Nitrogen is a vital component of proteins and nucleic acids, and thus is necessary for life. The atmosphere is composed of roughly 80% N2, but most organisms cannot directly utilize N2 for biosynthesis. Bacteria capable of "fixing" nitrogen ( ie converting N2 to a chemical form, such as NH3, which can be utilized in the biosynthesis of proteins and nucleic acids) are called diazatrophs. The ability of some plants like legumes to fix nitrogen is due to a symbiotic relationship between the plant and nitrogen-fixing bacteria that live in the plant’s roots.

Assume the hypothetical reaction for fixing nitrogen biologically is N2 (g) + 3H2O (l) → 2 NH3 (aq) + 3/2 O2 (g)

a. Calculate the standard enthalpy change for the biosynthetic fixation of nitrogen at T = 298 K. For NH3 (aq), ammonia dissolved in aqueous solution, ΔHof = - 80.3 kJ mol-1.

b. In some bacteria, glycine is produced from ammonia by the reaction NH3 (aq) + 2CH4 (g) + 5/2 O2 (g) → NH2CH2COOH (s) + 3H2O (l) Calculate the standard enthalpy change for the synthesis of glycine from ammonia. For glycine, ΔHof = - 537.2 kJ mol -1. Assume that T = 298 K.

c. Calculate the standard enthalpy change for the synthesis of glycine from nitrogen, oxygen, and methane.

Answer:

I. heating the helium

Explanation:

Final answer:

(I) Heating the helium and (IV) decreasing the pressure outside the container are the two processes that will cause the piston to move away from the base and decrease the pressure of the gas. Hence, (c) is the correct option.

Explanation:

Assuming ideal gas behavior, the question asks which processes will lead the piston to travel away from the base and lower the helium gas's pressure. Let's examine each of the scenarios that are presented:

I. Heating the helium: Heating increases the internal energy of the gas molecules, making them move faster. This increases the force they exert against the piston, pushing it outward and increasing the volume, thus reducing the pressure according to Boyle's Law (P₁V₁ = P₂V₂).

II. Removing some of the helium from the container: If you remove some of the helium, there will be fewer molecules to exert force on the piston, leading to a decrease in pressure. However, the piston will not necessarily move since the external pressure remains the same. Without a corresponding change in external pressure, there is no force to move the piston outward.

III. Turning the container on its side: Turning the container on its side will have no effect on the pressure or volume of the gas if the external conditions remain the same. The position of the piston is influenced by forces and pressures, not by its orientation in space.

IV. Decreasing the pressure outside the container: Decreasing the external pressure acting on the piston will allow the internal pressure of the gas to push the piston outward, increasing the volume and therefore decreasing the gas pressure, as explained by the ideal gas law (PV = nRT).

Therefore, the processes that will cause the piston to move away from the base and decrease the pressure of the gas are heating the helium and decreasing the pressure outside the container. So the correct answer to the question is (c) 2.

Answer:

0.23 L

Explanation:

Let's consider the following balanced equation.

3 Pb(NO₃)₂(aq) + 2 Na₃PO₄(aq) ⇄ Pb₃(PO₄)₂(s) + 6 NaNO₃(aq)

The moles of Pb(NO₃)₂ are:

[tex]0.18L\times \frac{5.0mol}{L} =0.90mol[/tex]

The molar ratio of Pb(NO₃)₂ to Na₃PO₄ is 3:2. The moles of Na₃PO₄ are:

[tex]0.90molPb(NO_{3})_{2}.\frac{2molNa_{3}PO_{4}}{3molPb(NO_{3})_{2}} =0.60molNa_{3}PO_{4}[/tex]

The volume of Na₃PO₄ required is:

[tex]\frac{0.60mol}{2.6mol/L} =0.23L[/tex]

the molal freezing point depression constant [tex](\(K_f\))[/tex] of liquid X is approximately [tex]\(4.13 \, \text{°C/molal}\)[/tex].

To calculate the molal freezing point depression constant (\(K_f\)) of liquid X, we can use the formula:

[tex]\[ \Delta T_f = K_f \times m \][/tex]

Where:

- [tex]\( \Delta T_f \)[/tex] is the freezing point depression (given as [tex]\(0.4^\circ \text{C} - (-0.5^\circ \text{C}) = 0.9^\circ \text{C}\)[/tex]),

- [tex]\( m \)[/tex] is the molality of the solution,

- [tex]\( K_f \)[/tex] is the molal freezing point depression constant.

First, we need to calculate the molality of the solution, which is defined as the number of moles of solute per kilogram of solvent.

Given:

- Mass of urea [tex](\( \text{NH}_2\text{CO} \))[/tex] = 5.90 g

- Mass of liquid X = 450.0 g

We need to find the moles of urea first:

[tex]\[ \text{moles of urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} \][/tex]

The molar mass of urea [tex](\( \text{NH}_2\text{CO} \))[/tex] is the sum of the molar masses of nitrogen, hydrogen, carbon, and oxygen:

[tex]\[ \text{molar mass of urea} = 14.01 + 2(1.01) + 12.01 + 16.00 = 60.03 \, \text{g/mol} \][/tex]

[tex]\[ \text{moles of urea} = \frac{5.90 \, \text{g}}{60.03 \, \text{g/mol}} = 0.0983 \, \text{mol} \][/tex]

Now, we can calculate the molality of the solution:

[tex]\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \][/tex]

[tex]\[ \text{molality} = \frac{0.0983 \, \text{mol}}{0.450 \, \text{kg}} = 0.218 \, \text{mol/kg} \][/tex]

Now, we can rearrange the formula for [tex]\(K_f\)[/tex] and solve for it:

[tex]\[ K_f = \frac{\Delta T_f}{m} \][/tex]

[tex]\[ K_f = \frac{0.9^\circ \text{C}}{0.218 \, \text{mol/kg}} \][/tex]

[tex]\[ K_f \approx 4.13 \, \text{°C/molal} \][/tex]

Therefore, the molal freezing point depression constant [tex](\(K_f\))[/tex] of liquid X is approximately [tex]\(4.13 \, \text{°C/molal}\)[/tex].

The molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X is 4.124 °C/(mol/kg).

To calculate the molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X, we can use the formula for freezing point depression:

[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]

where:

[tex]\( \Delta T_f \)[/tex] is the freezing point depression, which is the difference between the normal freezing point of the solvent and the freezing point of the solution.

i is the van 't Hoff factor, which is the number of moles of solute particles per mole of solute dissolved. For urea, i is typically 1 because urea does not dissociate in solution.

[tex]\( K_f \)[/tex] is the molal freezing point depression constant for the solvent.

m is the molality of the solution, which is the number of moles of solute per kilogram of solvent.

First, we calculate the freezing point depression [tex]\( \Delta T_f \)[/tex]:

[tex]\[ \Delta T_f = T_f^0 - T_f = 0.4C - (-0.5C) = 0.9C \][/tex]

Next, we need to calculate the molality m of the solution. The molar mass of urea [tex]\( NH_2CONH_2 \)[/tex] is 60.06 g/mol. The number of moles of urea is:

[tex]\[ n_{urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} = \frac{5.90 \text{ g}}{60.06 \text{ g/mol}} \] \[ n_{urea} = 0.0982 \text{ mol} \][/tex]

The molality m is:

[tex]\[ m = \frac{n_{urea}}{\text{mass of solvent in kg}} = \frac{0.0982 \text{ mol}}{0.450 \text{ kg}} \] \[ m = 0.2182 \text{ mol/kg} \][/tex]

Now we can rearrange the freezing point depression formula to solve for [tex]\( K_f \)[/tex]:

[tex]\[ K_f = \frac{\Delta T_f}{i \cdot m} \] Since \( i = 1 \) for urea, we have: \[ K_f = \frac{0.9C}{1 \cdot 0.2182 \text{ mol/kg}} \] \[ K_f = 4.124 \text{ C/(mol/kg)} \][/tex]

Therefore, the molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X is 4.124 °C/(mol/kg).

Answer:

At equilibrium, Kc = 0.058

Explanation:

Step 1: Data given

Mol SO3 = 12.6

Volume = 4.0 L

At equilibrium we have:

3.4 mol of SO2

Step 2: The balanced equation

2SO3(g) + ⇆ 2SO2(g) + O2(g)

Step 3: ICE-chart

The initial number of moles are:

SO3: 12.6 moles

SO2 : 0 mol

O2: 0 mol

There will react:

SO3: -2x

SO2: +2x

O2: +x

The number of moles at the equilibrium are:

SO3: 12.6 - 2x

SO2: 2x  = 3.4 mol

O2: x

Since at the equilibrium, we have 2x = 3.4 mol. x = 1.7 mol

This means at the equilibrium we have 1.7 mol of O2 and 12.6 -3.4 = 9.2 mol of SO3

Step 4: Calculate the equilibrium constant Kc

Kc = [3.4/4]² *[1.7/4] / [9.2/4]²

Kc = 0.058

At equilibrium, Kc = 0.058

Answer:

The correct answer is C)H2PO4-(aq) + H2O(l)--> H3O+(aq) + HPO42-(aq)

Explanation:

The acid dissociation equilibrium involves the loss of a proton of the acid to give the conjugated acid. In this case, the acid is H₂PO₄⁻ and it losses a proton (H⁺) to give the conjugated acid HPO₄²⁻ (without a proton and with 1 more negative charge). In the aqueous equilibrium, the proton is taken by H₂O molecule to give the hydronium ion H₃O⁺.

H₂PO₄⁻(aq) + H₂O(l)--> H₃O⁺(aq) + HPO₄²⁻(aq)

Answer:

C) H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)

Explanation:

For which of the following equilibria does Kc correspond to the acid-dissociation constant, Ka, of H₂PO₄⁻?

A) H₂PO₄⁻(aq) + H₃O⁺(aq) → H₃PO₄(aq) + H₂O(l)

NO. This is the inverse of the acid dissociation of H₃PO₄.

B) H₂PO₄⁻(aq) + H₂O(l) → H₃PO₄(aq) + OH⁻(aq)

NO. This is the basic dissociation of H₂PO₄⁻.

C) H₂PO₄⁻(aq) + H₂O(l) → H₃O⁺(aq) + HPO₄²⁻(aq)

YES. This is the acid dissociation of H₂PO₄⁻. The acid-dissociation constant is:

[tex]Ka=\frac{[H_{3}O^{+}].[HPO_{4}^{2-} ]}{[H_{2}PO_{4}^{-} ]}[/tex]

D) H₃PO₄(aq) + H₂O(l) → H₃O⁺(aq) + H₂PO₄⁻(aq)

No. This is the acid dissociation of H₃PO₄.

E) HPO₄²⁻(aq) + H₂O(l) → H₂PO₄⁻(aq) + OH⁻(aq)

NO. This is the basic dissociation of HPO₄²⁻.

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

The entropy change ΔSo for a reaction can be obtained by considering the sum of decompositions and formations of reactants and products. In this scenario, while we were only able to compute the enthalpy change ΔHxn of -136.80 kJ, the entropy change, ΔSo, cannot be calculated without additional data.

To find the entropy change ΔSo for the reaction 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g), we start by writing this as the sum of decompositions of 3NO₂(g) and 1H₂O(1) into their constituent elements. Similarly, we identify the formation of 2HNO3(aq) and 1NO(g) from their constituent elements. By summing the enthalpy changes obtained from standard enthalpy changes of formation (ΔHf) for these compounds, we find the result for ΔHxn, which in this case equates to -136.80 kJ. However, the original question asked for ΔSo, not ΔHxn. Without knowing the ΔSo for the individual substances in this reaction, we cannot calculate ΔSo for the entire reaction.

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Answer:

Respiration is responsible for removing carbon dioxide and providing oxygen to our body cells. Carbondioxide is eliminated from the body as a byproduct.  

Explanation:

Answer:

Hope this helps:)

Explanation:

The values for the table entries are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential.

Elemental fluorine, for example, is the strongest common oxidizing agent.

Lithium metal is therefore the strongest reductant (most easily oxidized) of the alkali metals in aqueous solution. The standard reduction potentials can be interpreted as a ranking of substances according to their oxidizing and reducing power

Answer:

S = 6.40 × 10⁻⁷ M

Explanation:

In order to calculate the solubility (S) of M(OH)₂ in pure water we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in concentration.

            M(OH)₂(s) ⇄ M²⁺(aq) + 2 OH⁻(aq)

I                                   0                  0

C                                 +S               +2S

E                                   S                 2S

The solubility product (Kps) is:

Kps = 1.05 × 10⁻¹⁸ = [M²⁺].[OH⁻]²=S.(2S)²

1.05 × 10⁻¹⁸ = 4S³

S = 6.40 × 10⁻⁷ M

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

Step 1: Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

Step 2: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

Step 3: Define the formula

Pressure is constant.  → the heat evolved from the reaction is equivalent to the enthalpy of reaction.  

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the  calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for  water (1g/mL and 4.184 J/g°C)

The increased temperature after the precipitation reaction of NaCl and AgNO3 in a coffee cup calorimeter indicates that heat is released. Calculation of the heat released depends on the specific heat capacity of the solution and its mass. Some assumptions made include the solution having the same heat capacity and density as water, and perfect insulation of the calorimeter.

The heat released by the precipitation reaction of NaCl and AgNO3 to form AgCl can be deduced from the increase of temperature observed in the coffee cup calorimeter. The process happens when Ag+ from AgNO3(aq) and Cl- ion from NaCl(aq) react to form the solid AgCl. This reaction is exothermic as it leads to an increase in the temperature of surroundings, that is, the solution in the calorimeter.

To calculate the heat produced, we need the specific heat capacity of the solution, which is assumed to be approximately equal to that of water (4.18 J/g°C), and the total mass of the solution which is calculated by adding the volumes of the two solutions since the density of the solution is assumed to be approximately 1 g/mL (same as water).

The approximations made in this calculation include treating the solution as having the same specific heat capacity and density as pure water, and assuming that the calorimeter perfectly insulates the solution so that no heat is lost to the environment.

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Final answer:

The standard enthalpy change (ΔH°) for the process C6H6(l) → C6H6(g) is 82.93 kJ/mol. The standard entropy change (ΔS°) for the process is 269.2 J/K*mol. The standard free energy change (ΔG°) for the process is 7361.033 kJ/mol.

Explanation:

The standard enthalpy change, ΔH°, for the process C6H6(l) → C6H6(g) can be calculated using the standard enthalpy of formation of benzene in the gaseous state and the liquid state. The equation for ΔH° is ΔH° = ΣΔH°(products) - ΣΔH°(reactants). In this case, since benzene is the only product and there are no reactants, the equation simplifies to ΔH° = ΔH°(C6H6(g)). Therefore, ΔH° = 82.93 kJ/mol.

The standard entropy change, ΔS°, for the process can be calculated using the standard entropy of benzene in the gaseous state and the liquid state. The equation for ΔS° is ΔS° = ΣΔS°(products) - ΣΔS°(reactants). Similar to the calculation for ΔH°, since benzene is the only product and there are no reactants, the equation simplifies to ΔS° = ΔS°(C6H6(g)). Therefore, ΔS° = 269.2 J/K*mol.

The standard free energy change, ΔG°, for the process can be calculated using the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. Since the temperature is given as 25.00°C, we need to convert it to Kelvin by adding 273.15. Therefore, T = 25.00°C + 273.15 = 298.15 K. Substituting the values into the equation, we get ΔG° = 82.93 kJ/mol - (298.15 K)(269.2 J/K*mol) = 7361.033 kJ/mol.

Answer:

Five electrons are gained.

Explanation:

Oxidation number or oxidation state of an atom in a chemical compound is  the number of electrons lost or gained. It is also defined as the degree of oxidation of the atom in the compound.

This is a theoretical number which can help to decipher the oxidation and reduction in a redox reaction.

Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.

The given reaction is shown below as:

[tex]MnO_4^-\rightarrow Mn^{2+}[/tex]

Manganese in [tex]MnO_4^-[/tex] has oxidation state of +7

Manganese in [tex]Mn^{2+}[/tex] has an oxidation state of +2

It reduces from +7 to +2 . It means that 5 electrons are gained.

Answer:

553 nm

Explanation:

When an electron from O absorbs radiation with an energy (E) of 3.6 × 10⁻¹⁹ J, it is excited from orbital 2p to orbital 3s. The wavelength (λ) associated with that radiation can be calculated using the Planck-Einstein equation.

E = h. ν = h . c . λ⁻¹

where,

h is the Planck's constant

c is the speed of light

ν is the frequency

[tex]\lambda = \frac{h.c}{E} =\frac{6.63 \times 10^{-34}J.s \times 3.00 \times 10^{8} m/s}{3.6 \times 10^{-19}J } .\frac{10^{9}nm }{1m} =553nm[/tex]

Answer:

C.

Explanation:

The entropy (S) is the measure of the randomness of a system, and ΔS = Sfinal - Sinitial. As higher is the disorder of the system, as higher is the entropy.

A. When KCl is fractionated in power, there'll be more portions of it, so, the disorder must be higher, then ΔS is positive.

B. As higher is the temperature, higher is the kinetic energy of the system, and because of that, the disorder is also higher, so ΔS is positive.

C. The decrease in the volume (compression) decreases the distance between the molecules, so the system will be more organized, then ΔS is negative.

D. The volume before the mixing will be higher, and the ethanol will dissociate, so it will be more particles, the disorder will increase, and ΔS is positive.

E. Sgas > Sliquid > Ssolid because of the disorder of the molecules, then ΔS is positive.

Final answer:

The process with a negative change in entropy (ΔS) is compressing 1 mol Ne at constant temperature from 1.5 L to 0.5 L, as it leads to a decrease in volume and increases the orderliness of the system.

Explanation:

The question asks for which process the change in entropy (ΔS) is negative. Entropy generally refers to the measure of disorder or randomness in a system. A negative ΔS indicates a decrease in entropy, meaning the system becomes more ordered. Considering the options:

A. grinding a large crystal of KCl to powder - Increases disorder by breaking down the crystal structure, so ΔS is positive.

B. raising the temperature of 100 g Cu from 275 K to 295 K - Increases thermal motion and disorder, so ΔS is positive.

C. compressing 1 mol Ne at constant temperature from 1.5 L to 0.5 L - Decreases the volume and increases order, so ΔS is negative.

D. mixing 5 mL ethanol with 25 mL water - Mixing increases disorder, so ΔS is positive.

E. evaporation of 1 mol of CCl4(l) - Changing from liquid to gas increases disorder, so ΔS is positive.

Therefore, the process with a negative ΔS is C. compressing 1 mol Ne at constant temperature from 1.5 L to 0.5 L.

Answer:

a. The central atom is sulfur

b. SF2

c. The central atom has two lone pairs

d. The ideal angle between the sulfur-fluorine bonds is 109.5°

e.  I expect the actual angle between the sulfur-fluorine bonds to be less than 109.5° because unbonded pairs repel bonded pairs more than bonded pairs repel other bonded pairs. So the bonds here will be pushed closer than normal

Explanation:

Question #1:  What is the central atom?

The central atom of this molecule is Sulfur, S.

Question #2:  Enter its chemical symbol.

The chemical symbol of the molecule is SF2 but the chemical symbol of the central atom is S.

Question #3:  How many lone pairs are around the central atom?

There are two lone pairs around the central atom of Sulfur.

Question #4:  What is the ideal angle between the sulfur-fluorine bonds?

The ideal angle between the Sulfur-Fluorine bonds is 109.5 degrees.

Question #5:  Compared to the ideal angle, you would expect the actual angle between the sulfur-fluorine bonds to be.

I would expect the actual angle between the Sulfur-Fluorine bonds to be less than 109.5 degrees since the unbonded pairs have a greater repulsion with bonded pairs than the repulsion that happens between two bonded pairs.  Therefore, the bonds would be closer to each other causing a smaller angle.

Answer:

B. 0.5M NaNO₂ and 0.5M HNO₂ .

C. 0.1M KC₆H₅COO and 0.05M C₆H₅COOH .

E. 0.05M NaOH and 0.1M HCHO₂.

Explanation:

A buffer is made of 2 components:

Select the following solutions that would form a buffer. Select all that apply.

A. 0.1M HNO₃ and 0.15M NH₄Cl. NO. HNO₃ is a strong acid

B. 0.5M NaNO₂ and 0.5M HNO₂. YES. HNO₂ is a weak acid and NO₂⁻ (coming from NaNO₂) its conjugate base.

C. 0.1M KC₆H₅COO and 0.05M C₆H₅COOH. YES. C₆H₅COOH  is a weak acid and C₆H₅COO⁻ (coming from KC₆H₅COO) its conjugate base.

D. 0.1M NH₃ and 0.1M KClO. NO. It does not have the components of a buffer system.

E. 0.05M NaOH and 0.1M HCHO₂. YES.  HCHO₂ is a weak acid that can react with NaOH to produce CHO₂⁻, its conjugate base.

F. 0.1M KF and 0.15M HCl. NO. HCl is a strong acid.

G. 0.1M HBr and 0.1M NaOH. NO. HBr is a strong acid.

The options that would form a buffer are B, C, and E.

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. Using this information, we can analyze the given options:

Based on this analysis, the options that would form a buffer are B, C, and E.

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Answer:

1.48x10⁻³ mol

Explanation:

The balanced equation is

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

The aqueous solution will disociate, so:

Mg(s) + 2H⁺(aq) + 2Cl⁻(aq) → Mg²⁺(aq) + 2Cl⁻(aq) + H₂(g)

Simplifying by eliminating the bolded substances that have the same amount in both sides, we have the net ionic equation:

Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

a) The number of moles of Magnesium is the mass divided by the molar mass (24.305 g/mol):

n = 0.0360/24.305

n = 1.48x10⁻³ mol

Answer:

11 A

Explanation:

The magnetic field inside a solenoid can be calculated by the equation:

B = μ*(N/L)*i

Where B is the magnetic field, μ is the magnetic permeability (which is 1.256x10⁻⁶ T/m.A at vacuum), N is the number os loops, L is the length of the solenoid (2 cm = 0.02 m), and i the current.

4 = 1.256x10⁻⁶ *(5653/0.02)*i

0.355i = 4

i ≅ 11 A

Answer:

2-methoxy-2-methylpropane

Explanation:

The first step for this reaction is the carbocation formation. In this step, a tertiary carbocation is formed. Also, we will have a good leaving group so bromide will be formed. Then the methanol acts as a nucleophile and attacks the carbocation. Next, a positive charge is generated upon the oxygen, this charge can be removed when the hydrogen leaves the molecule as [tex]H^+[/tex]. (See figure)

Predators grow along with their victims. Over time, prey animals are now developing and avoiding themselves to get eaten by their predators. Such tactics and modifications can take many forms that make their work easier, including disguise, mimicry, defense mechanisms, flexibility, distance, habits and even tool use.

Though fact an equilibrium appears to occur within an ecosystem between predators and prey, there are several factors which affect it, including the birth and death rates of predators and preys.

Answer:

a. 0.80V b. 2Ag⁺(aq) + H2(g) ⇄ 2Ag(s) +2H⁺(aq) c. i) 0.88 ii) 1.03 d. Cell is a ph meter with the potential being a function of hydrogen ion concentration

Explanation:

a. The two half cell reactions are

1. 2H⁺(aq) +2e⁻ → H₂(g) Eanode = 0.00V

2. Ag⁺(aq) + e⁻ → Ag(s) Ecathode = 0.80V

The balanced cell reaction is

2Ag(aq)⁺ + H₂(g) ⇄  2Ag(s) + 2H⁺(aq)

therefore Ecell = Ecathode - Eanode = 0.80 - 0.00 = +0.80V

b. Since the Ecell is positive, the spontaneous cell reaction under standar state conditions is

2Ag(aq)⁺ + H₂(g) ⇄  2Ag(s) + 2H⁺(aq)

c. Use Nernst Equation

E = Ecell - (0.0592/n)log([H⁺]/[Ag⁺]²[P H₂]), where n is the number of moles of Ag and P H₂= 1.0 atm

i) E = 0.80 - (0.0592/2)log(4.2x10^-2)/(1.0)²(1.0) = 0.88V

ii) E =  0.80 - (0.0592/2)log(9.6x10^-5)/(1.0)²(1.0) = 1.03V

d . From the above calculation we can conclude that the cell acts as a pH meter as a change in hydrogen ion concentration results in a change in the potential of the cell. A change of ph of 2.64 changes the E of cell by 0.15 V.

The correct order of decreasing acid strength in an aqueous solution among HCl, HOCl, HOBr, and HOI is HCl > HIO > HBrO > HClO. This order is based on the strength of the H-X bond and the stability of the X- ion, where HCl is the strongest acid.

The correct order of decreasing acid strength in aqueous solution: HCl, HOCl, HOBr, and HOI, is option C: HCl > HIO > HBrO > HClO. This order is determined by two main factors: the strength of the H-X bond and the stability of the X-ion. Hydrochloric acid (HCl) is the strongest because it can easily donate protons in solution. The other compounds are oxyacids, and their strength increases with increasing electronegativity and size of the halogen attached to oxygen. So, when comparing HOCl, HOBr, and HOI, the more polarizable (or larger in size) the halogen, the stronger the acid. Iodine is the largest here and thus HOI is more acidic than HOBr and HOCl.

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Answer:  2.00 V

Explanation:

The balanced redox reaction is:

[tex]2Al+3Cu^{2+}\rightarrow 2Al^{3+}+3Cu[/tex]

Here Al undergoes oxidation by loss of electrons, thus act as anode. Copper undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]Al^{3+}+3e^-\rightarrow Al[/tex] [tex]E^0_{[Al^{3+}/Al]}=-1.66V[/tex]

[tex]Cu^{2+}+2e^-\rightarrow Cu[/tex] [tex]E^0_{[Cu^{2+}/Cu]}=0.340V[/tex]

[tex]E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Al^{3+}/Al]}[/tex]

[tex]E^0=+0.34- (-1.66V)=2.00V[/tex]

Thus the standard cell potential is 2.00 V

Answer:

24.895 kJ/mol

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

[tex]\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c [/tex]

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314 J /mol K)

c is the constant.

For two situations and phases, the equation becomes:

[tex]\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)[/tex]

Given:

[tex]P_1[/tex] = 271.2 mmHg

[tex]P_2[/tex] = 641.8 mmHg

[tex]T_1[/tex] = 241.3 K  

[tex]T_2[/tex] = 259.3 K

So,  

[tex]\ln \:\left(\:\frac{271.2}{641.8}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{259.3}-\:\frac{1}{241.3}\:\right)[/tex]

[tex]\Delta \:H_{vap}=\ln \left(\frac{271.2}{641.8}\right)\frac{8.314}{\left(\frac{1}{259.3}-\:\frac{1}{241.3}\right)}\ J/mol[/tex]

[tex]\Delta \:H_{vap}=\left(-\frac{520199.41426}{18}\right)\left(\ln \left(271.2\right)-\ln \left(641.8\right)\right)\ J/mol[/tex]

ΔHvap = 24895.015 J/mol = 24.895 kJ/mol ( 1 J = 0.001 kJ )

Final answer:

This detailed answer explains how to calculate the enthalpy of vaporization for liquid SO2 using the Clausius-Clapeyron equation and vapor pressure data at different temperatures.

Explanation:

The enthalpy of vaporization (ΔHvap) can be calculated using the Clausius-Clapeyron equation:

In this case, the enthalpy of vaporization for SO2 can be calculated using the given vapor pressure data at different temperatures:

Convert temperatures to Kelvin.

Use the Clausius-Clapeyron equation to find ΔHvap.

By applying the equation with the provided data, the calculated enthalpy of vaporization for liquid SO2 is 24.895 kJ/mol.

Answer:

Fe(s) + Sn²⁺(aq) ⇒ Sn(s) + Fe²⁺(aq)

Explanation:

When solid Fe metal is put into an aqueous solution of Sn(NO₃)₂, solid Sn metal and a solution of Fe(NO₃)₂ result. The resulting molecular equation is:

Fe(s) + Sn(NO₃)₂(aq) ⇒ Sn(s) + Fe(NO₃)₂

The full ionic equation includes all the ions and the species that do not dissociate in water.

Fe(s) + Sn²⁺(aq) + 2 NO₃⁻(aq) ⇒ Sn(s) + Fe²⁺(aq) + 2 NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and species that do not dissociate in water.

Fe(s) + Sn²⁺(aq) ⇒ Sn(s) + Fe²⁺(aq)

Final answer:

The net ionic equation for the reaction of solid Fe metal with an aqueous solution of Sn(NO3)2 is Fe(s) + Sn2+(aq)
ightarrow Fe2+(aq) + Sn(s), a single displacement redox reaction.

Explanation:

When solid Fe metal is introduced to an aqueous solution of Sn(NO3)2, a single displacement reaction occurs where Fe displaces Sn from the stannous nitrate. The net ionic equation for this chemical reaction, considering that nitrates are soluble and metallic tin will precipitate out of the solution as a solid, can be written as follows:

Fe(s) + Sn2+(aq)
ightarrow Fe2+(aq) + Sn(s)

In this equation, solid iron (Fe) reacts with stannous ions (Sn2+) in solution to form ferrous ions (Fe2+) and solid tin (Sn). This type of reaction is also referred to as a redox reaction.

Answer:

0.477 V

Explanation:

When a substance is gaining electrons, it's reducing, and when the substance loses electrons, it's oxidizing. In a galvanic cell, one substance oxides giving electrons for the other, which reduces. Then, the substance with higher reduction potential must reduce and the other must oxide.

E°cell = E°red(red) - E°red(oxid)

Where, E°red(red) is the reduction potential of the substance that reduces, and E°red(oxid) is the reduction potential of the substance that oxides. For the value given, Cu⁺² reduces, so:

E°cell = +0.337 - (-0.140)

E°cell = 0.477 V

The standard cell potential for this galvanic cell is [tex]0.477 \ V[/tex].

To calculate the standard cell potential [tex](\( E^\circ_\text{cell} \))[/tex] for the galvanic cell using the given reduction potentials, we use the formula:

[tex]\[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \][/tex]

Given reduction potentials:

[tex]\( E^\circ_\text{red}(\text{Sn}^{2+} \rightarrow \text{Sn}) = -0.140 \, \text{V} \)[/tex]

[tex]\( E^\circ_\text{red}(\text{Cu}^{2+} \rightarrow \text{Cu}) = +0.337 \, \text{V} \)[/tex]

Since the reduction potential for [tex]\( \text{Cu}^{2+} \)[/tex] is more positive, it acts as the cathode:

[tex]\[ E^\circ_\text{cathode} = +0.337 \, \text{V} \][/tex]

And the reduction potential for [tex]\( \text{Sn}^{2+} \)[/tex] is less positive (more negative), so it acts as the anode:

[tex]\[ E^\circ_\text{anode} = -0.140 \, \text{V} \][/tex]

Now, calculate the standard cell potential:

[tex]\[ E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} \][/tex]

[tex]\[ E^\circ_\text{cell} = (+0.337 \, \text{V}) - (-0.140 \, \text{V}) \][/tex]

[tex]\[ E^\circ_\text{cell} = +0.337 \, \text{V} + 0.140 \, \text{V} \][/tex]

[tex]\[ E^\circ_\text{cell} = +0.477 \, \text{V} \][/tex]

Answer:

Kf = 1.11x10¹³

Explanation:

The value of Kf for a multistep process that involves an equilibrium at each step, is the multiplication of the constant of the equilibrium of each step.

Kf = K1xK2xK3xK4

Kf = 1.90x10⁴ x 3.90x10³ x 1.00x10³ x 1.50x10²

Kf = 1.11x10¹³

Final answer:

The overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ is 2.1 x 10^13.

Explanation:

The overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ can be calculated by multiplying the stepwise formation constants (K1, K2, K3, K4). In this case, the value of Kf is calculated as follows:

Kf = K1 * K2 * K3 * K4 = 1.90×10^4 * 3.90×10^3 * 1.00×10^3 * 1.50×10^2 = 2.1 x 10^13

Therefore, the overall formation constant (Kf) for the formation of [Cu(NH3)4]2+ from [Cu(H2O)4]2+ is 2.1 x 10^13.

Answer:

The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)

Explanation:

Step 1: Data given

2.0 M Al(NO3)3

Step 2:

Al(NO3)3 in water will dissociate as following:

Al(NO3)3 → Al^3+ + 3NO3^-

For 1 mol of Al(NO3)3 we will have 1 mol of Al^3+ and 3 moles of NO3^-

We know that the molarity of Al(NO3)3 = 2.0 M, this means 2.0 mol/ L

The mol ratio Al(NO3)3 and Al^3+ is 1:1 so the molarity of Al^3+ is 2.0 M

The mol ratio Al(NO3)3 and NO3^- is 1:3 so the molarity of NO3^- is 6.0M

The correct answer is: 2M Al3+(aq) and 6 M NO3-(aq)

Answer:

d

Explanation:

The complexes that involve metal having d10 electrons arrangement are usually colorless because:

A). There is no d electron that can be promoted via the absorption of visible light.

Thus, option A is the correct answer.

Learn more about 'Metals' here:

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The water molecule has an H-O-H bond angle of 104.5° which is due to the central oxygen atom having 2 shared pairs and 2 lone pairs of electrons, resulting in a bent molecular geometry due to the repulsion of lone pairs and sp³ hybridization.

The bond angle in a water molecule is closer to 104.5° rather than the stated 105°, and the distribution of electrons around the central oxygen atom that best explains this bond angle is 2 shared pairs, 2 lone pairs.

In the water molecule, the central oxygen atom is sp³ hybridized, with its four hybrid orbitals occupied by two lone pairs of electrons and two bonding pairs that form covalent bonds with hydrogen atoms.

Lone pairs require more space than bonding pairs, which leads to a repulsion that pushes the hydrogen atoms closer together, resulting in the H-O-H bond angle being slightly less than the ideal 109.5° angle of a tetrahedron.