g when the following skeletal equation is balanced under acidic conditions, what are the coefficients of the species shown? ni2 zn ni zn2 water appears in the balanced equation as a fill in the blank 5 (reactant, product, neither) with a coefficient of . (enter 0 for neither.) which species is the reducing agent?

They  perform a wide range of functions in the body such as catalyzing reactions, transporting molecules, and providing structural support to cells and tissues.

When answering questions on the platform Brainly, the following guidelines should be followed:- Always be factually accurate, professional, and friendly- Be concise and do not provide extraneous amounts of detail-

Ignore any typos or irrelevant parts of the question- Use the relevant terms in your answer as provided in the student question.The molecule that has a central carbon with an attached hydrogen atom, carboxyl group, and amino group is an amino acid.

Amino acids are the building blocks of proteins. They contain an amino group (-NH2), a carboxyl group (-COOH), and a side chain (-R) attached to the central carbon atom. Some amino acids have additional functional groups such as hydroxyl (-OH), sulfur (-SH), or a ring structure.

Amino acids are the building blocks of proteins. Amino acids can form peptide bonds with other amino acids to form proteins. Proteins are the basic functional and structural units of life.

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We learned that k is adsorbed to negatively charged soil colloids by electrostatic attraction to three types of exchange sites or binding positions. The cation exchange sites are the most readily available to the soil solution.

The binding position that is readily available to the soil solution is the cation exchange sites. This is because the negatively charged soil colloids attract positively charged ions, known as cations, through electrostatic attraction. These cations are then exchanged with other cations in the soil solution, leading to the term "cation exchange sites."

The other two types of exchange sites are the anion exchange sites, which attract negatively charged ions, known as anions, and the inner sphere binding sites, which involve a direct bond between the metal ion and the soil colloid surface. However, the cation exchange sites are the most readily available to the soil solution.

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a) The reduction reaction is the reaction that occurs at the zinc electrode, which is Zn²⁺(aq) + 2e⁻ → Zn(s).

b) The oxidation reaction is the reaction that occurs at the tin electrode, which is Sn(s) → Sn²⁺(aq) + 2e⁻.

c) The overall reaction is: Sn(s) + Zn²⁺(aq) → Sn²⁺(aq) + Zn(s).

d) The overall voltage for the cell is the difference between these two potentials is -0.62 V.

e) Zn electrode | Zn(NO₃)₂ || SnSO₄ | Sn electrode

f) As the reaction proceeds, the mass of the tin electrode will decrease as tin atoms are oxidized to form Sn²⁺ ions.

To calculate the voltage for this cell, we need to use the standard reduction potentials for the half reactions. The standard reduction potential for the reduction reaction is -0.76 V, and the standard reduction potential for the oxidation reaction is -0.14 V. The overall voltage for the cell is the difference between these two potentials: Ecell = Ered - Eox = (-0.76 V) - (-0.14 V) = -0.62 V.

In a galvanic cell, the reduction reaction occurs at the cathode, while the oxidation reaction occurs at the anode. In this case, the zinc electrode is the cathode, where Zn²⁺ ions are reduced to form zinc atoms. The tin electrode is the anode, where tin atoms are oxidized to form Sn²⁺ ions.

The overall reaction occurs spontaneously, as the standard reduction potential for the reduction reaction is more positive than the standard reduction potential for the oxidation reaction. This means that the electrons will flow from the anode to the cathode, generating an electrical current.

The salt bridge is used to maintain charge neutrality in the two half-cells, as the electrons flow from the anode to the cathode, and ions must move to balance the charges. Na⁺ ions move towards the anode, while NO³⁻ ions move towards the cathode. This allows for the flow of ions to maintain charge neutrality in both half-cells.

As the reaction proceeds, the mass of the tin electrode will decrease as tin atoms are oxidized to form Sn²⁺ ions. Conversely, the mass of the zinc electrode will increase as zinc ions are reduced to form zinc atoms. This is due to the conservation of mass, as the total mass of the system remains constant.

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[tex]\sf \underline{CuCl_2 +\pink{2NaNO_3} \longrightarrow \pink{ Cu(NO3)_2}+2NaCl}[/tex]

According to the equation, 1 mole of [tex]\sf CuCl_2 [/tex] reacts with 2 moles of [tex]\sf NaNO_3[/tex] to produce 1 mole of [tex]\sf Cu(NO_3)_2[/tex] and 2 moles of [tex]\sf NaCl [/tex]

Molar mass of [tex]\sf Cu(NO_3)_2[/tex] -

[tex] \:\:\:\:\:\:\longrightarrow \sf 63.5 + 2\times 14 + 16 \times 6 \\[/tex]

[tex] \:\:\:\:\:\:\longrightarrow \sf 187.5 \\[/tex]

The concentration of reactants and products in a system is determined by the equilibrium constant.

When the statement "carbon monoxide binds to hemoglobin 140 times more strongly than oxygen does" is taken into account, what does this tell you about the equilibrium constants for the two reactions of hemoglobin with carbon monoxide and oxygen?

The answer is: the equilibrium constant for the binding of CO is greater.Carbon monoxide (CO) is a highly toxic gas that is often found in confined spaces such as garages, buildings, and mines.

The gas is odourless, colourless, and tasteless, making it difficult to detect without special instruments.When carbon monoxide binds to hemoglobin, it binds to the same sites as oxygen, but it does so about 200 times more tightly.

As a result, a small amount of CO binding to hemoglobin can have a significant impact on the oxygen-carrying capacity of the blood. The equilibrium constant for CO is greater than that for oxygen due to the fact that CO has a stronger affinity for hemoglobin than oxygen does.

Hemoglobin binds to oxygen with high affinity, and as a result, oxygen binds to hemoglobin more tightly than carbon monoxide does.

The formation of an equilibrium is a common occurrence when a reaction is reversible. An equilibrium constant, in simple terms, is a measure of the extent to which a reaction favours the formation of products over reactants.

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The base dissociation (Kb) of the fluoride anion is 1.32 × 10⁻¹¹. The value indicates the strength of fluoride ions as a base. The lower the value of Kb, the weaker the base. The Kb of fluoride ions is quite small, indicating that it is a weak base.

To find the Kb of the fluoride anion, we need to use the relationship between the acid dissociation constant, Ka, and the base dissociation constant, Kb, for the conjugate acid-base pair.

The equation for this relationship is:

Ka x Kb = Kw

where Kw is the ion product constant for water, which is equal to 1.0 x 10⁻¹⁴ at 25°C.

In this case, the acid is HF and its conjugate base is F⁻ (fluoride). The student has determined that the Ka of HF is 7.6 x 10⁻⁴.

Therefore, we can use the above equation to solve for the Kb of F⁻:

Ka x Kb = Kw

Kb = Kw / Ka

Kb = 1.0 x 10⁻¹⁴ / 7.6 x 10⁻⁴

Kb = 1.32 x 10⁻¹¹

Therefore, the Kb of the fluoride anion is 1.32 x 10⁻¹¹.

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The pair of aqueous solutions that will result in a precipitation reaction is (i) potassium carbonate and barium hydroxide. When these two solutions are mixed, they will react to form solid barium carbonate and aqueous potassium hydroxide. The balanced chemical equation for this reaction is:

K2CO3(aq) + Ba(OH)2(aq) → BaCO3(s) + 2KOH(aq)

The other two pairs of solutions, (ii) aluminum nitrate and sodium phosphate, and (iii) ammonium bromide and potassium hydroxide, will not result in a precipitation reaction. When these two solutions are mixed, they will form aqueous solutions of the resulting products.

It is important to remember that the solubility rules can be used to predict whether a precipitation reaction will occur when two aqueous solutions are mixed. If one of the products formed in the reaction is insoluble in water, then a solid precipitate will form.

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At the atmospheric pressure on the summit of Mt. Everest (0.333 atm), water boils at a temperature of approximately 2710.39 °C.

In this specific question, we are being asked to calculate the temperature (in °C) at which water boils on the summit of Mt. Everest, given that the atmospheric pressure there is 0.333 atmospheres and ∆Hvap for water is 40.7 kJ/mol.

We can use the Clausius-Clapeyron equation to solve for the boiling point of water at this pressure. The equation is given by:ln(P2/P1) = (∆Hvap/R) * (1/T1 - 1/T2)

where:P1 = 1 atm (standard pressure)

P2 = 0.333 atm (pressure on the summit of Mt. Everest)

∆Hvap = 40.7 kJ/molR = 8.31 J/mol*K (universal gas constant)

T1 = 373 K (boiling point of water at standard pressure)

T2 = ? (boiling point of water at 0.333 atm pressure)

Solving for T2, we get:T2 = T1 * {∆Hvap/R * ln(P2/P1) + 1}T2 = 373 K * {40.7 kJ/mol / (8.31 J/mol*K) *

ln(0.333 atm / 1 atm) + 1}T2 = 373 K * {7.98}T2 = 2983.54 K

We can convert the boiling point of water at 0.333 atm pressure from Kelvin to Celsius by subtracting 273.15 from the result:T2 (in °C) = 2983.54 K - 273.15K = 2710.39 °C

Therefore, at the atmospheric pressure on the summit of Mt. Everest (0.333 atm), water boils at a temperature of approximately 2710.39 °C.

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To calculate the following for test tube 1 and test tube 2:

1. The number of moles of copper(II) sulfate used:

Test tube 1: 0.2 g of copper(II) sulfate was used, which is equivalent to 0.001255 moles (0.2 g / 159.60 g/mol).
Test tube 2: 0.4 g of copper(II) sulfate was used, which is equivalent to 0.002510 moles (0.4 g / 159.60 g/mol).

2. The heat absorbed by the water, in joules:

Test tube 1: Q = (10.0 g) x (4.186 J/g°C) x (20.0°C) = 837.2 J

Test tube 2: Q = (10.0 g) x (4.186 J/g°C) x (30.0°C) = 1257.9 J

3. The change in internal energy of the copper(II) sulfate:
Since the energy released by the copper(II) sulfate is absorbed by the water, the change in internal energy of the copper(II) sulfate is equal to the negative of the heat absorbed by the water.
Test tube 1: ΔU = -837.2 J
Test tube 2: ΔU = -1257.9 J
4. The reaction enthalpy, in joules/mole:
The reaction enthalpy can be calculated using the formula ΔH = ΔU + PΔV, where PΔV represents the work done by the system. Assuming that the reaction was carried out at constant pressure (i.e., atmospheric pressure), PΔV can be approximated to zero, and thus the reaction enthalpy is equal to the change in internal energy.
Test tube 1: ΔH = -837.2 J / 0.001255 mol = -666,876 J/mol
Test tube 2: ΔH = -1257.9 J / 0.002510 mol = -500,357 J/mol
Therefore, the results can be recorded in the following table:

|           | Moles of CuSO4 used | Heat absorbed by water (J) | Change in internal energy (J) | Reaction enthalpy (J/mol) |
|-----------|---------------------|-----------------------------|---------------------------------|---------------------------|
| Test tube 1 | 0.001255            | 837.2                       | -837.2                          | -666,876                 |
| Test tube 2 | 0.002510            | 1257.9                      | -1257.9                         | -500,357                 |

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The mass of scheelite that contains a million oxygen atoms is 0.478 femtograms.

Scheelite is the calcium tungstate mineral, with the chemical formula CaWO₄. To calculate the mass of scheelite that contains a million oxygen atoms, we need to use the Avogadro's number to convert the number of atoms to the number of moles, and then use the molar mass of CaWO₄ to calculate the mass.

The molar mass of CaWO₄ can be calculated as follows;

Molar mass of CaWO₄ = (molar mass of Ca) + (molar mass of W) + 4 x (molar mass of O)

Molar mass of Ca = 40.08 g/mol

Molar mass of W = 183.84 g/mol

Molar mass of O = 16.00 g/mol

Molar mass of CaWO₄ = 40.08 g/mol + 183.84 g/mol + 4 x 16.00 g/mol

Molar mass of CaWO₄ = 287.94 g/mol

Now, we can use Avogadro's number to convert the number of oxygen atoms to moles;

1 mole of oxygen atoms = 6.022 x 10²³ oxygen atoms

1 million oxygen atoms = 1 x 10⁶ / 6.022 x 10²³ moles of oxygen atoms

1 million oxygen atoms = 1.661 x 10⁻¹⁸ moles of oxygen atoms

Since there is 1 oxygen atom in 1 molecule of CaWO₄, the number of moles of CaWO₄ is also 1.661 x 10⁻¹⁸ moles.

Finally, we can calculate the mass of CaWO₄ using its molar mass.

Mass of CaWO₄ = number of moles x molar mass

Mass of CaWO₄ = 1.661 x 10⁻¹⁸ moles x 287.94 g/mol

Mass of CaWO₄ = 4.78 x 10⁻¹⁶ g or 0.478 femtograms (fg)

Therefore, the mass of scheelite contains a million oxygen atoms is 0.478 femtograms.

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0.056 mol of H₂ gas occupies a volume of 1.26 L at 25°C and 1.11 atm pressure. To solve this problem we make use of the expression of ideal gas law equation.

The ideal gas law is an equation that relates the pressure, volume, temperature, and number of moles of an ideal gas. It is expressed mathematically as:

PV = nRT

where P = pressure,

V = volume,

n = the number of moles,

R = 0.0821 L·atm/mol·K, and

T = the temperature in Kelvin.

First, we need to convert the temperature of 25°C to Kelvin:

T = 25°C + 273.15 = 298.15 K

From ideal gas law:

(1.11 atm) V = (0.056 mol) (0.0821 L·atm/mol·K) (298.15 K)

Simplifying the equation, we get:

V = (0.056 mol) (0.0821 L·atm/mol·K) (298.15 K) / (1.11 atm)

V = 1.26 L

Therefore, 0.056 mol of H₂ gas occupies a volume of 1.26 L at 25°C and 1.11 atm pressure.

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Answer:

When carbon is heated in air, it reacts with oxygen to form carbon dioxide. As the carbon reacts with oxygen to form carbon dioxide, the mass of the carbon decreases while the mass of the oxygen and carbon dioxide increases.

Answer:

When carbon is heated in air, it undergoes a process known as combustion or burning. During this process, carbon reacts with oxygen present in the air, resulting in the production of carbon dioxide gas. This reaction causes the carbon atoms to be lost in the form of carbon dioxide molecules. Hence, the original mass of carbon decreases due to the formation of lighter carbon dioxide molecules that are released into the atmosphere. Overall, the burning of carbon in air results in a reduction in its mass.

The volume of the concentrated 1.5 M solution required to prepare 25 mL of a 7.0 M solution is 0.117 L or 117 mL.

To solve the problem, we use the formula M1V1 = M2V2, which relates the initial concentration and volume of the concentrated solution (M1V1) to the final concentration and volume of the diluted solution (M2V2). In this case, we are given the final concentration (M2 = 7.0 M), the final volume (V2 = 25 mL), and the initial concentration (M1 = 1.5 M), so we can solve for the initial volume (V1) of the concentrated solution that we need to use.

We rearrange the formula to solve for V1, which gives us V1 = (M2 x V2) / M1. We substitute the given values into this equation, and we get V1 = (7.0 M x 0.025 L) / 1.5 M = 0.117 L or 117 mL.

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The pH of the solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH is 4.74.

To calculate the pH of the solution, we need to first determine the concentration of the remaining species in solution after the neutralization reaction between acetic acid and KOH is complete.

The balanced chemical equation for the neutralization reaction is:

CH3COOH + KOH → CH3COOK + H2O

The concentration of the potassium acetate can be calculated from the stoichiometry of the reaction:

moles of potassium acetate = moles of acetic acid = moles of KOH

moles of acetic acid = 0.10 mol/L × 0.0100 L = 0.0010

mol

moles of KOH = 0.10 mol/L × 0.0100 L = 0.0010 mol

moles of potassium acetate = 0.0010 mol

The volume of the final solution is 20.00 mL, so the concentration of the potassium acetate is:

[CH3COOK] = moles of potassium acetate / volume of solution

= 0.0010 mol / 0.0200 L

= 0.050 mol/L

The dissociation of potassium acetate can be written as:

CH3COOK ⇌ CH3COO- + K+

The equilibrium constant for this reaction is given by the expression:

Ka = [CH3COO-][H+]/[CH3COOH

At equilibrium, the concentration of CH3COOH is zero, so we can simplify this expression to:

Ka = [CH3COO-][H+]/[CH3COOK]

We know the value of Ka for acetic acid, which is 1.8 x

[tex] {10}^{ - 5} [/tex]

We can use this value to solve for the concentration of H+ in the solution:

1.8 x

[tex] {10}^{ - 5} [/tex]

= [H+][CH3COO-] / [CH3COOK]

To convert the concentration of H+ to pH, we use the expression:

pH = -log[H+]

= 4.74

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1-Moist air moves up as it encounters a mountain range: This condition can lead to the formation of clouds because as moist air rises, it cools, and the water vapor in the air condenses into liquid water droplets or ice crystals, which can form clouds.

2-High pressure is moving into an area: High-pressure systems are associated with clear and sunny weather, so this condition is less likely to lead to the formation of clouds.

3-The air over a large parking lot is warmer in the afternoon than the surrounding park is: This condition can lead to the formation of cumulus clouds as the warm air rises and cools, and water vapor condenses into visible clouds.

4-The horse latitudes are the areas of sinking air: This condition is associated with clear and dry weather, so it is less likely to lead to the formation of clouds.

5-Low pressure moves into your area: This condition can lead to the formation of clouds as the rising warm, moist air cools and condenses into clouds.

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Charles's Law-

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]

Where:-

As per question, we are given that -

We are given the initial temperature in °C.So, we first have to convert the temperature in Celsius to kelvin by adding 273-

[tex]\:\:\:\:\:\:\star\sf T_1[/tex] = 65+ 273 =338 K

Now that we have obtained all the required values, so we can put them into the formula and solve for T₂ :-

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\star\longrightarrow\sf \underline{\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}}\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{\dfrac{T_2}{V_2}=\dfrac{T_1}{V_1}}\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=\dfrac{T_1}{V_1} \times V_2}\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=\dfrac{338}{3.75} \times 6.52\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=90.13333...... \times 6.52\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=587.669.........\:K\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=(587.67 -273)°C\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf T_2=314.66933…....\:°C\\[/tex]

[tex]\:\:\:\:\:\: \:\:\:\:\:\:\longrightarrow\sf \underline{T_2=314.67\:°C}\\[/tex]

Therefore, the temperature will change from 65°C to 314.67°C, when the volume of a gas is changed from 3.75 L to 6.52 L.

The enthalpy is the principal driving force because it is a large enough negative value to ensure the free energy is negative, making the reaction spontaneous. Hence, option A is correct.

In order to determine the driving force for a reaction, we look at the Gibbs free energy change (ΔG). The Gibbs free energy is given by the equation:

ΔG = ΔH - TΔS

where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change. If ΔG is negative, the reaction is spontaneous, meaning that it will occur without any external input of energy. If ΔG is positive, the reaction is nonspontaneous and will only occur with an input of energy.

Assuming that ΔS° is also negative, option A is the best choice, The enthalpy is the principal driving force because it is a large enough negative value to ensure the free energy is negative, making the reaction spontaneous. This option assumes that ΔS° is negative, which would result in a negative ΔG value and a spontaneous reaction.

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--The complete question is, Consider the reaction CO2(g) + 2NH3(g) à CO(NH2)2(s) + H2O(l) ΔH° 298K = -134kJ b. Which factor, the change in enthalpy, ΔH°, or the change in entropy, ΔS° provides the principle driving force for the reaction at 298K? Explain.

A) The enthalpy is the principal driving force because it is a large enough negative value to ensure the free energy is negative, making the reaction spontaneous

B) The enthalpy is the principal driving force because it is a large enough negative value to ensure the free energy is positive, making the reaction nonspontaneous

C) The entropy is the principal driving force, because it is a large enough negative value to ensure the free energy is negative making the reaction spontaneous

D) The entropy is the principal driving force, because it is a large enough negative value to ensure the free energy is positive making the reaction nonspontaneous--

The answer is (a) MgO only, as all of the magnesium and oxygen react completely to form magnesium oxide, and there would be no unreacted magnesium or oxygen left in the reaction vessel.

What is Limiting Reagent?

The limiting reagent can be determined by comparing the mole ratios of the reactants and the coefficients in the balanced chemical equation for the reaction. The reactant that produces the least amount of product based on the stoichiometry of the balanced equation is the limiting reagent.

If 1g of magnesium (Mg) and 1g of oxygen (O2) reacted, they would combine to form magnesium oxide (MgO) according to the following chemical equation:

2Mg + O2 -> 2MgO

The molar mass of magnesium is 24.31 g/mol, while the molar mass of oxygen is 32.00 g/mol. Therefore, 1g of magnesium is equivalent to 0.041 mol, and 1g of oxygen is equivalent to 0.03125 mol.

According to the balanced equation, 2 moles of magnesium react with 1 mole of oxygen to form 2 moles of magnesium oxide. Thus, the limiting reactant in this reaction is oxygen because only 0.03125 mol of oxygen is available, whereas 0.082 mol of magnesium is available.

Using the limiting reactant, we can calculate the theoretical yield of magnesium oxide:

0.03125 mol O2 x (2 mol MgO / 1 mol O2) x (40.31 g MgO / 1 mol MgO) = 2.5 g MgO

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The position of the element in the periodic table influences the cation that it forms.

An ion is a particle that has either a positive or negative charge. Ions are formed by removing or adding electrons from/to an atom or molecule. They could be classified into two categories: cations and anions. Cations are positively charged ions that are formed when an atom loses one or more electrons.

Anions are negatively charged ions that are formed when an atom gains one or more electrons.The position of the element in the periodic table influences the cation that it forms. Because the number of valence electrons changes as we move through the periodic table from left to right or top to bottom, this is the case.

As a result, the chemical properties of the elements change as we move from left to right or top to bottom. This influences the type of cations that are formed because cations are formed by losing electrons. The most prevalent cations are those that result from the loss of one, two, or three electrons by an element. Cations such as H+, Na+, K+, and Ca2+ are formed by metals.

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The following are the predictions if whether each of the substances will conduct an electric current: a) Conducts b) Does not conduct c) Does not conduct hence the correct option is b.

a) Potassium hydroxide (KOH) is an ionic compound that dissolves in water to form potassium ions (K+) and hydroxide ions (OH-), which allows the solution to conduct an electric current.

b) Glucose (C6H12O6) is a covalent compound and does not dissociate into ions when dissolved in water, so the solution does not conduct an electric current.

c) Ethanol (C2H5OH) is also a covalent compound and does not dissociate into ions when dissolved in water, so the solution does not conduct an electric current.

The correct options are therefore a)conducts, b) does not conduct and c) does not conduct which corresponds to choice b.

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The consequence of having cis configuration in double bonds of unsaturated fatty acids is that it creates a kink or bend in the fatty acid chain, which affects the packing of molecules and the fluidity of the membrane.

The cis configuration of double bonds in unsaturated fatty acids results in a kink or bend in the fatty acid chain. This bend affects the overall structure and function of the fatty acid. It causes the molecules to pack less tightly together, making them more fluid and flexible at room temperature. This property is important for cell membrane function, as it allows the membrane to remain fluid and adaptable to changing environmental conditions.

Additionally, the cis configuration can affect the biological activity of unsaturated fatty acids, influencing their interactions with enzymes and other molecules in the body.

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There are approximately 4.92 x 10^21 nitrogen atoms in 1.20 grams of aspartame.

The two naturally occurring amino acids phenylalanine and aspartic acid, which are also parts of proteins in our bodies and food, are what makeup aspartame. Aspartame's sweet flavor comes from a small modification of the phenylalanine by the addition of a methyl group.

The molecular formula of aspartame is C14H18N2O5.

The molar mass of aspartame,

(14 x 12.01 g/mol) + (18 x 1.01 g/mol) + (2 x 14.01 g/mol) + (5 x 16.00 g/mol) = 294.30 g/mol

Aspartame's molecular weight in 1.20 grams can be computed as follows:

1.20 g / 294.30 g/mol = 0.00408 mol

Every aspartame molecule has two nitrogen atoms.

0.00408 mol x 2 = 0.00816 moles of nitrogen

So, we will use Avogadro's number in order to transform nitrogen moles to nitrogen atoms:

0.00816 mol x 6.022 x 10^23 atoms/mol = 4.92 x 10^21 atoms of nitrogen

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Answer:

Part 1:

To determine the total change in enthalpy (ΔH) from the diagram, we need to look at the difference between the potential energy of the products and the potential energy of the reactants. In this diagram, the potential energy of the products (point B) is higher than the potential energy of the reactants (point A), so ΔH is positive.

To determine the activation energy (Ea), we need to look at the difference between the potential energy of the reactants and the highest point on the curve, also known as the transition state (point C). In this diagram, the potential energy of the reactants (point A) is lower than the potential energy of the transition state (point C), so Ea is positive.

Part 2:

If the reaction was exothermic, the potential energy of the products would be lower than the potential energy of the reactants. This means that the curve would be inverted, with the potential energy decreasing as the reaction proceeds. The potential energy of the reactants would be higher than the potential energy of the products.

The enthalpy change (ΔH) for an exothermic reaction would be negative, as energy would be released during the reaction. The activation energy (Ea) would still be positive, as the reactants would still need to absorb energy to reach the transition state. However, the height of the curve would be lower for an exothermic reaction, indicating a lower activation energy.

Hope This Helps!

Answer:

a. diphenylmethanol

Explanation:

As benzaldehyde is used instead of benzophenone, the expected product is Diphenylmethanol.

Benzophenone is a molecule of the ketone type that is made up of two phenyl groups that are bonded to a carbonyl group. In polar solvents, it is a crystalline substance that is almost white, odorless, and has a faintly sweet taste. It is commonly used as a UV light stabilizer for various plastics and as a chemical intermediate in the manufacture of pharmaceuticals, fragrances, and dyes. Diphenylmethanol is an organic compound that belongs to the class of aromatic alcohol. It is a white crystalline substance that has a sweet, floral scent. It is produced by combining benzaldehyde with the Grignard reagent. The reaction will lead to the formation of Diphenylmethanol. Therefore, if benzaldehyde was used instead of benzophenone, the expected product would be Diphenylmethanol.

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The sodium ion concentration in the solution is 0.238 M, calculated by dividing the total moles of Na+ (0.690 mol) by the solution volume (2.90 L).

To work out the sodium particle fixation in an answer ready by blending 0.345 mol Na2SO4 in enough water to make 2.90 L of arrangement, we can involve stoichiometry and the equation for molarity.

In the first place, we really want to decide the all out number of moles of sodium particles in the arrangement. Every mole of Na2SO4 separates into 2 moles of sodium particles, so we can work out the quantity of moles of sodium particles as follows:

0.345 mol Na2SO4 x (2 mol Na+/1 mol Na2SO4) = 0.690 mol Na+

Then, we can work out the grouping of sodium particles in the arrangement by isolating the quantity of moles of sodium particles by the volume of the arrangement in liters:

0.690 mol Na+/2.90 L = 0.238 M Na+

Accordingly, the sodium particle fixation in the arrangement is 0.238 M. This intends that for each liter of arrangement, there are 0.238 moles of sodium particles present. This estimation can be valuable in different synthetic applications, including deciding the ionic strength of an answer, working out response rates, and planning compound responses.

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The concentration (M) of the hydroxide ion when 25ml of saturated  Ca(OH)₂ is titrated with 0.028 ml of HCl is  0.054 M.

The concentration of the hydroxide ions can be calculated using the following formula:

[OH⁻] = ([tex]V_{B}[/tex] × [tex]M_{B}[/tex])/ ([tex]V_{S}[/tex] × n)

where [tex]V_{B}[/tex] is the volume of HCl used, [tex]M_{B}[/tex] is the molarity of HCl, [tex]V_{S}[/tex] is the volume of Ca(OH)₂ solution used and n is the number of OH⁻ ions per molecule of Ca(OH)₂ which is 2.

Here, [.] denotes the concentration of an entitled ion or molecule.

The concentration of a chemical species, specifically a solute in a solution, is measured by its molarity. It is described as the quantity of solute in one liter of solution, expressed in moles. The letter M stands for molarity.

After substituting the values provided in the question, we get:

[OH⁻] = (38.1 ml × 0.028 M) / (25 ml × 2)

[OH⁻] = 0.054 M

Therefore, the concentration of hydroxide ion in the saturated Ca(OH)2 solution is 0.054 M.

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The net ionic equation for the dissolution of solid cobalt(III) phosphate, CoPO₄(s), in an acid solution can be written as; CoPO₄(s) + 3H₃O⁺(aq) ↔ Co₃⁺(aq) + H₂PO₄⁻(aq) + 3H₂O(l)

In this reaction, CoPO₄(s) reacts with hydronium ions, H₃O⁺(aq), to form cobalt(III) ions, Co₃⁺(aq), and hydrogen phosphate ions, H₂PO₄⁻(aq), along with water molecules, H₂O(l). The hydrogen phosphate ion, H₂PO₄⁻, is the product of the reaction and is formed by the reaction of H₃O⁺ with PO₄³⁻ in CoPO₄(s), which acts as a weak acid.

The net ionic equation only shows the species that are directly involved in the chemical reaction, so the spectator ions, such as counterions, are omitted. In this case, the counterion for CoPO₄ is not included in the net ionic equation since it does not participate in the reaction.

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