Part 1: Write the complete balanced MOLECULAR equation (including all states of matter) for the precipitation reaction that occurs between aqueous copper(II) chloride and aqueous sodium hydroxide.

Part 2: Write the NET IONIC EQUATION for the REVERSE reaction in Part 1 (include all states of matter).

Part 3: Write the equilibrium expression for the net ionic equation written in Part 2.

Answer:

The answer is flattening

Explanation:

A physical change is generally something that affects the shape of form of the matter and a chemical change results from a chemical reaction. Flames are caused by chemical reactions, as is rust, and the process of a fruit becoming ripe. Thus, the answer is “flattening”.

In the realm of chemistry, only flattening out of the listed options is considered a physical change as it alters the condition of a substance without modifying its underlying chemical structure.

In the context of Chemistry, physical changes are processes that change the form or appearance of a substance, but not its chemical composition. Of the processes you listed: burning, rusting, flattening, and ripening, flattening is an example of a physical change. For instance, if you have a piece of aluminum foil and you flatten it, it is still aluminum foil - no new substance is created. On the other hand, burning, rusting, and ripening are chemical changes because they result in new substances being formed.

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The given question is incomplete. The complete question is :

Predict the products of the reaction below. That is, complete the right-hand side of the chemical equation. Be sure your equation is balanced and contains state symbols after every reactant and product

[tex]HNO_3(aq)+H_2O(l)\rightarrow[/tex]

Answer: The complete equation is [tex]HNO_3(aq)+H_2O(l)\rightarrow NO_3^-(aq)+H_3O^+(aq)[/tex]

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

[tex]HNO_3[/tex] being a strong acid dissociates to give [tex]H^+[/tex] ions an [tex]H_2O[/tex] will act as base and accept [tex]H^+[/tex] to form [tex]H_3O^+[/tex]

Thus the complete equation is [tex]HNO_3(aq)+H_2O(l)\rightarrow NO_3^-(aq)+H_3O^+(aq)[/tex]

Answer:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Explanation:

Step 1: Data given

gaseous methane = CH4(g)

Combustion reaction is adding O2. The products will be carbondioxide (CO2) and water vapor (H2O)

Step 2: The unbalanced equation

CH4(g) + O2(g) → CO2(g) + H2O(g)

Step 3: Balancing the equation

CH4(g) + O2(g) → CO2(g) + H2O(g)

On the left side we have 4x H (in CH4), on the right side we have 2x H (in H2O). To balance the amount H on both sides, we have to multiply H2O by 2.

CH4(g) + O2(g) → CO2(g) + 2H2O(g)

On the left side we have 2x O (in O2), on the right side we have 4x O (2x in CO2 and 2x in 2H2O). To balance the amount of O on both sides, we have to multiply O2 on the left side, by 2. Now the equation is balanced.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

The balanced chemical equation for the combustion of gaseous methane (CH₄) with gaseous oxygen (O₂) to form gaseous carbon dioxide (CO₂) and gaseous water (H₂O) is:

CH₄ + 2O₂ → CO₂ + 2H₂O

The combustion of gaseous methane (CH₄) with gaseous oxygen (O₂) to form gaseous carbon dioxide (CO₂) and gaseous water (H₂O) is a fundamental chemical reaction that occurs in natural gas combustion and many other combustion processes. To write a balanced chemical equation for this reaction, we must ensure that the number of atoms of each element on both sides of the equation is the same.

The unbalanced equation for the combustion of methane is:

CH₄ + O₂ → CO₂ + H₂O

Now, let's balance the equation:

Balance the carbon (C) atoms:

There is one carbon atom on the left and one on the right, so carbon is already balanced.

Balance the hydrogen (H) atoms:

There are four hydrogen atoms on the left (in CH₄) and two on the right (in H₂O). To balance hydrogen, we need to place a coefficient of 2 in front of H₂O on the right side.

CH₄ + O₂ → CO₂ + 2H₂O

Balance the oxygen (O) atoms:

On the left side, there are two oxygen atoms in CH₄ and two in O₂, making a total of four oxygen atoms. On the right side, there are two oxygen atoms in CO₂ and four in 2H₂O, making a total of six oxygen atoms. To balance the oxygen atoms, we need to adjust the coefficient of O₂ on the left side.

CH₄ + 2O₂ → CO₂ + 2H₂O

Now, the equation is balanced with an equal number of atoms of each element on both sides:

CH₄ + 2O₂ → CO₂ + 2H₂O

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Answer:

Option C. The concentration of base is higher than the concentration of the acid.

Explanation:

The reaction between a strong acid and a strong base follows the equation:

HA + OH⁻ ⇆  A⁻ + H₂O

The pH is:

[tex] pH = -log [H_3O^{+}] [/tex]

If we have 100 mL of a strong acid and 100 mL of a strong base, for the pH to be more than 7, that means that the concentration of the base is higher than the concentration of the acid. This is because, the number of moles that remains in the solution after the reaction between the acid and the base will be the moles of the base. The number of moles of the reaction above is:

[tex] n_{T} = n_{a} - n_{b} [/tex]    (1)

Where na: is the moles of acid, nb: the moles of the base, and nT is the total number of moles.

Case A) If the concentration of acid is the same as the concentration of the base since the volume of the acid and the base are the same, the number of moles of acid is the same as the number of moles of the base, hence, they neutralize, so the pH = 7. This is not the correct option.

Case B) If the concentration of acid is higher than the base since the volume of the acid and the base are the same, the number of moles of acid is also higher than the number of moles of the base and the total moles in equation (1) results in an excess of moles of the acid, so the pH is < 7. This is not the correct option.

Case C) If the concentration of base is higher than the concentration of the acid since the volume of the acid and the base are the same, the number of moles of the base is also higher than the number of moles of the acid and the total moles in equation (1) results in an excess of moles of the base, so the pH is > 7. This is the correct option.

Case D) The water autoionizes to give [OH-] ions. The autoionization of the water produces the same concentration of acid that the base, so this is not the correct option.

From all of the above, the correct option is C. For the pH to be more than 7, the concentration of the base is higher than the concentration of the acid.

I hope it helps you!

The Organisms that formed them likely lived during the Paleozoic era.

Final answer:

When trilobite fossils are found alongside other species in the same rock layer, a scientist can infer that these organisms lived during the same time period and may have had similar environmental conditions. Trilobites are used as index fossils which can determine the age of the rocks they are in due to their distinct presence in the geological record from 500 to 600 million years ago.

Explanation:

When a scientist discovers fossilized remains of other species in the same rock layer as a trilobite, they can draw some important conclusions. Since trilobites were widespread marine animals that lived between 500 and 600 million years ago, any other fossils found within the same stratum are likely from the same geological period. This is due to the principle of superposition, which states that in any undisturbed sequence of rocks deposited in layers, the youngest layer is on top and the oldest on bottom, each layer being younger than the one beneath it and older than the one above it.

Furthermore, trilobites are known as index fossils because they were widespread, rapidly evolving, and limited in geological time. Therefore, they serve as indicators of the age of the rock layer in which they are found. Discovering other species in conjunction with trilobite fossils suggests that these organisms also lived during the same time period and may have shared similar environments. This can be used to infer that these species are from the same age and geological period as the trilobites.

The study of these fossils can provide invaluable insight into the biodiversity and ecological conditions of prehistoric marine environments. Moreover, these findings contribute to our understanding of evolutionary history and may help to identify patterns of mass extinction or other significant pale ontological events.

Answer:

31.5mL

Explanation:

The following were obtained from the question:

C1 (concentration of stock solution) = 2M

V1 (volume of stock solution) =.?

C2 (concentration of diluted solution) = 0.630M

V2 (volume of diluted solution) = 100mL

Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:

C1V1 = C2V2

2 x V1 = 0.630 x 100

Divide both side by 2

V1 = (0.630 x 100) /2

V1 = 31.5mL

Therefore, 31.5mL of 2M solution of FeCl2 required

Answer:

We need 31.5 mL of the 2.0 M FeCl2 solution

Explanation:

Step 1: Data given

Molarity of a FeCl2 solution = 2.0 M

Initial volume of FeCl2 = 100 mL

Initial molarity of FeCl2 = 0.630 M

Step 2: Calculate volume of the stock solution

C1V1 = C2V2

⇒with C1 = the initial molarity FeCl2 = 0.630 M

⇒with V1 = the initial volume = 100 mL = 0.100 L

⇒with C2 = the new molarity FeCl2 = 2.0 M

⇒with V2 = the new volume = TO BE DETERMINED

0.630M * 0.100 L = 2.0 M * V2

V2 = (0.630 * 0.100) / 2.0

V2 = 0.0315 L = 31.5 mL

We need 31.5 mL of the 2.0 M FeCl2 solution

Answer:

992.302 K

Explanation:

V(rms) = 750 m/s

V(rms) = √(3RT / M)

V = velocity of the gas

R = ideal gas constant = 8.314 J/mol.K

T = temperature of the gas

M = molar mass of the gas

Molar mass of CO₂ = [12 + (16*2)] = 12+32 = 44g/mol

Molar mass = 0.044kg/mol

From

½ M*V² = 3 / 2 RT

MV² = 3RT

K = constant

V² = 3RT / M

V = √(3RT / M)

So, from V = √(3RT / M)

V² = 3RT / M

V² * M = 3RT

T = (V² * M) / 3R

T = (750² * 0.044) / 3 * 8.314

T = 24750000 / 24.942

T = 992.302K

The temperature of the gas is 992.302K

Note : molar mass of the gas was converted from g/mol to kg/mol so the value can change depending on whichever one you use.

The temperature of a gas can be calculated from the root-mean-square speed of its molecules using the kinetic theory of gases. For CO2 gas with the rms speed of 750 m/s, the derived temperature is approximately 485.4 K.

The temperature of a gas can be calculated from the root-mean-square (rms) speed of its molecules using the kinetic theory of gases. This theory develops a relationship between the average kinetic energy of the gas molecules and the temperature of the gas through the equation K = 3/2kBT = mv²/2, where kB is Boltzmann’s constant, T is the temperature, m is the mass of a gas molecule, and v is the rms speed. Thus, temperature can be derived as T = mv² / (3kB).

For CO2, the molar mass is 0.044 kg/mol. Knowing that the number of molecules (n) is the number of moles times Avogadro's number (6.022 × 10²³), we derive the molecular mass m = 0.044 kg/mol / (6.022 × 10²³) = 7.3 × 10^-26 kg. Plugging in the values for v (750 m/s), m, and kB (1.38 × 10^-23 J/K), we get an approximate temperature of 485.4 K.

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The dissolution of potassium hydroxide (KOH) in water is an exothermic process, as indicated by the increase in temperature when KOH is dissolved in water.

The dissolution of potassium hydroxide (KOH) in water is an exothermic process. This can be determined based on the observation that the temperature of the water increased from 23.0°C to 34.0°C when 11.1 g of KOH was dissolved in 250.0 g of water. The positive change in temperature indicates that heat was released by the KOH dissolving in water, resulting in an increase in temperature.

The reaction is exothermic as the temperature of the water rises from 23.0°C to 34.0°C, indicating the release of heat.

To determine if the reaction is exothermic or endothermic, we need to look at the temperature change during the dissolution of potassium hydroxide (KOH).

Since the temperature of the water rises from 23.0 °C to 34.0 °C, which is an increase of 11.0 °C, this indicates that the solution absorbs heat.

This increase in temperature demonstrates that the reaction releases heat into the surroundings, thus it is an exothermic reaction. In an exothermic reaction, the temperature of the surroundings rises because energy is released.

Example Calculation:

Mass of water (m): 250 g

Specific heat capacity (C) of water: 4.184 J/g°C

Temperature change (ΔT): 34.0 °C - 23.0 °C = 11.0 °C

Heat (q) absorbed by the solution: q = m × C × ΔT = 250 g × 4.184 J/g°C × 11.0 °C = 11, 506 J or 11.506 kJ

Since the heat is released by the dissolution of KOH, the reaction is said to be exothermic.

Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

Answer:

b) 0.963

Explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

        HSO4- (aq)    +    H2O        ⇌      SO42- (aq)     +     H3O+ (aq)

I          0.1                  solid &                   0                          0.1

C         -x                     liquid                 + x                            + x

E         0.1 - x          are ignored              x                          0.1 + x

Calculate x

Ka = products/reactants

  = [tex]\frac{[SO42-] [H3O+]}{[HSO4-]}[/tex]

0.011 = [tex]\frac{x (0.1 + x)}{0.1 - x}[/tex]

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x = [tex]{ -b +,-\sqrt{b^2 - 4ac[/tex] / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x = [tex]\sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }[/tex]  / 2(-1)

x = 0.111 +,- [tex]\sqrt{0.012321 + 0.0044}[/tex] / -2

x = 0.111 +,- [tex]\sqrt{0.016721}[/tex] / -2

x = [tex]\frac{0.111 +, - 0.1293}{-2}[/tex]

x = [tex]\frac{0.111 + 0.1293}{-2}[/tex]   , x = [tex]\frac{0.111 - 0.1293}{-2}[/tex]

x = [tex]\frac{0.2403}{-2}[/tex]    , x = [tex]\frac{0.0183}{-2}[/tex]

x = - 0.12015  , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963

Final answer:

The pH of a 0.100 M H2SO4 solution is approximately 1.00, considering the complete ionization of the first proton and the partial ionization of the second proton, which is less significant due to its lower Ka value.

Explanation:

The question asks about the pH of a 0.100 M H2SO4 solution, taking into account the ionization of both protons. Sulfuric acid (H2SO4) is a strong diprotic acid that dissociates completely for the first proton, yielding a concentration of 0.100 M H+ and 0.100 M HSO4-. The second proton dissociation is less extensive, with a given Ka of 1.1x10-2, which we need to include in our calculation of pH.

First step ionization (complete dissociation):
H2SO4 → H+ + HSO4-

Second step ionization (partial dissociation):
HSO4- ↔ H+ + SO42- (Ka = 1.2 x 10-2)

To calculate the pH, we first consider the complete ionization of the first proton, which directly gives us 0.100 M of H+. The pH contribution from this ionization is pH = -log(0.100) = 1.00. Then we consider the second ionization of HSO4-. Given the Ka and the initial HSO4- concentration of 0.100 M, we can set up an equilibrium expression to find the additional contribution of H+ from the second ionization. However, because the ionization is low, the change in concentration of H+ due to the second ionization can be negligible for this approximate calculation. Therefore, we can assume the pH of the solution largely results from the first dissociation.

Therefore, the answer is that the pH of the 0.100 M H2SO4 solution is approximately 1.00.

Answer:

First, Second, Second-to-last, and last choice

Explanation:

Chemistry student. The other options are incorrect. If you would like a more thorough explanation, please reply to this comment.

pH according to Arrhenius definition is the measure of hydrogen ion concentration in solution:

[tex]pH=-log[H^{+}][/tex] Acids release H+ ions into solution, while bases release OH- ions into solution. This explains why second-to-last choice is correct.

Litmus paper are thin strips of paper that have been manufactured with indicators, examples being red cabbage or phenolphthalein. Indicators react to changes in pH with an according color change. This explains why the last choice is also correct.  

Acid strength is better the lower the number it is, hence why pH 2 is stronger acid than 5, but in turn is a weaker base. This explains why second choice is correct.

The first option can be explained by the pH diagram. HCl is acidic, and will turn blue (basic) litmus paper more acidic, and move towards an acidic pH, hence more "red."

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thanks,

Answer:

I dont know if im correct number 1 is 2

Explanation:

ps im not good at this subject

Final answer:

Only statement b, which says that chlorine is reduced from +7 to -1 in both reactions, is correct. Statements a, c, and d cannot be confirmed as correct based on the given information.

Explanation:

The student has presented two chemical reaction equations involving the thermal decomposition of NH4ClO4 (ammonium perchlorate) and powdered aluminum. Let's address the statements provided:

a. All nitrogen atoms lose three electrons in both reactions. This statement is incorrect. In the provided reactions, nitrogen goes from an oxidation state of -3 in NH4+ to 0 in N2, which means each nitrogen atom gains three electrons.

b. Chlorine is reduced from +7 to -1 in both reactions. This statement is correct. In NH4ClO4, chlorine starts with an oxidation state of +7 and is reduced to -1 in HCl.

c. Reaction 2 produces more energy per gram of Al. Without details of the mass of aluminum involved in Reaction 2, we cannot determine which reaction produces more energy per gram of aluminum. However, if the reactions involve the same mass of aluminum, then Reaction 1 is more energetic since the absolute value of ΔH is greater.

d. The thrust produced by the formation of gaseous products is greater per mole of ammonium perchlorate in Reaction 2. This statement cannot be evaluated without more information about the moles of gaseous products formed in Reaction 2.

Therefore, based on the information provided, the correct statement is b. Chlorine is reduced from +7 to -1 in both reactions.

Answer:

2) HClO3 is stronger because chlorine is more electronegative than iodine.

Explanation:

The more electronegative the element is the more strong or acidic it becomes.

Chlorine being more electronegative than Iodine makes it easier for it to pull the electron of hydrogen more strongly and hence has a higher tendency to release a H+ unit. Hence that makes it stronger.

HClO3 is stronger than HIO3 because chlorine is more electronegative than iodine.

Chloric acid (HClO3) is more stronger than Iodic acid (HIO3) because chlorine is more electronegative than iodine. As we go from top to bottom in the periodic table, the atomic size increases and electronegativity decreases.

Electronegativity is the ability of an atom to attract electron pair towards itself. Chlorine atom comes on the top whereas iodine is present lower than chlorine in the periodic table so the atomic size of chlorine is smaller and higher value of electronegativity as compared to iodine so we can conclude that Chloric acid (HClO3) is more stronger than Iodic acid (HIO3).

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Answer:

[tex]\large \boxed{\text{a) HBr; b) K; c) 0.0503 g}}[/tex]

Explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r:   39.10    80.41               2.016  

            2K  +  2HBr ⟶ 2KBr + H₂

m/g:     5.5       4.04

a) Limiting reactant

(i) Calculate the moles of each reactant  

[tex]\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}[/tex]

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:  

The molar ratio of H₂:K is 1:2.

From HBr:  

The molar ratio of H₂:HBr is 3:2.  

[tex]\text{Moles of H}_{2} = \text{0.049 93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{2 mol HBr}} = \text{0.024 97 mol H}_{2}[/tex]

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

[tex]\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\\text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$}[/tex]

Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

Explanation:

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = ?

n = number of moles = 0.684

R = gas constant = [tex]0.0821Latm/Kmol[/tex]

T =temperature =[tex]273K[/tex]   (at STP)

[tex]V=\frac{nRT}{P}[/tex]

[tex]V=\frac{0.684\times 0.0821L atm/K mol\times 273K}{1atm}[/tex]

[tex]V=15.3L[/tex]

Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L

Answer:

3,5-dimethyl-2-octene

Explanation:

Please note that there is no H at carbon 3 less carbon becomes penta hydra.

The compound is:

CH3

I

C = CH2 - CH3

I

CH2

I

CH - CH3

I

CH2

I

CH2

I

CH3

To name the above compound, do the following:

1. Locate the longest continuous chain i.e octene

2. Start counting from the side that gives the double bond the lowest low count since the double bond is the functional group. In doing this, the double bond is at carbon 2.

3. Locate the substituent groups attached and their position in the parent chain. In doing so, you will see that there are two CH3 group attached and they are at carbon 3 and carbon 5. Since the substituents attached are the same, we'll name them as 'dimethyl' indicate that they are two methyl groups

Now, we'll combine the above findings in order to obtain the name. Therefore, the name of the compound is:

3,5-dimethyl-2-octene

Answer:

Water of crystallization, X = 4.

Explanation:

Molar mass of [tex]CuSO_{4}.XH_{2} O[/tex]

64 + 32 + (4x18) + x ( 1 × 2 + 16)

= 160 + 18x

Given: % water of crystallization (decrease in mass after heating) = 30%

⇒ [tex]\frac{18x}{160 + 18x} =\frac{31}{100}[/tex]

1800x = 31 (160 + 18x)

58.0645x = 160 + 18x

(58.0645 - 18)x = 160

x = [tex]\frac{160}{40.0645}[/tex] = 3.99 ≅ 4.

Water of crystallization, X = 4.

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1  O ₂N N H ⁻ + H ⁺ ( f a s t  e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ]  / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

The observed rate constant k is related to the individual rate constants of the mechanism by the equation k = k2 (k1/k-1), where k2 is the rate constant of the rate-determining slow step and k1/k-1 is the equilibrium constant of the first fast equilibrium step.

The relationship between the observed rate constant k (in the rate law) and the rate constants for the individual steps (k1, k-1, k2, k3) in the proposed mechanism for the decomposition of nitramide can be determined by examining the rate-determining step (RDS). In a reaction mechanism, the slowest step controls the overall reaction rate. For the given mechanism:

the observed rate law is rate = k [O2NNH2] [H+]. Because the second step is slow, it is the RDS. The equilibrium of the first step means that the concentration of the intermediate O2NNH−(aq) can be expressed in terms of the concentrations of the reactants O2NNH2 and H+. Therefore, the observed rate constant k is a function of the rate constants of the individual steps, particularly k2 and the equilibrium constant (K = k1/k−1) from the first step. Hence, we can conclude that k is equal to k2 multiplied by the equilibrium constant of the first step, where k = k2 (k1/k−1)

Answer:

Mole Fraction of O2 --> 0.42

Mole Fraction of Ar --> 0.037

Explanation:

Answer:

Mole Fraction of O2 --> 0.42

Mole Fraction of Ar --> 0.037

Explanation:

Answer:

9.12x10¹³ Hz

Explanation:

The vibrational frequency (ω) of a molecule is given by:

[tex] \omega = \sqrt{\frac{k}{\mu}} [/tex]

Where:

k: is the spring constant

μ: is the reduced mass

The reduced mass of a diatomic molecule is:

[tex] \frac{1}{\mu} = \frac{1}{m_{a}} + \frac{1}{m_{b}} [/tex]

Where ma and mb are the atomic masses of the atoms a and b, respectively, of the diatomic molecule.

Hence, the vibrational frequency of the hydrogen molecule is:

[tex]\omega_{H_{2}} = \sqrt{\frac{k}{\mu_{H_{2}}}}[/tex]   (1)

From equation (1) we can find k:

[tex] k = \omega_{H_{2}}^{2}*\mu_{H_{2}} [/tex]    (2)

The vibrational frequency of the deuterium molecule is:

[tex] \omega_{D_{2}} = \sqrt{\frac{k}{\mu_{D_{2}}}} [/tex]    (3)

By entering equation (2) into equation (3) we can calculate the vibrational frequency of the deuterium molecule:

[tex] \omega_{D_{2}} = \sqrt{\frac{\omega_{H_{2}}^{2}*\mu_{H_{2}}}{\mu_{D_{2}}}} [/tex]

[tex] \omega_{D_{2}} = \sqrt{\frac{\omega_{H_{2}}^{2}*\mu_{H_{2}}}{2*\mu_{H_{2}}}} [/tex]

[tex] \omega_{D_{2}} = \frac{\omega_{H_{2}}}{\sqrt{2}} = \frac{1.29 \cdot 10^{14} Hz}{\sqrt{2}} = 9.12 \cdot 10^{13} Hz [/tex]

Therefore, the vibrational frequency of the deuterium molecule is 9.12x10¹³ Hz.

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The vibrational frequency of D₂ is :   9.12 * 10¹³ Hz

Given that:

Vibrational frequency ( w ) = [tex]\sqrt{\frac{k}{u} }[/tex]

u = reduced mass

The reduced mass of a diatomic molecule is expressed as

[tex]\frac{1}{u} = \frac{1}{m_{a} } + \frac{1}{m_{b} }[/tex]

Where : Ma and Mb are the atomic masses of mass A and mass B

First step : expressing the vibrational frequency of the hydrogen molecule

wH₂ = [tex]\sqrt{\frac{k}{uH_{2} } }[/tex]   ----- ( i )

from the equation

k = ( wH₂ )² * uH₂ ---- ( ii )

Next step : expressing the vibrational frequency of the deuterium molecule.

wD₂ = [tex]\sqrt{\frac{k}{uD_{2} } }[/tex]  ---- ( iii )

Insert equation ( ii ) into equation ( iii )

wD₂ = [tex]\frac{wH_{2} }{\sqrt{2} }[/tex]  = ( 1.29 * 10¹⁴ ) / ( √2 ) = 9.12 * 10¹³ Hz

Hence we can conclude that The vibrational frequency of D₂ is :   9.12 * 10¹³ Hz.

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Answer:

p3=0.36atm (partial pressure of NOCl)

Explanation:

2 NO(g) + Cl2(g) ⇌ 2 NOCl(g)  Kp = 51

lets assume the partial pressure of NO,Cl2 , and NOCl at eequilibrium are P1 , P2,and P3 respectively

[tex]Kp=\frac{[NOCl]^{2} }{[NO]^{2} [Cl_2] }[/tex]

[tex]Kp=\frac{[p3]^{2} }{[p1]^{2} [p2] }[/tex]

p1=0.125atm;

p2=0.165atm;

p3=?

Kp=51;

On solving;

p3=0.36atm (partial pressure of NOCl)

The reaction,

Given values,

Pressure,

Value of Kp,

Now,

→ [tex]K_p = \frac{[NOCl]^2}{[NO]^2[Cl_2]^2}[/tex]

or,

→ [tex]K_p = \frac{[P_3]^2}{[P_1]^2[P_2]}[/tex]

By substituting the values,

   [tex]51 = \frac{[P_3]^2}{[0.125]^2[0.165]}[/tex]

   [tex]P_3 = 0.36 \ atm[/tex]

Thus the response above is appropriate.  

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Answer:

B) 1-propanol has more hydrogen bonding than ethyl methyl ether.

Explanation:

1-propanol and ethyl methyl ether look very similar but have very different boiling points. They both have three carbon atoms but ethyl methyl ether has an oxygen atom.

Ethyl methyl ether has both London dispersion forces and dipole-dipole interactions.

1-Propanol has London dispersion forces, dipole-dipole interactions and hydrogen bonding.

1-propanol has more hydrogen bonding than ethyl methyl ether which explains why they both have different boiling points.

Answer:

V= 9.45L

Explanation:

P=3.75atm, V=?, n= 1.5, R= 0.082, T= 15+ 273= 288K

Applying

PV= nRT

Substitute and Simplify

3.75*V= 1.5*0.082*288

V= 9.45L

The approximate volume of a 1.50 mol sample of gas at 15.0°C and a pressure of 3.75 atm is calculated using the Ideal Gas Law to be about 9.20 liters.

To calculate the approximate volume of a 1.50 mol sample of gas at 15.0°C and a pressure of 3.75 atm, we can use the Ideal Gas Law, which is PV = nRT. First, we must convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature. Then we can solve for V, the volume of the gas.

The conversion from Celsius to Kelvin: T(K) = 15.0 + 273.15 = 288.15 K

Using the Ideal Gas Law constants, R = 0.0821 L·atm/K·mol. Substituting the known values into the Ideal Gas Law equation:

PV = nRT
(3.75 atm) × V = (1.50 mol) × (0.0821 L·atm/K·mol) × (288.15 K)

V = (1.50 mol × 0.0821 L·atm/K·mol × 288.15 K) / 3.75 atm

V = 9.2029 L

Therefore, the approximate volume of the gas sample is about 9.20 liters.

Answer:

base and it's in solution

Explanation:

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Answer:

The amount that oxidized to NiO(OH) is = 1.46 gm

Explanation:

Given data

Current I = 0.35 A

Time taken = 145 min

We know that charge

Q = I t

Q = 0.35 × 145 × 60

Q = 3045 C

Faraday's constant = 96500 C

No. of  moles of electron

[tex]N = \frac{3045}{96500}[/tex]

N = 0.03155

1 mol of [tex]NI(OH)_{2}[/tex] is oxidized by 2 moles of electrons,  so no. of moles can be oxidized is

[tex]\frac{0.03155}{2}[/tex] = 0.015775 moles

Now convert this moles into gm by multiplying 92.708 [tex]\frac{gm}{mol}[/tex]

0.015775 × 92.708 = 1.46 gm

Therefore the amount that oxidized to NiO(OH) is = 1.46 gm

Answer:

Yes

Explanation:

It causes excess bromine to be given off as a gas. -It is used to absorb excess heat generated by the exothermic reaction.

Answer

The desired product should contain acidic group like -COOH group.

Explanation:

Saturated sodium bicarbonate solution will react with a acid compound which contain in reaction mixture and produce CO2 gas.

We have the reaction has;  

NaHCO3 (aq) + R-COOH (aq) ..............> R-COONa (aq) + H2O (l) + CO2 (g)

Therefore.

The desired product will contain acidic group like -COOH (carboxylic acid group).

Answer:

we are given the 3-methyl2 butanone and upon the reduction with LiAIH4 there is formed alcohol, there are two possible side attack,

therefore, whenever the front side attacks then the -CH3 moves back  the plain and the H will be above the plane. More so, when  attack from the back side the H moves below the plane and -CH3 moves above the plane. -OH is evident in the plane. see the attachment below to view the structure.

Explanation:

Answer : The molarity of the resulting solution is, 0.29 M

Explanation :

When NaCl dissolved in water then it dissociates to give sodium ions and chloride ions.

Given,

Moles of NaCl = 0.87 mol

Volume of solution = 3.00 L

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

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

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

[tex]\text{Molarity}=\frac{0.87mol}{3.00L}=0.29mole/L=0.29M[/tex]

Therefore, the molarity of the resulting solution is, 0.29 M

Answer:

Molality = 6.0 m

Explanation:

The molecular weight of tartaric acid = 150.087 g/mol

Given that:

Density  of the solution = 1.023 g/mL

Molarity =  3.23M

Density is given as : [tex]Molarity ( \frac{1}{molality } +\frac{mol.wt}{1000} )[/tex]

∴

[tex]1.023 = 3.23 (\frac{1}{molality } +\frac{150.087}{1000} )[/tex]

[tex]\frac{1}{molality } =( \frac{1.023}{3.23} - \frac{150.087}{1000} )[/tex]

[tex]\frac{1}{molality } =0.3167 - 0.1500[/tex]

[tex]\frac{1}{molality } = 0.1667[/tex]

Molality  = [tex]\frac{1}{0.1667}[/tex]

Molality = 5.999 m

Molality ≅ 6.0 m

Answer:

We should weigh 72.1 of sodium sulfate

Explanation:

The solution must be made of sodium sulfate. This salt can be dissociated like this:

Na₂SO₄ →  2Na⁺  +  SO₄⁻²

From this dissociation we can say that, 2 moles of sodium cation are obtained from 1 mol of salt.

Therefore 0.513 moles of sodium cation will be obtained from 0.2565 moles of salt (0.513 . 1) / 2. The thing is that this number are the moles contained in 1 L of solution, and we need to prepare such 1.98 L

Molarity = mol / Volume (L) → Molarity . volume (L) = mol

0.2565 mol/L . 1.98L = 0.508 moles

These are the moles we should weigh. Let's convert them to moles:

0.508 mol . 142.06 g / 1 mol = 72.1 g

Answer:

We need 72.1 grams of Na2SO4

Explanation:

Step 1: Data given

sodium ion concentration = 0.513 mol/L

Volume = 1.98 L

Step 2: The balanced equation

Na2SO4 → 2Na+ + SO4^2-

Step 3: Calculate moles Na+

Moles Na+ = molarity Na+ * volume

Moles Na+ = 0.513 * 1.98 L

Moles Na+ = 1.01574 moles

Step 4: Calculate moles Na2SO4

For 1 mol Na2SO4 we need 2 moles Na+ and 1 mol SO4^2-

For 1.01574 moles Na+ we'll need 1.01574/2 = 0.50787 moles Na2SO4

Step 5: Calculate moles Na2SO4

Mass Na2SO4 = moles Na2SO4 * molar mass Na2SO4

Mass Na2SO4 = 0.50787 moles * 142.04 g/mol

Mass Na2SO4 = 72.1 grams

We need 72.1 grams of Na2SO4