Identify the housing option that has the lowest maintenance cost a single-family home b condominium c mobile home or d apartment

Answer: The concentration of sulfuric acid is 0.219 M

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.

We are given:

[tex]n_1=2\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=43.74mL[/tex]

Putting values in above equation, we get:

[tex]2\times M_1\times 50.00=1\times 0.500\times 43.74\\\\M_1=\frac{1\times 0.500\times 43.74}{2\times 50.00}=0.219M[/tex]

Hence, the concentration of sulfuric acid is 0.219 M

The correct option is C.

The gas in question can most likely identified as SO2 (Sulfur Dioxide), based on a calculated molar mass of approximately 44.12 g/mol.

To find the identity of this gas, we can calculate the molar mass using the ideal gas law equation, which is PV = nRT, where P is the pressure in atm, V is the volume in L, n is the number of moles of gas, R is the ideal gas constant (0.0821 L.atm/K.mol) and T is the temperature in Kelvin.

However, we also know that the density of a gas can be represented as the molar mass divided by the molar volume (22.4 L at STP), so we can rearrange and solve the ideal gas law equation for the molar mass. Doing so gives us molar mass = density * R * T / P.

Substituting the given values into this equation we get molar mass = 1.4975g/L * 0.0821 L.atm/K.mol * 349.6 K / 2.384 atm = 44.12 g/mol.

Looking at the options you've provided, the molar mass closest to our calculated molar mass is SO2 or Sulfur Dioxide, which has a molar mass of 64.07 g/mol.

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The complete question is given below:

What is the identity of a gas that has a density of 1.4975 g/L and a volume of 8.64 L at a pressure of 2.384 atm and with a temperature of 349.6 Kelvin?

A) H2O B) N2 C) SO2 D) NO3 E) Cl2

the term for a solid when two solutions are mixed is  a percipitate

Answer:

Cl⁻

Explanation:

A potassium ion K+ would bond with chloride ion (Cl⁻) than any other ion or element listed. Because potassium ion (K+) is electro positive and has lost an electron and having a low charge denisty while chloride ion Cl⁻ has a high charge density and is electro negative and has accepted an electron. Magnesium ion (Mg+) and sodium ion (Na+) can't bond with potassium ion (K+) because they have the same charge and has each lost an electron. Same charges or like charges repel while unlike charges or opposite charges attract towards each other.

Oxygen atom (O) can't bond with Potassium ion (K+) because oxygen is in an elemental state while potassium is in an ionic state and hence can't bond with each other. However of oxygen changes into ionic state (O²⁻), it'll definitely bond with potassium ion (K+) to form a compound (K₂O)

Final answer:

To find the molar mass of sucrose based on the freezing point depression, we first determine the molality of the solution from the observed freezing point depression and the freezing point depression constant of water. Then we calculate the number of moles of sucrose and finally divide the mass of sucrose dissolving by the number of moles to obtain the molar mass, which is 343.0 g/mol.

Explanation:

To calculate the molar mass of sucrose using freezing point depression, we utilize the formula ΔTf = i * Kf * m, where ΔTf is the change in freezing point, i is the van't Hoff factor (for sucrose, i = 1 since sucrose does not dissociate into ions), Kf is the freezing point depression constant for the solvent (water in this case), and m is the molality of the solution.

First, we need to determine the molality (m) of the sucrose solution. The freezing point depression observed is 0.77 degrees Celsius, and the freezing point depression constant (Kf) for water is 1.86 K/kg/mol. The molality is calculated using the observed ΔTf and Kf:

m = ΔTf / (Kf * i)

m = 0.77°C / (1.86°C*kg/mol)

m = 0.414 mol/kg

Now, convert the mass of the water to kilograms:

100g water = 0.1kg water

Next, calculate the number of moles of sucrose using the molality and mass of solvent:

n (moles of sucrose) = molality * mass of water (in kg)

n = 0.414 mol/kg * 0.1 kg

n = 0.0414 mol

Finally, calculate the molar mass of sucrose (M) by dividing the mass of sucrose used by the number of moles:

M = mass of sucrose / number of moles

M = 14.2g / 0.0414 mol

M = 343.0 g/mol

Therefore, the molar mass of sucrose is 343.0 g/mol.

The answer is B.) Water molecules at the surface experience fewer hydrogen bonds than water molecules within the liquid.

Hope this helped!

Answer:

Pollutant

Explanation:

took test

The name of the compound P₄O₁₀  is Phosphorus pentoxide.

Phosphorus pentoxide can be described as a chemical compound with the molecular formula  P₄O₁₀   and the empirical formula, P₂O₅. This white crystalline solid can be described as the anhydride of phosphoric acid. It acts as a desiccant and dehydrating agent.

Phosphorus pentoxide can be crystallized in at least four forms or polymorphs. The most familiar one, a metastable form comprises molecules of P₄O₁₀. Weak van der Waals forces hold molecules together in a hexagonal lattice.

The structure of the P₄O₁₀  cage is similar to adamantane with a tetrahedral symmetry point group. The density of P₄O₁₀  is 2.30 g/cm³ and boils at 423 °C under atmospheric pressure if heated making it sublimate.

P₄O₁₀ can be made by condensing the vapor of phosphorus pentoxide rapidly and is prepared by burning tetraphosphorus with the supply of oxygen.

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The question is incomplete, here is the complete question:

The chemistry of fresco painting is the chemistry of limestone and lime plaster. limestone is calcium carbonate, an abundant, naturally occurring mineral. two key reactions, shown below, are involved in the process of converting limestone to lime plaster (calcium hydroxide). classify each of these reactions as one of the four types of reactions listed in this experiment (decomposition, synthesis, single replacement, or double replacement).

(a)  [tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]

(b)  [tex]CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(aq.)[/tex]

Answer: Equation (a) is a decomposition reaction and equation (b) is a combination reaction.

Explanation:

Decomposition reaction is defined as the reaction in which a single large substance breaks down into two or more smaller substances.

[tex]AB\rightarrow A+B[/tex]

Combination reaction is defined as the reaction in which smaller substances combine to form a larger substance.

[tex]A+B\rightarrow AB[/tex]

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

[tex]A+BC\rightarrow AC+B[/tex]

Double displacement reactions are defined as the reactions in which exchange of ions takes place.

[tex]AB+CD\rightarrow AD+CB[/tex]

For the given equations:

(a)  [tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]

This equation is a type of decomposition reaction in which calcium carbonate breaks down into calcium oxide and carbon dioxide

(b)  [tex]CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(aq.)[/tex]

This equation is a type of combination reaction in which calcium oxide reacts with water to form calcium hydroxide.

Final answer:

Adding silver nitrate to the Fe/SCN equilibrium introduces a common ion, Ag+, which reacts with SCN- to form a precipitate, reducing SCN- concentration and shifting the equilibrium to compensate, thus affecting the system.

Explanation:

The addition of silver nitrate to the Fe/SCN equilibrium affects the equilibrium even though silver ion or nitrate ion isn't part of the equilibrium reaction because it leads to the formation of a precipitate, AgSCN, thus reducing the concentration of free SCN− in the solution.

This is an example of the common ion effect, where the addition of a common ion shifts the position of equilibrium according to Le Chatelier's Principle. As AgSCN is removed from the solution, the equilibrium shifts to the left to restore balance, decreasing the concentration of Fe(SCN)2+ and lightening the color of the solution.

Adding silver nitrate to the Fe/SCN equilibrium affects the equilibrium by removing SCN− ions through precipitation, causing a shift in the equilibrium to the left. This results in a lower concentration of Fe[tex](SCN)^2^+[/tex] and a lighter solution color.

When silver nitrate is added to the Fe/SCN equilibrium, it affects the equilibrium even though neither silver ion nor nitrate ion is directly part of the equilibrium reaction. This is because silver ion (Ag+) reacts with thiocyanate ion (SCN−) to form a precipitate of AgSCN: [tex]Ag^+ _(_a_q_)[/tex] + [tex]SCN^-_(_a_q_)[/tex] = AgSCN (s)

This reaction removes SCN− ions from the solution, thereby reducing its concentration. According to Le Chatelier's principle, the equilibrium will shift to counteract this decrease by shifting to the left, thus decreasing the concentration of Fe(SCN[tex])^2^+[/tex] and causing the solution to become lighter in color.

They can form a delta, because the river location B leads into the Gulf of Mexico.

Some metals lose their shine because of corrosion, a galvanic process involving oxidation by substances like oxygen.

The reason some metals lose their shine over time is due to a process known as corrosion, which is a galvanic process that leads to the deterioration of metals through oxidation. Metals like iron rust and silver tarnish when exposed to air because of their reaction with oxygen, forming oxides on the surface. However, gold does not corrode easily due to its resistance to oxidation by common substances.

Aluminum, although reactive, forms an aluminum oxide coating that protects it from further corrosion, while copper reacts with carbon dioxide to form a green patina that serves as a protective layer. Precious metals such as gold and platinum, known for their corrosion resistance and durability, defy normal oxidation and maintain their luster over time. They are impervious to most elements and can be corroded by only a few special fluids.

The activation energy for the reverse reaction between NO2(g) and CO(g) is 125 kJ/mol. This is calculated using the formula EaR = EaF + ΔH, where EaR is the activation energy of the reverse reaction, EaF is the activation energy of the forward reaction, and ΔH is the change in enthalpy for the reaction.

The activation energy for the reverse reaction can be calculated using the relationship between the activation energies of the forward and reverse reactions and the enthalpy of the reaction. The formula to calculate the activation energy of the reverse reaction (EaR) is given by EaR = EaF + ΔH, where EaF is the activation energy of the forward reaction, and ΔH is the change in enthalpy for the reaction. Here, the activation energy for the forward reaction (EaF) = 375 kJ/mol and the change in enthalpy (ΔH) = -250 kJ/mol. Substituting these values in the equation gives EaR = 375 kJ/mol - 250 kJ/mol = 125 kJ/mol. Hence, the activation energy for the reverse reaction is 125 kJ/mol.

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The result is an activation energy of 125 kJ/mol for the reverse reaction.

To find the activation energy for the reverse reaction, we use the formula:

Activation Energy (reverse) = Activation Energy (forward) + ΔH

Given:

Therefore:

Activation Energy (reverse) = 375 kJ/mol + (-250 kJ/mol) = 375 kJ/mol - 250 kJ/mol = 125 kJ/mol

So, the activation energy for the reverse reaction is 125 kJ/mol.

A) The direction in which the mixture must shift to achieve equilibrium is;

Left Direction

B) The direction in which the mixture must shift to achieve equilibrium is;

Left Direction

C) The direction in which the mixture must shift to achieve equilibrium is;

Right Direction

We are given the balanced equation reaction at equilibrium as;

N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)

We are given;

Kp value at equilibrium = 4.51 × 10⁻⁵

A) Formula to find Kp in an equilibrium equation is;

Kp =  [P(Product)]ⁿ/[P(Reactant 1)]ⁿ * [P(Reactant 2)]ⁿ

Where;

n is the coefficient attached to the respective product or reactant

P is the pressure

At 98 atm of NH₃, 45 atm N₂, 55 atm H₂

Thus;

Kp = [P(NH3)]²/ [P(N₂)] × [P(H2)]³

Kp = 98²/(45 × 55³)

Kp = 1.28 × 10⁻³

This calculated Kp value is greater than the given Kp value at equilibrium and thus the mixture is not equilibrium but it will shift to the left direction towards the reactants to achieve equilibrium.

B) At 57 atm NH₃, 143 atm N₂, No H₂

Thus;

Kp = [P(NH₃)]²/ [P(N₂)]

Kp = 57²/143

Kp = 22.7

This calculated Kp value is greater than the given Kp value at equilibrium and thus the mixture is not equilibrium but it will shift to the left direction towards the reactants to achieve equilibrium.

c) At 13 atm NH₃, 27 atm N2, 82 atm H₂

Thus;

Kp =  [P(NH₃)]²/ [P(N₂)] × [P(H₂)]³

Kp =  13²/(27 × 82³)  

Kp = 1.14 × 10⁻⁵

This calculated Kp value is less than the given Kp value at equilibrium and thus the mixture is not equilibrium but it will shift to the right direction towards the product to achieve equilibrium.

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Answer: Option (B) is the correct answer.

Explanation:

When vapors of a flammable or combustible liquid are mixed together with air in the presence of a source of ignition in proportions rapid combustion or an explosion can occur. The proper proportion is called the flammable or explosive range.

Boiling range is defined as the range of temperature which is involved distillation of oil from the starting time to the time till it evaporates.

The lowest temperature at which an ignitable mixture can be formed from a liquid in air near its surface.

Answer:

The molar mass of the unknown gas is 200 g/mol. R: [tex]2*10^2 g/mol[/tex]

Explanation:

We have an effusion experiment with oxygen and another unknown substance.

Oxygen effuses (v1) through a tiny hole 2.5 times faster than unknown substances (v2). It means

v1 = 2.5v2.

Molar mas of Oxygen is M1 = 32.0 g/mol.

This process can be studied by using Graham law.

[tex]\frac{v1}{v2} = \sqrt[2]{\frac{M2}{M1} }[/tex]

Where M2 is the unknown molecular mass, all the other data are given in the problem. Replacing and isolating M2. we can fin its value:

[tex]\frac{v1}{v2} = \sqrt[2]{\frac{M2}{M1} } \\v1 = 2.5 v2\\\frac{2.5v2}{v2} = \sqrt[2]{\frac{M2}{M1} } \\\\2.5 =  \sqrt[2]{\frac{M2}{M1} } \\\frac{M2}{M1} = (2.5)^2\\M2 = (2.5)^2 M1 = (2.5)^2*32.0 \frac{g}{mol}\\ M2 = 6.25* 32.0 \frac{g}{mol} = 200 \frac{g}{mol}\\M2 = 200 \frac{g}{mol}[/tex]

The molar mass of the unknown gas is 200 g/mol. R: [tex]2*10^2 g/mol[/tex]

The boiling point of water can be calculated by the equation:

Where:

P = Pressure in mm Hg

Po = Atmospheric pressure in mm Hg

ΔH= heat of vaporization in kJ/mol

R = Ideal Gas Constant (J/mol-K)

To = normal boiling point in Kelvin

T = boiling point of water (K)

Our known values are:

P = 630 mm Hg

Po = 760 mm Hg

ΔH = 40.66 kJ/mol = 40.66×1000 =40660

R = 8.314 J mol⁻¹ K ⁻¹

To = 373 K

Putting these values in the equation,

[tex] ln \frac{P_{0}}{P}= \frac{\Delta H}{R}(\frac{1}{T}-\frac{1}{T_{0}})[/tex]

[tex] ln \frac{760}{630}= \frac{40660}{8.314}(\frac{1}{T}-\frac{1}{373})[/tex]

Solving the equation will give:

T=370K

so, the boiling point of water is 370 K.

Answer:

295.3744 grams of NaBr will produce 244 grams of [tex]NaNO_3[/tex].

Explanation:

[tex]Pb(NO_3)_2(aq)+2 NaBr(aq)\rightarrow PbBr_2(s)+2 NaNO_3(aq)[/tex]

Moles of sodium nitarte= [tex]\frac{244 g}{85 g/mol}=2.8705 mol[/tex]

According to reaction, 2 moles of sodium nitrate is obtained from 2 moles of sodium bromide.

Then 2.8705 mol of sodium nitrate will be obtained from :

[tex]\frac{2}{2}\times 2.8705mol=2.8705 mol[/tex] of sodium nitrate

Mass of 2.8705 moles of sodium nitrate:

[tex]2.8705 mol\times 102.9 g/mol=295.3744 g[/tex]

295.3744 grams of NaBr will produce 244 grams of [tex]NaNO_3[/tex].

Answer:

Both are formed by gravitational forces.

Explanation:

I took the test!!

The equilibrium concentrations of PCl5(g) and PCl3(g) can be calculated using an ICE table and the equilibrium constant Kc for the reaction PCl5 = PCl3 + Cl2 at a given temperature.

To calculate the equilibrium concentrations of PCl5(g) and PCl3(g) after 0.366 mol of PCl5(g) is injected into a 4.45 L vessel held at 250 0C, we use the equilibrium constant (Kc) and perform an ICE (Initial, Change, Equilibrium) table calculation. Since we are not given the Kc for this particular reaction at 250 0C, let's assume we know this value or it is provided in the actual context of the problem.

For example, if the Kc was 0.0211, we would set up the equilibrium as follows: PCl5(g) = PCl3(g) + Cl2(g). With the initial concentration of PCl5 being 0.366 mol / 4.45 L, we would solve for the equilibrium concentrations using the Kc value, and the stoichiometry of the reaction.

At equilibrium, the concentrations are:

[tex]- \( [\text{PCl}_5] \approx 0.0032 \text{ M} \)\\- \( [\text{PCl}_3] \approx 0.079 \text{ M} \)[/tex]

To calculate the equilibrium concentrations of [tex]\( \text{PCl}_5(\text{g}) \)[/tex]and [tex]\( \text{PCl}_3(\text{g}) \)[/tex] in a reaction vessel, we need to consider the equilibrium reaction and the equilibrium constant (if provided). For the decomposition of [tex]\( \text{PCl}_5(\text{g}) \)[/tex]:

[tex]\[ \text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \][/tex]

Let's denote the equilibrium constant for this reaction as [tex]\( K_c \)[/tex].

### Step-by-Step Calculation

1. **Initial conditions and setup:**

  - Initial moles of [tex]\( \text{PCl}_5 \)[/tex] = 0.366 mol

  - Volume of the vessel = 4.45 L

  - Initial concentration of[tex]\( \text{PCl}_5 \)[/tex]:

    [tex]\[ [\text{PCl}_5]_0 = \frac{0.366 \text{ mol}}{4.45 \text{ L}} = 0.0822 \text{ M} \][/tex]

  Initial concentrations of[tex]\( \text{PCl}_3 \) and \( \text{Cl}_2 \)[/tex] are 0 M since they are not present initially.

2. **Change in concentrations:**

  - Let ( x ) be the change in concentration of[tex]\( \text{PCl}_5 \)[/tex]that decomposes.

  - At equilibrium:

   [tex]\[ [\text{PCl}_5] = 0.0822 - x \] \[ [\text{PCl}_3] = x \] \[ [\text{Cl}_2] = x \][/tex]

3. **Expression for the equilibrium constant [tex]\( K_c \)[/tex]:**

  [tex]\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{x \cdot x}{0.0822 - x} = \frac{x^2}{0.0822 - x} \][/tex]

  If[tex]\( K_c \)[/tex] is provided, we can solve for \( x \). If not, let's assume a hypothetical value for[tex]\( K_c \)[/tex]to illustrate the process (e.g.,[tex]\( K_c = 1.80 \)[/tex] at 250 °C, but you should use the actual value given or found from tables).

4. **Solve for ( x ):**

[tex]\[ 1.80 = \frac{x^2}{0.0822 - x} \] \[ 1.80 (0.0822 - x) = x^2 \] \[ 0.14796 - 1.80x = x^2 \] \[ x^2 + 1.80x - 0.14796 = 0 \][/tex]

  Solving this quadratic equation for ( x ):

 [tex]\[ x = \frac{-1.80 \pm \sqrt{(1.80)^2 + 4 \cdot 0.14796}}{2} \] \[ x = \frac{-1.80 \pm \sqrt{3.24 + 0.59184}}{2} \] \[ x = \frac{-1.80 \pm \sqrt{3.83184}}{2} \] \[ x = \frac{-1.80 \pm 1.9574}{2} \][/tex]

  This gives two possible solutions:

[tex]\[ x = \frac{-1.80 + 1.9574}{2} \approx 0.079 \] \[ x = \frac{-1.80 - 1.9574}{2} \approx -1.879 \] (discard this negative value)[/tex]

5. **Equilibrium concentrations:**

  [tex]\[ [\text{PCl}_5] = 0.0822 - 0.079 \approx 0.0032 \text{ M} \] \[ [\text{PCl}_3] = x \approx 0.079 \text{ M} \] \[ [\text{Cl}_2] = x \approx 0.079 \text{ M} \][/tex]

### Conclusion

At equilibrium, the concentrations are:

[tex]- \( [\text{PCl}_5] \approx 0.0032 \text{ M} \)\\- \( [\text{PCl}_3] \approx 0.079 \text{ M} \)[/tex]

The specific values of the concentrations will depend on the actual [tex]\( K_c \)[/tex]value at 250 °C.

The electron pair geometry around the Sb atom in SbF3 is trigonal pyramidal.

The electron pair geometry around the Sb (antimony) atom in SbF3 is trigonal pyramidal. This means that there are three bonding pairs of electrons and one lone pair of electrons around the central atom. The molecular structure of SbF3 is also trigonal pyramidal, which means that the lone pair of electrons forms the apex of the pyramid.

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The equation from which heat can be calculated is represented as:

q=mc\Delta T

Where, q=Heat (J)

m=mass of aluminium(g)

Delta T=Change in temperature(⁰C)

c=specific heat (J/g ⁰C)

Here ,

m=mass of aluminium(g)=0.3 g

Delta T=Change in temperature(⁰C)=150-30⁰C=120⁰C

c=0.9J/gC

Putting all the values in the above equation,

q=mc\Delta T

q=0.3×0.9×120

q=32.4 J

So, heat required will be 32.4 J.

The missing 500 joules of energy when burning 1,000 joules of coal are lost to inefficiencies in the energy conversion process, including factors like heat loss.

When 1,000 joules of coal is burned and only 500 joules of electricity is produced, the missing 500 joules of energy are lost primarily due to inefficiencies in the energy conversion process. This is a result of the Second Law of Thermodynamics, which states that when energy is transformed or transferred, part of it assumes a form that cannot be used to do work (often manifested as heat). In the context of a coal-fired power station, the energy undergoes several conversions: chemical energy from the coal is converted to thermal energy (heat), which is then converted to mechanical energy (turbines spinning), and finally converted to electrical energy.

The lost 500 joules of energy in the process are dissipated as heat into the environment, sound, and other forms of energy that are not useful in generating electricity. This comes from the inefficiency of the power plant, where some of the energy is unavoidably lost due to factors like friction, heat loss through conduction, convection, and radiation, and other resistive processes within the power generation system. Therefore, the efficiency of the power station is around 50% as it is only able to convert half of the energy content of the coal into electricity. Ex. In the case provided, a coal power plant with 1,000 joules of input energy from burning coal and an output of 500 joules of electrical energy would have an efficiency of 50%. This percentage is a simplified representation of many real-world power stations' efficiency, which tends to be around 30%-40%. It is also important to note that efforts to increase the efficiency of power stations can lead to more sustainable energy practices by reducing both waste energy output into the environment, which has ecological consequences, and the amount of coal required, which affects the release of CO2 emissions.

The stoichiometric ratio between Fe(NH4)2(SO4)2·6H2O and KMnO4 in the given redox reaction is 5:1. Therefore, for each mole of Fe(NH4)2(SO4)2·6H2O, 0.2 moles of KMnO4 are required.

This problem deals with the stoichiometry of a redox reaction between Fe(NH4)2(SO4)2·6H2O and KMnO4. The balanced redox reaction is:

8H+ + 5Fe+2 + MnO4 - --> Mn+2 + 5Fe+3 + 4H2O

From this balanced chemical equation, we see a stoichiometric ratio of 5:1 between Fe+2 (iron in the compound Fe(NH4)2(SO4)2·6H2O) and MnO4- (manganese in KMnO4). Thus, for each mole of Fe(NH4)2(SO4)2·6H2O, you require 1/5 (or 0.2) moles of KMnO4. In order to fully answer your question, you would need to know the amount (in moles) of Fe(NH4)2(SO4)2·6H2O present.

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