Calculate Δ Hrxn for the following reaction: CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g) given these reactions and their ΔH values: C(s)C(s)H2(g)+++2H2(g)2Cl2(g)Cl2(g)→→→CH4(g),CCl4(g),2HCl(g),ΔH=−74.6 kJΔH=−95.7 kJΔH=−184.6 kJ Express the enthalpy in kilojoules to one decimal place.

Final answer:

The percent-by-mass concentration of acetic acid in the vinegar solution is approximately 5.15%, calculated by dividing the acetic acid mass by the total solution mass and then multiplying by 100.

Explanation:

To find the percent-by-mass concentration of acetic acid in vinegar, the mass of acetic acid is divided by the total mass of the solution and then multiplied by 100. First, convert the solution volume to mass using the given density. The density of the solution is 1.005 g/mL, which means 1.000 L (or 1000 mL) of solution has a mass of 1005 g (1000 mL × 1.005 g/mL). The mass of acetic acid is given as 51.80 g. Thus, the percent-by-mass concentration is calculated as (51.80 g / 1005 g) × 100%.

The calculation gives a percent-by-mass concentration of approximately 5.15%. This value represents the mass of acetic acid expressed as a percentage of the total mass of the solution.

Answer:

The electrochemical reaction of the battery must be reversible.

Explanation:

An electrochemical cell is device in which chemical reactions produce electricity. There are two main types of electrochemical cells:  

Primary cells: In primary cells, the chemical reaction through which electric current is generated is irreversible and hence cannot be re-charged.  

Secondary cells: are those in which chemical reactions through which electricity is generated is reversible. Hence they can be re-charged.

First you need to calculate the number of moles of aluminium and copper chloride.

number of moles = mass / molecular weight

moles of Al = 512 / 27 = 19 moles

moles of CuCl = 1147 / 99 = 11.6 moles

From the reaction you see that:

if        2 moles of Al will react with 3 moles of CuCl

then  19 moles of Al will react with X moles of CuCl

X = (19 × 3) / 2 = 28.5 moles of CuCl, way more that 11.6 moles of CuCl wich is the quantity you have. So the copper chloride is the limiting reagent.

Answer:

Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds.

Explanation:

When water boils, hydrogen bonds are broken between adjacent water molecules. The hydrogen bond is an intermolecular bond between adjacent oxygen and hydrogen atoms of water molecules.  

During electrolysis, water dissociates in the presence of electric current. Here, ions are formed in the process. Therefore, covalent bonds are broken here.

Boiling water breaks the intermolecular hydrogen bonds between water molecules, causing it to change from liquid to gas. Electrolysis on the other hand, breaks the covalent bonds within individual water molecules, splitting them into hydrogen and oxygen.

There is a distinct difference between the type of interactions broken when water is boiled and when it's electrolyzed. Boiling water primarily breaks the intermolecular attractions, specifically hydrogen bonds that exist between water molecules, thereby allowing water molecules to escape into the air as steam or water vapor. On the other hand, electrolysis of water involves breaking of the covalent bonds within individual water molecules, thereby splitting water into its constituent hydrogen and oxygen atoms. This process requires more energy compared to boiling water due to the strength of covalent bonds when compared to hydrogen bonds.

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Answer:

I think that depends on the type of gas and the volume of the container.

Answer:

a) 52.8

Explanation:

M1V1 = M2V2

(7.91 M)(x ml) = (2.13 M) (196.1 ml)

(7.91M) (xml) = 417.693 M.ml

x ml = 417.693/ 7.91

x   =  52.8

Answer:

A. 52.8

Explanation:

We make use of the dilution formula to solve this

[tex]M_1\times V_1 =M_2\times V_2[/tex]

Where ([tex]M_1[/tex] and [tex]M_2[/tex] are the initial molarity and final molarity and [tex]V_1[/tex] and [tex]V_2[/tex] are  the initial volume and final volume)

Plugging into the values in the formula  

[tex]M_1\times V_1 =M_2\times V_2[/tex]

[tex]7.91M \times V_1=2.13M \times 196.1mL[/tex]

[tex]V_1=\frac {(2.13M \times 196.1mL)}{7.91M}[/tex]

= 52.8 mL is the Answer

Answer:

B. Pulverizing the zinc metal into a fine powder

Answer: B. Pulverizing the zinc metal into a fine powder

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]

A. Decreasing the concentration of [tex]HCl[/tex] and [tex]Zn[/tex]: Thus rate of the reaction would decrease on decreasing the concentration of hydrochloric acid and zinc.

B.  Pulverizing the zinc metal into a fine powder : If the solute particles is present in smaller size, more it can take part in the chemical reaction due to large surface area, hence increasing the rate of reaction.

C. Performing the reaction at a lower temperature. Decreasing the temperature means the energy of the particles is less and thus lesser reactants would cross the energy barrier and thus lesser will be the rate.

Answer:

A. Arrhenius acid    

B. Bronsted-Lowry acid

Explanation:

An acid base reaction invoves an acid and a base and yields salt and water as products.

A neutralization is a reaction between an acid and a base to yield salt and water only. The highlighted compound is not shown here however we we shall tell what is compound is in this reaction.

Hence, this an acid base reaction in the Arrhenius sense.

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Answer:

Option 3. 4p

Explanation:

The greater the principal quantum number, i.e. the main energy level, means that the electron is occupying a larger orbital (bigger radius) thus you must check which wavefunction has the bigger principal quantum number. When two electrons have the same principal quantum number, then you must check the orbital shape because the rank of the radii of the orbitals in increasing order is: s < p < d < f.

The list of choices is:

Hence, the wavefunction 4p (third option) is the largest one and, so, it is for it that you mostlikely would find the electron the farthest from its nucleus.

In the 4p wavefunction is mostlikely to find the electron farthest from thenucleus.

The electronic configuration of an atom determines the orbital location of electrons, based on energy level and orbital type.

The higher the energy level, the farther that electron is from the nucleus.

Thus, the wavefunction that demonstrates the electron furthest from the nucleus is 4p, which has a higher energy level.

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Answer:

a. 52.8

Explanation:

To find the number of moles of HCl we use the relation M₁V₁=M₂V₂

where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.

M₁=7.91 M

M₂=2.13 M

V₁=?

V₂=196.1 mL

Replacing these values in the relationship.

M₁V₁=M₂V₂

7.91 M× V₁=2.13 M×196.1 mL

V₁=(2.13 M×196.1 mL)/7.91 M

=52.8 mL

Answer: T2= 962.2 K

Explanation:

The ideal gases is often written like PV=nRT, where P is pressure, V is volume, n is moles, R is the universal constant of the gases and T is Temperature.

So, in this problem there is a container that is a closed system, therefore n is constant and volume too. The initial point is 1 and the final point is 2, so

V1=V2  ⇒

[tex]\frac{n1RT1}{P1} = \frac{n2RT2}{P2} \\\\\frac{T1}{P1} = \frac{T2}{P2} \\\\T2= \frac{T1P2}{P1} =\frac{170 K 283KPa}{50 KPa} =962.2 K[/tex]

Answer:

acetyl CoA

Explanation:

The starting molecule for the krebs cycle is acetyl CoA.

In a bond, Sulfur (S) is most likely to attract electrons due to its higher electronegativity. This is based on the trend in the periodic table where electronegativity increases across a group from left to right and decreases down a group.

In a bond, elements that have high electronegativity are more likely to attract electrons. Electronegativity is a measure of the ability of an atom to attract the electrons in a bond. It increases across a period from left to right, and decreases down a group in the periodic table. Based on this premise, among the atoms listed, Sulfur (S) and Selenium (Se) have higher electronegativities than Potassium (K) and Sodium (Na).

However, between Sulfur and Selenium, Sulfur (S) is more likely to attract electrons in a bond because it is higher up in the table, given that electronegativity decreases down a group. Therefore, the answer to your question is C) Sulfur (S).

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Answer: 2.796 grams

Explanation:

[tex]Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4[/tex]

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

[tex]\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles[/tex]

[tex]\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles[/tex]

According to stoichiometry:

1 mole of [tex]BaCl_2[/tex] reacts with 1 mole of [tex]Na_2SO_4[/tex]

0.012 moles of [tex]BaCl_2[/tex] will react with=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]Na_2SO_4[/tex]

Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product. [tex]Na_2SO_4[/tex]  is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

1 mole of [tex]BaCl_2[/tex] produces 1 mole of [tex]BaSO_4[/tex]

0.012 moles of [tex]BaCl_2[/tex] will produce=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]BaSO_4[/tex]

Mass of [tex]BaSO_4=moles\times {\text {Molar mass}}=0.012\times 233=2.796g[/tex]

Thus 2.796 grams of [tex]BaSO_4[/tex]  are produced.

Answer:

The molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

Explanation:

Consider a slightly soluble solid with formula M₃X₂. Its solubility product expression is

[tex]\begin{array}{rcccc}M_{3}X_{2}(s) & \rightleftharpoons&3M^{2+}(aq) & + & 2X^{3-}(aq)\\& & 3s & &2s\\\\K_\text{sp}& = & [3s]^{3}[2s]^{2}&= & 108s^{5}\\\\\end{array}[/tex]

Thus, the molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

A is wrong. The solubility product constant is a constant. It does not change in the presence of a common ion.

B is wrong. It is correct only for compounds with formula MX.

C is wrong. Ksp does not equal the concentration of the compound in solution.

The value of Ksp is related to the molar solubility in that molar solubility helps calculate the ion concentrations in a solution, which are used to calculate Ksp. The presence of a common ion affects both solubility and Ksp through the common ion effect.

The value of Ksp, or the solubility product constant, is closely related to the molar solubility of a compound, which is a measure of how much of the compound can dissolve in a given amount of solvent to form a saturated solution. The connection between Ksp and molar solubility can be understood through a series of steps:

Furthermore, if a common ion is present in the solution, it affects the solubility of the compound and thus the Ksp value. This is a demonstration of the common ion effect, where the presence of a common ion decreases the solubility and Ksp of the compound according to Le Chatelier's principle.

Therefore, understanding the relationship between Ksp and molar solubility is essential for predicting the solubility of compounds and determining the possible concentrations of ions in solution.

Answer:

The correct answer is option A) generally free of technical jargon

Explanation:

Hello!

Let's solve this!

When we read popular magazines or newspapers, we see scientific findings. They generally do not use scientific or technical vocabulary so that more people can understand it. They are also not the most reliable sources, since to look for reliable information, we have to look for scientific journals.

After the analysis we conclude that the correct answer is option A) generally free of technical jargon

Answer: The volume of chlorine gas produced in the reaction is 2.06 L.

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol[/tex]

To calculate the moles of hydrochloric acid, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

[tex]6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol[/tex]

[tex]2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O[/tex]

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = [tex]\frac{2}{16}\times 0.27=0.033moles[/tex] of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = [tex]\frac{5}{16}\times 0.27=0.0843moles[/tex] of chlorine gas.

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]40^oC=[40+273]K=313K[/tex]

Putting values in above equation, we get:

[tex]1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L[/tex]

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

Final answer:

The question involves a stoichiometry problem in chemistry, where we calculate the volume of chlorine gas from the reaction of potassium permanganate with hydrochloric acid. We identify the limiting reactant and then use the ideal gas law to find the volume under the given conditions of temperature and pressure.

Explanation:

The student's question concerns a chemical reaction between potassium permanganate (KMnO4) and hydrochloric acid (HCl) to produce chlorine gas (Cl2). This is a stoichiometry problem where we will calculate the volume of chlorine gas generated at specific conditions using the ideal gas law. To find the number of moles of chlorine gas that can be produced, we first determine the limiting reactant. Then, using the ideal gas law (PV=nRT), we can calculate the volume at the given temperature and pressure.

First, we must find the reaction equation:
KMnO4 + 16HCl → 2KCl + 5Cl2 + 2MnCl2 + 8H2O

Next, we calculate the moles of KMnO4 and HCl. Following this, we determine the limiting reactant and use it to calculate moles of Cl2 produced. Lastly, we use the ideal gas law (PV=nRT, with R=0.0821 L·atm/(mol·K)) to find the volume of Cl2 at 40°C and 1.05 atm.

Answer:

0.006 48 km/s

Explanation:

1. Convert miles to kilometres

14.5 mi × (1.609 km/1 mi) = 23.33 km

2. Convert hours to seconds

1 h × (60 min/1h) × (60 s/1 min) = 3600 s

3. Divide the distance by the time

14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s

Answer:

2

Explanation:

In balancing nuclear reactions the mass number and atomic numbers are usually conserved. This implies that from the given equation, the sum of the number of the subscript on the right hand side must be equal to that on the left hand side. This also applies to the superscript:

   For the mass numbers(superscript):

   235 + 1 = 1 + 139 + 95

       236 = 235

This is not balanced

  For the atomic number:

  92 + 0 = 0 + 53 + 39

      92 = 92

This is balanced.

We simply inspect to see how to balance the mass number.

By putting a coefficient of 2 behind the neutron atom, the equation becomes balanced.

Answer:

1.5 M.

Explanation:

M = (no. of moles of LiBr)/(Volume of the solution (L).

∵ no. of moles of LiBr = (mass/molar mass) of LiBr = (97.7 g)/(86.845 g/mol) = 1.125 mol.

Volume of the solution = 750.0 mL = 0.75 L.

∴ M = (no. of moles of luminol)/(Volume of the solution (L) = (1.125 mol)/(0.75 L) = 1.5 M.

Hello!

Determine the molarity of a solution formed by dissolving 97.7 g LiBr in enough water to yield 750.0 ml of solution.

We have the following data:  

M (Molarity) =? (in mol / L)

m1 (mass of the solute) = 97.7 g

V (solution volume) = 750 ml → V (solution volume) = 0.75 L

MM (molar mass of LiBr)

Li = 6.941 u

Br = 79.904 u

---------------------------

MM (molar mass of LiBr) = 6.941   + 79.904

MM (molar mass of LiBr) = 86.845 g/mol

Now, let's apply the data to the formula of Molarity, let's see:

[tex]M = \dfrac{m_1}{MM*V}[/tex]

[tex]M = \dfrac{97.7}{86.845*0.75}[/tex]

[tex]M = \dfrac{97.7}{65.13375}[/tex]

[tex]M = 1.49999... \to \boxed{\boxed{M \approx 1.5\:mol/L}}\:\:\:\:\:\:\bf\green{\checkmark}[/tex]

________________________

________________________

*** Another way to solve is to find the number of moles (n1) and soon after finding the molarity (M), let's see:

[tex]n_1 = \dfrac{m_1\:(g)}{MM\:(g/mol)}[/tex]

[tex]n_1 = \dfrac{97.7\:\diagup\!\!\!\!\!g}{86.845\:\diagup\!\!\!\!\!g/mol}[/tex]

[tex]n = 1.12499.. \to \boxed{n_1 \approx 1.125\:mol}[/tex]

[tex]M = \dfrac{n_1\:(mol)}{V\:(L)}[/tex]

[tex]M = \dfrac{1.125\:mol}{0.75\:L}[/tex]

[tex]\boxed{\boxed{M = 1.5\:mol/L}}\:\:\:\:\:\:\bf\blue{\checkmark}[/tex]

_____________________

[tex]\bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]

Answer:

The melting points of organic compounds are usually lower than those of inorganic compounds.

The boiling points of organic compounds are usually lower than those of inorganic compounds.

Explanation:

Organic compounds are compounds of carbon with exception of simple ones like oxides, carbides and carbonates. Inorganic compounds are other compounds excluding organic compounds.

Inorganic compounds due to the nature of their bonds typically have higher melting and boiling points compared to organic compounds.

Organic compounds are also highly flammable and would easily burn compared to inorganic ones.

The order of increasing dipole moment is II < III < I. BCl3 has no dipole moment, BIF2 has a dipole moment, and BClF2 has a greater dipole moment compared to BIF2.

The order of increasing dipole moment is II < III < I.

The dipole moment depends on the electronegativity difference of the bonded atoms and the bond length. In this case, BCl3 has no dipole moment since the molecule is symmetrical and the individual dipole moments cancel out. BIF2 has a dipole moment due to the electronegativity difference between B and I atoms, and BClF2 has a greater dipole moment compared to BIF2 due to the additional electronegativity difference between B and Cl atoms.

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The correct answer is Option A. The order of increasing dipole moment is I < II < III.

BCl₃ is a symmetrical molecule with no net dipole moment because its individual B-Cl bonds cancel out. Hence, BCl₃ has the smallest dipole moment.

For BIF₂, the asymmetry is introduced as iodine and fluorine have different electronegativities, creating a net dipole moment. However, with only one I atom, the effect is moderate.

BClF₂ is more polar than BIF₂ because the different electronegativities of Cl and F combined in an asymmetrical structure increase the net dipole moment.

The number of moles of [tex]NO_2[/tex] that should be added in order to form 1 mol  [tex]SO_3\\[/tex] at equilibrium is 0.263 mol.

Given,

Initial Conditions:

Moles of [tex]SO_\\2[/tex] initially = 2.50 mol

Moles of [tex]SO_3[/tex] at equilibrium = 1.00 mol

Moles of [tex]NO_2[/tex] initially = 0 mol

Moles of NO initially = 0 mol

Equilibrium constant = 3.80

Changes in Moles:

Change in moles of [tex]SO_2[/tex] = 2.50 mol (initial moles of [tex]SO_2[/tex])

Change in moles of [tex]SO_3[/tex]= 1.00 mol (moles of [tex]SO_3[/tex]at equilibrium)

The equilibrium constant expression for this reaction is:

[tex]K_c = \frac{[SO_3][NO]}{[SO_2][NO_2]}[/tex]

Where the square brackets denote the concentration of each species.

On using the equilibrium constant expression to solve for the change in moles of [tex]NO_2[/tex]:

[tex]\Delta n_{NO_2} = \frac{[SO_2][NO_2]}{[SO_3]}\times \frac{[NO]}{K_c}[/tex]

Substituting the values:

[tex]\rm \Delta n_{NO_2} = \frac {2.50\times 0}{1} \times \frac{1}{3.80}\\\Delta n_{NO_2} = 0.263 \ mol[/tex]

The number of moles that should be added in order to form 1 mol  [tex]SO_3\\[/tex] at equilibrium is 0.263 mol.

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According to the given chemical equation, for every mole of SO3 formed, a mol of NO2 is used up. Therefore, to form 1.00 mol of SO3 from 2.50 mol of SO2, you need to add 1.00 mol of NO2 to reach equilibrium.

This calculation, in essence, requires an application of Le Chatelier's Principle. For the chemical equation SO2(g)+NO2(g)↽−−⇀SO3(g)+NO(g) the equilibrium constant, Kc, at a certain temperature is 3.80. This implies that at equilibrium, the ratio of the concentrations of products to reactants (when raised to their stoichiometric coefficients) is 3.80.

To form 1.00 mol of SO3(g) from 2.50 mol of SO2(g), it means that according to the equation, 1 mol NO2(g) will be required. Furthermore, for the NO(g) produced, we do not have to worry about it since it is not involved in the later equilibrium calculation.

With Le Chatelier's Principle, when you increase the amount of one of the reactant (in this case NO2(g)), the equilibrium will shift to the right side to try and 'use up' that extra NO2 added, this will cause more SO3(g) to be produced and reach a new equilibrium state.

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1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?

Using the graph form the picture you count 8 times the halving of C¹⁴ and you arrive at 45600 years.

2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the half-life.

3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.

What the problem is telling you is that at 276 days only 25% original sample remains. If you divide the number of days by two the quantity of original sample will be multiplied by two, and you will have 138 days and 50% of original sample. This is the answer because the the half-life of a isotope is the time in which 50% of original quantity of radioactive atoms will disintegrate.

Answer:

The ratio acid to water in the mixture is 2:3

Explanation:

Let the volume of 20% acid solution used to make the mixture = x units

So, the volume of 50% acid solution used to make the mixture = 2x units

Total volume of the mixture = x + 2x = 3x units

For 20% acid solution:

C₁ = 20% , V₁ = x

For 50% acid solution :

C₂ = 50% , V₂ = 2x

For the resultant solution of sulfuric acid:

C₃ = ? , V₃ = 3x

Using

C₁V₁ + C₂V₂ = C₃V₃

20×x + 50×2x = C₃×3x

So,

20 + 50×2 = C₃×3

Solving

120 = C₃×3

C₃ = 40 %

Thus, for the resultant mixture,

Acid percentage = 40%

Water percentage = (100 - 40)% = 60%

Ratio acid to water in the mixture = 40:60 = 2:3

Answer:

2:3

Explanation:

To calculate the partial pressure of SO3 at equilibrium for the given reaction, we can use the equilibrium constant (Kp) and the known partial pressures of SO2 and O2. By substituting values into the equation Kp = (P(SO3))^2 / (P(SO2))^2 * P(O2), we can find the partial pressure of SO3.

Given the equilibrium constant (Kp) of 0.345 and the partial pressures of SO2 and O2 at equilibrium as 36.9 atm and 16.8 atm respectively, we can use the equation Kp = (P(SO3))^2 / (P(SO2))^2 * P(O2). Let's substitute the known values into the equation:

Kp = (P(SO3))^2 / (36.9 atm)^2 * (16.8 atm)

Now we can solve for the partial pressure of SO3:

(P(SO3))^2 = Kp * (36.9 atm)^2 * (16.8 atm)

P(SO3) = sqrt(Kp * (36.9 atm)^2 * (16.8 atm))

Calculating this value will give us the partial pressure of SO3 at equilibrium.

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Answer : The rate constant at 785.0 K is, [tex]1.45\times 10^{-2}s^{-1}[/tex]

Explanation :

According to the Arrhenius equation,

[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]K_1[/tex] = rate constant at [tex]600.0K[/tex] = [tex]6.1\times 10^{-8}s^{-1}[/tex]

[tex]K_2[/tex] = rate constant at [tex]785.0K[/tex] = ?

[tex]Ea[/tex] = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]600.0K[/tex]

[tex]T_2[/tex] = final temperature = [tex]785.0K[/tex]

Now put all the given values in this formula, we get:

[tex]\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}][/tex]

[tex]K_2=1.45\times 10^{-2}s^{-1}[/tex]

Therefore, the rate constant at 785.0 K is, [tex]1.45\times 10^{-2}s^{-1}[/tex]

The rate constant at the new temperature is 1.81 ×10^−2 s−1.

We have the reaction written as follows; C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g).

We have to use the formula;

ln(k2/k1) =Ea/R(1/T1 - 1/T2)

Now;

k2 = ?

k1 = 6.1×10−8 s−1

Ea = 262 ×10 ^3 J/mol

T1= 600.0 K

T2 = 785.0 K

R = 8.314 J/K. mol

ln (k2/6.1×10−8) = 262 ×10 ^3 /8.314 (1/600.0 - 1/785.0)

ln (k2/6.1×10−8) = 12.6

k2/6.1×10−8 = e^12.6

k2 = 6.1×10−8 (e^12.6)

k2 = 1.81 ×10^−2 s−1.

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Answer:

Law of Conservation of Mass

Explanation:

The Law of Conservation of Matter forms the basis for balancing chemical equations, ensuring that the number of each element is equal on both sides of the equations. This principle also serves to describe a reaction's stoichiometry, where amounts of reactants and products in a reaction are considered.

The scientific principle which forms the basis for balancing chemical equations is the Law of Conservation of Matter. According to this principle, the same number of each element must be represented on the reactant (input) and product (output) sides of an equation. This ensures that equations accurately reflect the reality that matter is not created or destroyed in a chemical reaction.

For instance, consider the following chemical equation: PC15 (s) + H₂O(1) →→→ POC13 (1) + 2HCl(aq). In this equation, we balance the equation by ensuring that for each of the elements involved (P, C, and H), the number of atoms of that element is equal on both sides of the equation.

Beyond simply balancing the equation, this principle also helps describe a reaction's stoichiometry, which involve the relationships between amounts of reactants and products. Coefficients from the balanced equation can be used in computations relating to reactant and product masses, molar amounts, and other quantitative properties.

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