Write the complete ionic equation and the Net ionic equation for: K2C2O4(aq)+Pb(OH)2(aq) -> 2KOH(aq)+PbC2O4(s)

Answer: [tex]Rate=k[A]^1[B]^5[C]^6[/tex]

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.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions are defined as the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical reaction.

[tex]A+5B+6C\rightarrow 3D+3E[/tex]

[tex]Rate=k[A]^1[B]^5[C]^6[/tex]

k= rate constant

1 = order with respect to A

5 = order with respect to B

6 = order with respect to C

Thus rate law is [tex]Rate=k[A]^1[B]^5[C]^6[/tex]

Final answer:

To neutralize the excess 16 M HNO₃, 5.33 mL of 6 M NaOH is required.

Explanation:

The task is to calculate the volume of 6 M NaOH needed to neutralize excess HNO₃ used in dissolving copper. The student used 6.0 mL of 16 M HNO₃, which is 2.0 mL more than the required amount.

First, we need to find the amount of excess HNO₃ in moles, as only 4.0 mL was required. The excess volume is 6.0 mL - 4.0 mL = 2.0 mL. The moles of excess HNO₃ is calculated as 2.0 mL × 16 M / 1000 mL/L = 0.032 moles.

To neutralize the acid, we need the same number of moles of OH-. Since the NaOH is 6 M, the volume V in liters needed is calculated using the molarity equation: moles = Molarity × Volume. Thus, 0.032 moles = 6 M × V, which gives V = 0.032 moles / 6 M. So, the volume of NaOH required is 0.00533 L, or 5.33 mL.

Final answer:

Sodium hydride donates a hydride ion to ethanol, resulting in the formation of hydrogen gas and the ethoxide ion in a sodium ethoxide complex.

Explanation:

The proton transfer reaction between sodium hydride (NaH) and ethanol (CH3CH2OH) involves sodium hydride acting as a base, donating a hydride ion (H-) to the proton (H+) of the ethanol. This reaction results in the formation of hydrogen gas (H2) and the ethoxide ion (CH3CH2O-), which remains in the solution complexed with the sodium ion (Na+). The balanced equation for this reaction is NaH + CH3CH2OH → H2 + Na+ + CH3CH2O-. This reaction utilizes the hydride ion from the sodium hydride as a nucleophile that abstracts a proton from the ethanol, leading to the evolution of hydrogen gas.

This question involves the application of the Arrhenius equation in chemistry, particularly in calculating the temperature dependence of reaction rates. Given the rate constant and the activation energy at a certain temperature, one can find the rate constant at a different temperature

The question pertains to the use of the Arrhenius equation, which calculates the temperature dependence of reaction rates. The equation is k = A * e^(-Ea/RT), where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

The given reaction has a rate constant (k) of 1.35 x 10² s-1 at 25.0°C (or 298.15K). The activation energy (Ea) is given as 85.6 kJ/mol. To find the rate constant at another temperature, rearrange the Arrhenius equation to solve for A, substitute the given values for Ea, k, R and T to find A. Then, plug the calculated A, given Ea, the new temperature in Kelvin (75.0°C or 348.15K), and R into the Arrhenius equation to solve for the new rate constant, k.

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Explanation:

It is known that number of moles present in a substance is equal to the mass divided by molar mass.

Mathematically,      No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

As it is given that number of moles are 0.432 moles and molar mass of [tex]C_{8}H_{9}O_{4}[/tex] is 169.15 g/mol.

Hence, calculate the mass of [tex]C_{8}H_{9}O_{4}[/tex] is as follows.

                 No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                     0.432 mol = [tex]\frac{mass}{169.15 g/mol}[/tex]

                         mass = 73.072 g

Thus, we can conclude that the mass of 0.432 moles of [tex]C_{8}H_{9}O_{4}[/tex] is 73.072 g.

Final answer:

Rubbing alcohol molecules as a gas have the same structure as in the liquid but with more kinetic energy and less intermolecular attraction, which upon evaporation causes a cooling effect through evaporative cooling.

Explanation:

When rubbing alcohol evaporates from your hand, it leaves a cooling sensation because the molecules in the liquid state require a certain threshold of kinetic energy to overcome intermolecular forces and escape into a gas state. The correct statement regarding how the molecules of gas compare to the molecules as a liquid is:

In the gaseous state, particles move faster and are further apart compared to when they are in the liquid state, where particles are closer together and have stronger intermolecular attractions. This process of evaporation involves evaporative cooling, where the molecules with higher kinetic energy escape, leaving behind those with lower kinetic energy, which results in a decrease in temperature.

Ionic compounds are generally formed from a metal cation (positively charged ion) and a nonmetal anion (negatively charged ion), so the correct answer to your question is option C.

Ionic compounds are typically composed of a metal cation and nonmetal anion. This means the correct answer to your question is option C. A cation is a positively charged ion, and in this context, it is typically formed by an element from the left side of the periodic table, or a metal. An anion, on the other hand, is a negatively charged ion, usually formed by an element from the right side of the periodic table, or a nonmetal. When these ions combine, they create an ionic compound, such as NaCl (sodium chloride), where sodium is the metal cation and chloride is the nonmetal anion.

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The activation energy can be determined using the Arrhenius equation. To find the activation energy at a different temperature, rearrange the equation and plug in the given rate constant. The frequency factor is approximately 1.529 * 10^9 /s.

The activation energy can be determined using the Arrhenius equation:

k = Ae^(-Ea/RT)

where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

To find the activation energy at a different temperature, we can rearrange the equation as:

Ea = -ln(k/A) * RT

Using the given rate constant at 683 K and the activation energy of 212 kJ, we can calculate the frequency factor as follows:

A = k * e^(Ea/RT)

Plugging in the values:

A = (6.06×10^(-4) /s) * e^(212000 J / (8.3145 J/mol*K * 683 K))

A = 1.529 * 10^9 /s

Therefore, the frequency factor is approximately 1.529 * 10^9 /s.

The activation energy for a reaction is the minimum amount of energy required for the reaction to occur. To find the activation energy at a different temperature, we can use the Arrhenius equation. The activation energy will be 2.71 kJ/mol at the desired condition.

The activation energy for a reaction is the minimum amount of energy required for the reaction to occur. It represents the energy barrier that the reactants must overcome before they can form products. In the given question, the activation energy for the gas phase decomposition of 2-bromopropane is 212 kJ.

To find the activation energy at a different temperature, we can use the Arrhenius equation:

k = A * e^(-Ea/RT)

where:
- k is the rate constant
- A is the frequency factor
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin

By rearranging the equation, we can solve for the activation energy at a different temperature:

Ea2 = -ln(k2/k1) * (R/T2 - R/T1)

Substituting the given values:

Ea2 = -ln(5.06x10^-3 / 6.06x10^-4) * (8.314 J/(mol·K) / 683 K - 8.314 J/(mol·K) / 298 K)

Ea2 = -ln(8.333) * (0.0122 - 0.0278) = -ln(8.333) * (-0.0156) = 0.1739 * 0.0156 = 0.00271 J/mol = 2.71 kJ/mol

Therefore, the activation energy will be 2.71 kJ/mol at the desired condition.

The final concentration (C_2) is 20 times less than the initial concentration (C_1), illustrating a dilution factor of 20. For instance, if C_1 were 2 mol/L, C_2 would be 0.1 mol/L.

Calculate the concentration of the diluted solution:

1. Identify the given information:

Initial volume of stock solution (V_1) = 13.0 mL

Final volume of diluted solution (V_2) = 0.260 L = 260 mL

Initial concentration of stock solution (C_1) is unknown (this is what we need to find)

2. Understand the key concept:

When a solution is diluted, the amount of solute (the substance dissolved in the solution) remains constant. Only the volume of the solution changes.

This means that the product of concentration and volume before dilution is equal to the product of concentration and volume after dilution.

3. Apply the dilution formula:

The formula is C_1V_1 = C_2V_2, where:

C_1 = initial concentration

V_1 = initial volume

C_2 = final concentration

V_2 = final volume

4. Solve for the unknown concentration (C_1):

Rearrange the formula to isolate C_1: C_1 = C_2V_2 / V_1

Plug in the known values: C_1 = (unknown) * 260 mL / 13.0 mL

Simplify: C_1 = 20 * C_2

5. Interpret the result:

The final concentration (C_2) will be 20 times less than the initial concentration (C_1). This means the solution has been diluted by a factor of 20.

6. Example:

If the initial concentration (C_1) was 2 mol/L, then the final concentration (C_2) would be 0.1 mol/L (20 times less).

The biggest elctronegative difference is between silicon and sulfur. So  Si-S will be most polar bond.

The polarity between the two atoms is determined by their relative difference in electronegativity.

The Electronegativity of ,

Silicon= 1.9

Phosphorus= 2.19

Sulfur= 2.58

The direction of polarity,

[tex]\rm \bold{ \delta^+Si\rightarrow \delta^-P}\\\rm \bold{ \delta^+Si\rightarrow \delta^-S}\\\rm \bold{ \delta^+P\rightarrow \delta^-S}[/tex]

Since, the biggest elctronegative difference is between silicon and sulfur (Si-S).

Hence we can say that Si-S will be most polar bond.

To know more about  electronegativity, refer to the link:

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[H⁺]=3.608.10⁻¹²

Weak acid ionization reaction occurs partially (not ionizing perfectly as in strong acids)

The ionization reaction of a weak acid is an equilibrium reaction

HA (aq) ---> H⁺ (aq) + A⁻ (aq)

The equilibrium constant for acid ionization is called the acid ionization constant, which is symbolized by Ka

The values ​​for the weak acid reactions above:

[tex]\rm Ka=\dfrac{[H][A^-]}{[HA]}[/tex]

The greater the Ka, the stronger the acid, which means the reaction to the right is also greater

Where Kb is the base ionization constant

LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)

[tex]\rm Kb=\dfrac{[L][OH^-]}{[LOH]}[/tex]

Kb of Ethylamine (C₂H₅NH₂) : 6.4.10⁻⁴

The ethylamine ionization reactions occur in water as follows:

C₂H₅NH₂ + H₂O ⇒ C₂H₅NH₃⁺ + OH⁻

with a Kb value:

[tex]\rm Kb=\dfrac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]}[/tex]

for example x = number of moles / concentration that reacts

Initial concentration of  Ethylamine (C₂H₅NH₂) : 1.2.10⁻²

Concentration at equilibrium = 1.2.10⁻²  -x

Initial concentration of C₂H₅NH₃ = 0

Concentration at equilibrium = x

Initial concentration OH⁻ = 0

Concentration at equilibrium = x

so the value of Kb =

[tex]\rm Kb=\dfrac{[x][x]}{[1.2.10^{-2}-x]}\\\\assumption\:x=so\:small\:then\\\\6.4.10^{-4}=\dfrac{x^2}{1.2.10^{-2}}\\\\x^2=7.68.10^{-6}\\\\x=2.771.10^{-3}[/tex]

x = [OH⁻] = 2.771.10⁻³

Ka x Kb = [H⁺] [OH-]

a water equilibrium constant value (Kw) of 1.10⁻¹⁴ at 25 °C

Ka x Kb = [H +] [OH-] = 1.10⁻¹⁴

1.10⁻¹⁴ = [H⁺] . 2.771.10⁻³

[H⁺]=3.608.10⁻¹²

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Calculate the value of the equilibrium constant, Kc

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Concentration of hi at equilibrium

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The order of the reaction with respect to clo2 is 1.

The order of the reaction with respect to clo2 can be determined by comparing the initial rates of different experiments and analyzing the effect of changing the concentration of clo2 on the reaction rate. By comparing the rates of Experiment 1 and Experiment 2, we can see that when the concentration of clo2 is halved, the rate of the reaction is also halved. This indicates that the reaction rate is directly proportional to the concentration of clo2. Therefore, the order of the reaction with respect to clo2 is 1.

Final answer:

To find the molecular formula, calculate the empirical formula mass, divide the molar mass by this number to find a multiplier, and apply the multiplier to the subscripts in the empirical formula.

Explanation:

Finding the Molecular Formula

To determine the molecular formula of a compound for which we know the molar mass is 110 grams/mole, and have the empirical formula from problem 1, we follow these steps:

For example, assuming our empirical formula is CH₂ (not given in the question), we would get the empirical formula mass as 12 (for C) + 2×1 (for H) = 14 g/mol. Then, 110 g/mol divided by 14 g/mol gives us approximately 7.86, which we round to 8 since it should be a whole number. Multiplying the subscripts in CH₂ by 8 gives us C₈H₁₆ as the molecular formula.

Answer : The correct option is, 0.2 mole Y left over .

Explanation : Given,

Moles of X = 4.8 mole

Moles of Y = 3.4 mole

The balanced chemical reaction is,

[tex]3X+2Y\rightarrow X_3Y_2[/tex]

From the balanced reaction, we conclude that

As, 3 moles of X react with 2 moles of Y

So, 4.8 moles of X react with [tex]\frac{2}{3}\times 4.8=3.2[/tex] moles of Y

From this we conclude that, the reactant Y is an excess reagent and X is a limiting reagent.

The moles of excess reagent left over at the end of the reaction = Given moles of X - Required moles of X

The moles of excess reagent left over at the end of the reaction = 3.4 - 3.2 = 0.2 mole

Therefore, the moles of excess reagent left over at the end of the reaction is, 0.2 mole Y left over.

Answer:

The correct answer is : '0.2 mol Y left over'.

Explanation:

[tex]3X + 2Y \rightarrow  X_3Y_2[/tex]

Moles of X = 4.8 moles

Moles of Y = 3.4  moles

According to reaction, 3 moles of X react with 2 moles of Y .

Then 4.8 moles of X react with :

[tex]\frac{2}{3}\times 4.8=3.2 [/tex]moles of Y

Moles of Y reacted = 3.2 moles

Moles of Y left unreacted = 3.4 moles - 3.2 moles = 0.2 moles

As we can see that X is in limiting amount and y is present in an excessive amount.And the left over amount of Y is 0.2 moles.

Final answer:

In gas chromatography, compounds elute based on their boiling points, with those having lower boiling points eluting first. Since anisole has a lower boiling point than d-limonene, anisole is expected to elute first.

Explanation:

The question asks which compound, anisole or d-limonene, would elute first in gas chromatography (GC) based on their boiling points. In gas chromatography, compounds generally elute in order of increasing boiling points because compounds with lower boiling points have lower retention times on the GC column. Also, the elution order correlates with the strength of intermolecular forces (IMFs) affecting the compounds; compounds with stronger IMFs tend to have higher boiling points and adhere more to the stationary phase, thus eluting later. Although the specific boiling points of anisole and d-limonene are not provided in this answer, it is known that anisole has a boiling point of about 154°C, and d-limonene has a boiling point around 176°C. Therefore, one would expect anisole to elute first in gas chromatography due to its lower boiling point compared to d-limonene.

Answer:

The amount of water that evaporates from the earth is approximately equal to the amount that falls as precipitation.

Explanation:

Water evaporation is critical to climate because it is directly related to precipitation formation. Water that evaporates from rivers, lakes, oceans and even our bodies helps to form rain. This occurs when the temperature cools. Under these climatic conditions, water vapor returns to its liquid form (condensation) and falls through rainfall. The amount of water evaporated is basically equal to the amount of water that comes back to land in precipitation.