At equilibrium at 1200°C, [I2] = 9.5 ×10–2 M and [I] = 3.2 × 10–2 M. What is the value of Keq for this system?

Fe2O3(s) + 3CO(g) -----> 3CO2(g) + 2Fe(s)

mass of iron = 198.5 g of iron

atomic mass of iron = 55.845g/mol

moles of iron = mass/Atomic mass

= 198.5g/55.845  

= 3.5545 moles Fe

moles of carbon monoxide

= 3.5545 moles Fe x    3 moles CO    

                                       2 moles Fe

= 5.3317 moles CO

molar mass of CO = 12.01 + 16 = 28.01g/mol

mass of carbon monoxide reacted

= moles x molar mass

= 5.3317 x 28.01

= 149.34g

grams of carbon monoxide needed to react with an excess of iron(II)oxide to produce 198.5g of iron is     149.34g CO

The electropositive elements or metals elements tend to lose electrons and form positive ions, while the  electronegative elements or non-metals tend to gain electrons and form negative ions.

Metals are electropositive elements with extra electrons in their atom. Hence, they easily loss electron to achieve octet. Metals donate electrons to electron deficient non-metals forming ionic compounds.

Non -metal are located in right side of the periodic table. Except group 18 elements all the non-metals are electron deficient. Some of them such as oxygen, nitrogen, fluorine are highly electronegative and easily gain electrons from metal atoms forming ionic compounds or share electrons from other non-metals forming covalent compounds.

Therefore,  metallic elements tend to lose electrons and form positive ions, while the  electronegative elements or non-metals tend to gain electrons and form negative ions.

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Metals lose electrons and form positive ions, while non-metals gain electrons and form negative ions due to their atomic structure.

The metals elements tend to lose electrons and form positive ions, while the non-metals elements tend to gain electrons and form negative ions. This is due to the atomic structure of these elements. Metals have fewer electrons in their outer shell, and thus they can easily lose them to become stable, which leads to the formation of positively charged ions. On the other hand, non-metals have more electrons in their outer shell and thus they need more electrons to become stable. As a result, they tend to receive electrons, forming negatively charged ions.

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Answer:

e

Explanation:

Answer:

It is a double replacement reaction, and all four compounds are different.

Step-by-step explanation:

The equation for the reaction is

[tex]\underbrace{\hbox{ ionic compound + ionic compound}}_{\hbox{reactants}} \longrightarrow  \underbrace{\hbox{ ionic compound + ionic compound}}_{\hbox{products}}[/tex]

This must be a double replacement reaction, in which the metal cations trade partners and all four compounds are different.

(A) and (C) are wrong, because single replacements involve elements in reactions of the type

element + compound ⟶ element + compound

(D) is wrong because, if each compound retained the same set of ions, there would be no reaction.  

Answer:

It is a double replacement reaction, and all four compounds are different.

Explanation:

Just took the FLVS 4.03 Single and Double Replacement Reactions quiz. I got a 100.

The total number of moles of H+ in the mixed solution is 0.079 in the mixed volume of 0.4 L. Thus the number of moles in 1 l solution is 0.197.

Molarity of a solution is the ratio of number of moles to the volume of solution in liters. Molarity is a temperature dependant term. This is the most common concentration term.

Here, the number of moles in 110 ml of 0.340 M HCl is calculated as follows:

number of moles =  molarity × volume

                             = 0.340 ×  0.1 l

                             = 0.034 moles.

Number of moles in 330 ml of 0.15 m nitric acid is 0.3 ×  0.15 = 0.045.

Total number of moles in the mixed solution = 0.034 + 0.045 = 0.079. This is the number of moles in the mixed solution of 0.4 L

Hence, the number of moles in 1 liter is 0.4 × 0.079 = 0.197 moles.

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

The molar relationship says that every 1 mole of an element or compound weighs equal to its molecular weight.

As 1 mole of molecular hydrogen [tex]H_2[/tex] weighs 2 g.

2 moles of molecular hydrogen will weigh[tex]=\frac{2}{1}\times {2}=4g[/tex]

1 mole of molecular oxygen [tex]O_2[/tex] weighs 32 g

1 mole of water [tex] H_2O[/tex] weighs = 18g

2 moles of water [tex] H_2O[/tex] weigh=[tex]\frac{18}{1}\times {2}=36g[/tex]

As can be seen from the balanced chemical equation above, 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.

Thus 4g of [tex]H_2[/tex] combines with 32 g of [tex]O_2[/tex] to give 36 moles of [tex] H_2O[/tex].

The study of chemicals and bonds is called chemistry. There are different types of elements, and these are metals and nonmetals.

The correct answer is 36 mole

According to the question, the 2 moles of water react with the mass of water:-

[tex]\frac{18}{1}*2 = 36g[/tex].

Hence, 36 g is used to react with hydrogen and oxygen.

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Answer: The balanced chemical reactions:

[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]

Explanation:

[tex]aKClO_3\rightarrow bKCl+cO_2[/tex]

The given reaction is an unbalanced reaction

Let 'a','b' and 'c' be the coefficient of [tex]KClO_3, KCl \text{ and } O_2[/tex] respectively.

So, in order to balance this given chemical equation:

a = 2, b = 2, c = 3

[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]

Answer:

2

2

3

in that order

Explanation:

FeCO3  ==> Fe + produsi minoritari.

m Fe impur= 700 kg

puritatea (p) = masa pura (mp)/ masa impura (mi) x 100

mp= p x mi / 100 sau mp = p/100 x mi => mp Fe = 90/100 x 700 = 630 kg Fe pur.

M FeCO3= 115.85 kg/kmol

115.85 kg FeCO3 .... 55.85 kg Fe

     x kg FeCO3  ........630 kg Fe

x= 630 * 115.85 /55.85 = 1306.81 kg FeCO3 (mp in formula puritatii)

p=mp/mi x 100

mi FeCO3 = 1500 kg

mp FeCO3=1306.81 kg

p=1306.81 / 1500 x 100 = 87.12% puritate Siderit

the answer is D 0.0556 mol

divide the given number of atoms 3.35x10^22 by 6.02x10^23 Avogadro's number(the number of atoms in one mol of anything) to get the total number of Moles. It's merely a fraction of a whole. 3.35/60.2 if you cancel out the tens.   1.0 mol mercury(the molar mass is the mass number on a periodic table) It couldn't possibly be so few atoms is to be gold or atoms of mercury as there are 6.02x10^23 atoms in a 0.0556 mol.  

have a nice day hope this helps !!!!!!!

Melting point of ethanol = -114°c  

Enthalpy of fusion = 5.02KJ/mol  

Specific heat of solid ethanol = 0.97J/gk  

specific heat of liquid ethanol = 2.3 j/gk.  

mass of ethanol = 25.0g

moles of ethanol = mass/Molar mass = 25.0/46.07 = 0.543 moles

step 1:

solid ethanol at -135 C changing to solid ethanol at -114 C

Q1 = mass x specific heat of solid ethanol x change in Temperature

 = 25.0 x 0.97 x [-114 - (-135)]  

  = 25.0 x 0.97 x [-114+135]  

 = 25.0 x 0.97 x 21  

Q1 = 509.25 J  

Step 2: solid ethanol at -114 C is changing to liquid ethanol at -114 C  

Q2 = moles x deltaHfusion  

= 0.543 x 5.02 KJ  

= 2.72586 x 1000 J  

Q2 = 2725.86 J  

Step 3: liquid ethanol at -114 C is changing to liquid ethanol at -50 C

Q3 = mass x specific heat of liquid ethanol x change in T  

= 25.0 x 2.3 x [-50 -(-114)]  

= 25.0 x 2.3 x [-50+114]  

= 25.0 x 2.3 x 64

Q3 = 3680J    

Total heat = Q1 + Q2 + Q3  

= 509.25 + 2725.86 + 3680  

= 6915.11 J      

Total heat = 6.915 KJ

Total heat required to convert 25.0g of solid ethanol at -135 C to liquid ethanol at -50 C is 6.9KJ.

To convert 25.0 g of solid ethanol at -135°C to liquid ethanol at -50°C, the heat required can be calculated through three main steps: heating solid ethanol to its melting point (-114°C), melting the solid ethanol to liquid form, and heating this liquid from -114°C to -50°C. With calculated heat for all three steps, the total heat required comes out to 6.913 kJ.

The calculation for this problem is divided into three parts: heating the solid ethanol from -135°C to its melting point (-114°C), melting the solid ethanol at -114°C to liquid ethanol, and finally heating the liquid ethanol from -114°C to -50°C.

Step 1 - Heating solid ethanol to its melting point: The heat required for this can be calculated using the formula q=m*CΔT, where m is the mass of the ethanol, C is the specific heat, and ΔT is the change in temperature. The calculation is q=(25.0 g)*(0.97 J/g°C)*(-114°--135°) which equals 507.75 J, or 0.50775 kJ.

Step 2 - Melting solid ethanol to liquid ethanol: We use the given enthalpy of fusion of ethanol 5.02 kJ/mol. However, first, we need to convert the mass of ethanol to moles, which gives us 0.542 moles. Therefore, the total heat for this step will be (0.542 mol)*(5.02 kJ/mol) = 2.72 kJ.

Step 3 - Heating liquid ethanol from -114°C to -50°C: We use the same q=m*CΔT formula, replacing C with the specific heat of liquid ethanol. The calculation becomes q=(25.0 g)*(2.3 J/g°C)*(-50°--114°) which gives us 3,685 J or 3.685 kJ.

In total, the heat required is the sum of the heat from all three steps: 0.50775 kJ + 2.72 kJ + 3.685 kJ = 6.913 kJ.

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Answer: D. They are made up of hard spheres that are in random motion.

Explanation:

A gas is a state of aggregation of matter in which, under certain conditions of temperature and pressure, its molecules interact weakly with each other, without forming molecular bonds, adopting the shape and volume of the container that contains them and tending to separate everything possible because of its high concentration of kinetic energy.

The molecules of a gas are practically free and have the ability to be distributed throughout the space in which they are contained because the gravitational forces and attraction between them are practically negligible compared to the speed at which they move. .

Therefore, gas molecules do not travel specific trajectories or vibrate in a stationary position, instead they move quickly and randomly through the entire space of the container that contains them.

The true statement about gases is that they are made up of hard spheres that are in random motion, which is option D.

Among the given options, statement D is true. Gases are composed of particles, such as atoms or molecules, that are in constant random motion. These particles move in straight lines until they collide with other particles or the walls of the container, where they undergo elastic collisions. The random motion of gas particles is responsible for their ability to fill the entire volume of the container and to exert pressure on the container walls.

Hence, the correct option is option d.

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3MgCl2 + 2Na3P => Mg3P2 + 6NaCl

The greater amount is the solvent and the lesser amount is the solute.

Hence ethanol(200g) which is the greater amount is the solvent here.

And water (145g) which is lesser is the solute here.

the molecularity of C12H22O11 is = 12x12+22x1+16x11 = 342 g/mol

molarity (M) = moles / V. solution

moles = mass / 342 = 15,6/342 = 0,0456 moles

=> molarity (M) = 0,0456/0,0655 = 0,696 M

To find the molarity of a solution, convert the mass of the solute to moles, convert the volume of the solution to liters, and divide the number of moles by the volume in liters. The molarity of a solution containing 15.6 g of sucrose in 65.5 mL of solution is 0.696 M.

The molarity (M) of a solution is calculated using the formula M = mol/L, where mol represents the number of moles of the solute and L represents the volume of the solution in liters. First, we must convert the mass of sucrose to moles. Sucrose, C12H22O11, has a molar mass of approximately 342.3 g/mol. Therefore, 15.6 g / 342.3 g/mol = 0.0456 mol. Next, we convert the volume of the solution to liters: 65.5 mL = 0.0655 L. Finally, we can calculate the molarity of the solution: M = 0.0456 mol / 0.0655 L = 0.696 M. Therefore, the correct answer is 0.696 M.

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Answer: If the number and kinds of atoms on both sides of the chemical equation are same, the chemical equation is called as balanced chemical equation.

If the number and kinds of atoms on both sides of the chemical equation are different, the chemical equation is called as skeletal chemical equation.

[tex]CH_4+O_2\rightarrow CO_2+H_2O[/tex]

As the number of atoms on both sides of the chemical equation are different, the above equation represents a skeletal chemical equation.

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

As the chemical equations must follow the law of conservation of mass, the number of atoms on both sides of the above chemical equation must be same so as the mass is same on both sides of the equation. So the equation was balanced.

Answer:

In the reactant side of the chemical equation, there are 4 carbon atoms, 12 hydrogen atoms, and 14 oxygen atoms. In the product side of the chemical equation, there are 4 carbon atoms, 12 hydrogen atoms, and 14 oxygen atoms. The number and kinds of atoms on both sides of the chemical equation are the same . So, this chemical equation is balanced

Explanation:

The reactants have 4 carbon atoms, 14 oxygen atoms, and 12 hydrogen atoms. The product has the same numbers and kinds of atoms. So, the equation is balanced.

In a titration, a molar mass lower than the actual one could be obtained by spilling part of the acid before titration or recording a dark red color instead of a light pink as the endpoint. These errors would lower the calculated molar mass because either too little acid or too much NaOH is assumed for the reaction.

In a titration experiment with a carboxylic acid and a standardized NaOH solution, there are factors that could lead to the measurement of a molar mass that is smaller than its actual value.

Error I: If some of the carboxylic acid is spilled when being transferred into the titration flask, this will decrease the amount of acid available for titration. This leads to an underestimation of the amount of NaOH needed for neutralization and therefore a smaller calculation for molar mass.

Error II: If the endpoint of the titration is recorded when the color is dark red rather than light pink, this means that the end-point has been overshot – too much NaOH has been added. This excess amount would lead to the calculation of a smaller molar mass since an increased volume of NaOH is wrongly assumed to have neutralized the acid.

Titration analysis therefore requires real accuracy both in process and observation to avoid such errors. It is important to determine and record the right endpoint based on a distinct, recognized color change with the suitable indicator, in this case, a change from colorless to light pink with phenolphthalein.

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The correct answer is that both errors I and II will lead to a molar mass that is smaller than the actual molar mass.

I. Some of the acid is spilled when being transferred into the titration flask.

When some of the acid is spilled, the mass of the acid that is actually titrated is less than the mass that was intended to be titrated.

Since molar mass is calculated by dividing the mass of the substance by the number of moles of that substance, a smaller mass will result in a smaller calculated molar mass if the number of moles is not adjusted accordingly.

The molar mass (M) is given by:

[tex]\[ M = \frac{m}{n} \][/tex]

where m is the mass of the acid and n is the number of moles of the acid.

If the mass m is smaller due to spillage, and n remains the same (because the titre does not change), then the calculated molar mass M will be smaller.

II. The endpoint is recorded when the solution is dark red in color rather than light pink.

Phenolphthalein is a common acid-base indicator that is colorless in acidic solutions and pink in basic solutions.

The endpoint of the titration is reached when the solution changes from colorless to light pink, indicating that the acid has been neutralized by the NaOH.

If the endpoint is recorded when the solution is dark red, it means that the titration has continued past the true endpoint, adding excess NaOH.

This excess NaOH will not react with the acid (since it's already been neutralized), but it will be included in the calculation of the moles of NaOH used.

The number of moles of acid [tex](\( n_{acid} \))[/tex] is calculated using the moles of NaOH [tex](\( n_{NaOH} \))[/tex] used to reach the endpoint:

[tex]\[ n_{acid} = n_{NaOH} \][/tex]

If [tex]\( n_{NaOH} \)[/tex] is overestimated due to continuing the titration past the endpoint, then [tex]\( n_{acid} \)[/tex] will also be overestimated. Since the mass of the acid m remains unchanged, the calculated molar mass M will be smaller:

[tex]\[ M = \frac{m}{n_{acid}} \][/tex]

In conclusion, both spilling some of the acid and recording the endpoint too late will result in a calculated molar mass that is smaller than the actual molar mass of the carboxylic acid.

Answer : The correct option is, (D) it is oxidized.

Explanation :

On the basis of electrons, there are two types of reactions :

(1) Oxidation reaction     (2) Reduction reaction

Oxidation reaction : When an element loses electrons in a chemical reaction then that element gets oxidized in the chemical reaction and its oxidation number increases.

Reduction reaction : When an element gains electrons in a chemical reaction then that element gets reduced in the chemical reaction and its oxidation number decreases.

Hence, the correct answer is (D).

Answer:

a) [tex]\text{no of moles}=\frac{\text{given mass}}{\text{Molecular mass}}[/tex]

[tex]\text{no of moles of sucrose}=\frac{1.202g}{342g/mol}=0.0035moles[/tex]

b) Moles of carbon in 1 mole of sucrose [tex]C_{12}H_{22}O_{11}[/tex]= 12 moles

Moles of carbon in  0.0035 moles of sucrose [tex]C_{12}H_{22}O_{11}=\frac{12}{1}\times 0.0035=0.042moles[/tex]

Moles of hydrogen in 1 mole of sucrose [tex]C_{12}H_{22}O_{11}[/tex]= 22 moles

Moles of hydrogen in  0.0035 moles of sucrose [tex]C_{12}H_{22}O_{11}=\frac{22}{1}\times 0.0035=0.077moles[/tex]

Moles of oxygen in 1 mole of sucrose [tex]C_12H_22O_11[/tex]= 11 moles

Moles of oxygen in  0.0035 moles of sucrose [tex]C_{12}H_{22}O_{11}=\frac{11}{1}\times 0.0035=0.042moles[/tex]

c)  1 mole of carbon contains [tex]=6.023\times 10^{23}atoms[/tex]

0.042 moles of carbon contain [tex]=\frac{6.023\times 10^{23}}{1}\times 0.042=0.25\times 10^{23}atoms[/tex]

1 mole of hydrogen contains [tex]=6.023\times 10^{23}atoms[/tex]

0.077 moles of hydrogen contain [tex]=\frac{6.023\times 10^{23}}{1}\times 0.077=0.46\times 10^{23}atoms[/tex]

1 mole of oxygen contains [tex]=6.023\times 10^{23}atoms[/tex]

0.042 moles of oxygen contain [tex]=\frac{6.023\times 10^{23}}{1}\times 0.042=0.25\times 10^{23}atoms[/tex]

Answer:

46. mmol  

Step-by-step explanation:

The equation for the equilibrium is:

HA + H₂O ⇌ A⁻ + H₃O⁺

The solution is a buffer, so we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻]/[HA])

   5.40 = 4.74 + log([A⁻]/(10.)     Subtract 4.74 from each side

   0.66 = log([A⁻]/(10.)                 Take the antilog of each dide

[A⁻]/10. = 10 ^0.66

[A⁻]/10. = 4.57                              Multiply each side by 10.

    [A⁻] = 46. mmol

You will add 46. mmol of sodium acetate.

The amount of acetate needed to add to the solution is ; 46 mmol

Given data:

pH of buffer solution = 5.4

volume of  Solution ( H A ) = 10 mmol

pKa of acetic acid = 4.74

Given that the solution is a buffer solution we will apply  Henderson-Hasselbalch equation:

pH = [tex]pKa + log ( [A^-] / [HA] )[/tex]  -------- ( 1 )

Insert values into equation 1

[tex]5.40 = 4.74 + log([A^-]/(10 )[/tex]

0.66 = [tex]log([A^-]/(10.)[/tex] -------- ( 2 )  ( after subtracting pKa value from both side )

∴  [tex][A^-]/10. = 10^{0.66}[/tex]  ----- ( 3 )  ( antilog )

Multiply both sides of equation 3 by  10

[tex][A^-] = 46 mmol[/tex].

Hence we can conclude that the amount of acetate needed to add to the solution is  46 mmol

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Equation 1, because Zn being more reactive, replaces Ag from AgNO3

Zn being highly reactive can replace Ag from AgNO3 and produce Ag metal.

Equation 1: 2AgNO3(aq) + Zn(s) → 2Ag(s) + Zn(NO3)2

This equation is the right one to extract silver metal from silvernitrate solution as its single displacement reaction.

Answer:

Equation 1, because Zn being more reactive, replaces Ag from AgNO3

Explanation:

The preferential activity of one metal to displace the other in a chemical reaction can be deduced based on the reactivity or activity series.

The elements at the top of the series are highly reactive and can displace metals that fall below them. Both Zn and Mg appear above Ag in the reactivity  series and can displace silver from silver nitrate.

1) AgNO3 + Zn → ZnNO3 + Ag

Metallic silver can be extracted through the above single displacement reaction.

2) 2AgNO3 + MgCl2 → 2AgCl + Mg(NO3)2

This is double displacement reaction which instead forms a precipitate of AgCl

Answer:

ΔH = + 2.4 kcal

Step-by-step explanation:

2HI(g) + 2.4 kcal ⇌ H₂(g) + I₂(g)

Energy is on the left-hand side of the equation, so it is being absorbed by the system.

The thermodynamic convention is that energy going into a system is positive. Thus,

ΔH = + 2.4 kcal

Answer: 2.4 kcal

Explanation:

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and [tex]\Delta H[/tex] for the reaction comes out to be positive.

For the given reaction:

[tex]2HI+2.4kcal\rightarrow H_2(g)+I_2(g)[/tex]

Enthalpy change is the net heat absorbed or released during a chemical reaction.

As heat is added to reactants, energy is being absorbed and thus enthalpy for the reaction will be positive and the value will be +2.4 kcal.

 N and H covalently bond to form the correct formula NH3.

A. N3H

B. NH3

C. NH2

D. NH

Answer: 2 g

Explanation: [tex]2Pb(NO_3)_2(s)\rightarrow 2PbO(s)+4NO_2(g)+O_2(g)[/tex]

As can be seen from the balanced chemical equation, 2 moles of lead nitrate produce 4 moles of nitrogen dioxide.

[tex]2\times 331.2g=662.4g[/tex] of lead nitrate produces [tex]4\times 46=184g[/tex] of nitrogen dioxide.

184 g of nitrogen dioxide will be produced by 662.4 g of lead nitrate

So 11.5 g of nitrogen dioxide will be produced by=[tex]\frac{662.4}{184}\times {11.5}=41.4 g[/tex] of lead nitrate

As can be seen from the balanced chemical equation, 2 moles of lead nitrate produce 1 mole of oxygen.

[tex]2\times 331.2g=662.4g[/tex] of lead nitrate produces 32 g of oxygen.

41.4 g of lead nitrate produces =[tex]\frac{32}{662.4}\times {41.4}=2g[/tex] of oxygen.

The grams  of   carbon monoxide  is 148. 68 g

calculation

Fe₂O₃(s)  + 3CO (g) →   3 CO₂(g)  + 2Fe (s)

 Step 1: find  the moles  of  CO

   moles = mass÷ molar mass

 from periodic table the  molar mass of Fe =56 g/mol

 moles =  198.5 g÷ 56 g/mol =3.54  moles

Step 2 : use the mole ratio to determine the  moles of  Co

 from given equation  Co: Fe is 3: 2

therefore the  moles of CO = 3.54 moles  x 3/2= 5.31 moles

Step 3 Find mass of Co

mass  =  moles × molar mass

from periodic table the  molar mass of CO = 12+ 16 = 28 g/mol

= 5.31  moles × 28 g/mol = 148.68  g

The accepted name for N2O5 is dinitrogen pentoxide.

The accepted name for the compound N2O5 is dinitrogen pentoxide. It is formed by the combination of two nitrogen atoms and five oxygen atoms. The prefix 'di' indicates that there are two nitrogen atoms, and the prefix 'pento' indicates that there are five oxygen atoms.

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Answer:

-12 258 kJ  

Step-by-step explanation:

Reaction 1: C₃H₈(g)+5O2(g) ⟶ 3CO₂(g)+ 4H₂O(g);          ΔH₁  = - 2 043 kJ  

Reaction 2: 6C₃H₈(g)+30O2(g) ⟶ 18CO₂(g)+ 24H₂O(g); ΔH₂ = -12 258 kJ

Reaction 2 is Reaction 1 multiplied by 6.

ΔH is an extensive property, so you must also multiply ΔH by 6.

Final answer:

The enthalpy change for Reaction 2 is calculated by multiplying the given enthalpy change for Reaction 1 by the factor of six. Since enthalpy is an extensive property, Reaction 2's enthalpy change is –12258 kJ.

Explanation:

ΔH, or change in enthalpy, is an extensive property, which means it is proportional to the amounts of reactants and products in a reaction. In the student's question, we need to calculate the enthalpy change for Reaction 2 based on the given enthalpy change for Reaction 1. We are given that the enthalpy change for Reaction 1, which is the combustion of propane [tex](C_3H_8)[/tex], is [tex]\Delta H_1 = -2043 kJ[/tex]:

[tex]C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g), \Delta 2 \Delta H_1 = -2043 kJ[/tex]

For Reaction 2, the equation is the same but scaled up by a factor of six:

[tex]6C_3H_8(g) + 30O_2(g) \rightarrow 18CO_2(g) + 24H_2O(g), \Delta 2 \Delta H_2 = ?[/tex]

To find ΔH2, we multiply ΔH1 by the scaling factor, which is six, because enthalpy is an extensive property:

[tex]\Delta H_2 = 6 \times \Delta H_1 = 6 \times (-2043 kJ) = -12258 kJ[/tex]

Therefore, the enthalpy change for Reaction 2 is ΔH2 = –12258 kJ.