{A}}{{\text{g}}^+}[/tex], [tex]{\text{P}}{{\text{b}}^{2+}}[/tex] and [tex]{\left({{\text{H}}{{\text{g}}_2}}\right)^{{\text{2+}}}}[/tex].

8. The chlorides, bromides, and iodides of all the metals are soluble in water, except for silver, lead, and mercury (II). Mercury (II) iodide is water-insoluble. Lead halides are soluble in hot water.

9. The perchlorates of group 1A and group 2A are soluble in nature.

10. All sulfates of metals are soluble, except for lead, mercury (I), barium, and calcium sulfates.

KCl and [tex]{\text{PbC}}{{\text{l}}_{\text{2}}}[/tex] both are the chloride salts that are white in color. According to the solubility rules, KCl is a soluble salt whereas [tex]{\text{PbC}}{{\text{l}}_{\text{2}}}[/tex] is an insoluble one and forms precipitate.

KCl and [tex]{\text{PbC}}{{\text{l}}_{\text{2}}}[/tex] can be distinguished by dissolving both the salts separately in water. The salt that forms precipitates in water is [tex]{\mathbf{PbC}}{{\mathbf{l}}_{\mathbf{2}}}[/tex] while the one that dissolves completely in water is KCl. This way, both solids can be distinguished.

Learn more:

1. Identify the precipitate in the reaction: https://brainly.com/question/8896163

2. What type of reaction occurs by mixing barium sulfide and sulfuric acid? https://brainly.com/question/5464325

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Grade: Senior School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: KCl, PbCl2, white solids, precipitate, water, solubility rules, soluble, insoluble, dissolving, salts, chlorides, sulfates, bromide, iodide, carbonates, hydroxides.

The equilibrium constant K and the forward rate constant k1 and backward rate constant k2 has the following relation:

K = k1 / k2

So from the equation, we can say that yes it is possible to have large K even if k1 is small given that k2 is very small compared to k1: (k2 very less than 1)

k2 << k1

The condition which makes the equilibrium constant very large with the smaller value of the forward and backward rate constant is,

[tex]\boxed{{{\mathbf{k}}_{\mathbf{f}}}{\mathbf{>>}}{{\mathbf{k}}_{\mathbf{b}}}}[/tex]

Further explanation:

Equilibrium constant:

The equilibrium constant is expressed as the equilibrium concentration of products divided by the equilibrium concentration of reactants raised to power of their stoichiometric coefficient in the balanced chemical equation.

For the given general reaction,

[tex]{\text{aA + bB}}\to{\text{cC + dD}}[/tex]

The expression for equilibrium constant is,

[tex]K=\frac{{{{\left[{\text{C}}\right]}^{\text{c}}}{{\left[{\text{D}}\right]}^{\text{d}}}}}{{{{\left[{\text{A}}\right]}^{\text{a}}}{{\left[{\text{B}}\right]}^{\text{b}}}}}[/tex]

Here K is equilibrium constant of the reaction.

Another definition:

The equilibrium reaction consists the forward reaction and reverse reaction and both reaction have their respective rate constant, therefore, equilibrium constant can also be defined as the ratio of rate constant of forward reaction to the rate constant of backward reaction.

[tex]K=\frac{{{k_{\text{f}}}}}{{{k_{\text{b}}}}}[/tex]

Here, [tex]{k_{\text{f}}}[/tex] is rate constant of forward reaction and [tex]{k_{\text{b}}}[/tex] is rate constant of backward reaction.

In the given case equilibrium constant should be larger with smaller values of rate constants. Therefore, the rate constants of backward reaction must be very small from the rate constant of backward reaction so that the value of the ratio is very large.

Hence the condition which makes the equilibrium constant very large with the smaller value of forward and backward rate constant is,

[tex]{{\mathbf{k}}_{\mathbf{f}}}{\mathbf{>>}}{{\mathbf{k}}_{\mathbf{b}}}[/tex]

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: https://brainly.com/question/3467841

2. Complete equation for the dissociation of na2co3(aq): https://brainly.com/question/5425813

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Equilibrium

Keywords: Equilibrium, reactions, reverse reaction, forward reaction, backward reaction, rate constant, reverse rate constant, forward rate constant, larger value of equilibrium constant, kf>kb.