What will be the pH of 1.0 mol dm-3 of NH4OH, which is 1% dissociated

Answer:

[tex]224.5 m^3[/tex]

Explanation:

By using the first law of thermodynamics, we can find the work done by the gas:

[tex]\Delta U=Q-W[/tex]

where in this problem:

[tex]\Delta U=-69 kJ[/tex] is the change in internal energy of the gas

[tex]Q=+260 kJ[/tex] is the heat absorbed by the gas

W is the work done by the gas (positive if done by the gas, negative otherwise)

Therefore, solving for W,

[tex]W=Q-\Delta U=+260-(-69)=+329 kJ = +3.29\cdot 10^5 J[/tex]

So, the gas has done positive work: it means it is expanding.

Then we can rewrite the work done by the gas as

[tex]W=p(V_f-V_i)[/tex]

where:

[tex]p=7400 Pa[/tex] is the pressure of the gas

[tex]V_i=180 m^3[/tex] is the initial volume of the gas

[tex]V_f[/tex] is the final volume

And solving for Vf, we find

[tex]V_f=V_i+\frac{W}{p}=180+\frac{3.29\cdot 10^5}{7400}=224.5 m^3[/tex]

Final answer:

To determine the final volume of the gas, the first law of thermodynamics was used, considering the process as isobaric due to constant pressure. The final volume was calculated to be approximately 45045.9 m³ after applying the formula and accounting for the work done and heat supplied.

Explanation:

To find the final volume of the gas, we can use the first law of thermodynamics, which is given by ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. Since the question states the pressure remains constant (isobaric process), the work done by the gas can be found by W = pΔV, where p is the pressure and ΔV is the change in volume.

From the given data, we have:

Using ΔU = Q - W and substituting W with pΔV, we get:

-69 kJ = 260 kJ - (7400 Pa × ΔV)

Convert kJ to J by multiplying by 1000 (since 1 kJ = 1000 J):

-69000 J = 260000 J - (7400 Pa × ΔV)

We can then solve for ΔV:

ΔV = (260000 J + 69000 J) / 7400 Pa

ΔV = 44865.9 m³ (rounded to one decimal place)

To find the final volume (V₂), we add the change in volume to the initial volume:

V₂ = V₁ + ΔV

V₂ = 180 m³ + 44865.9 m³

V₂ = 45045.9 m³

So, the final volume of the gas after the process is approximately 45045.9 m³.

Answer:

See explanation below

Explanation:

The anode reaction is :

2I^-(aq) -------> I2(g) +2e

Cathode reaction

F2(g) + 2e------> 2F^-(aq)

In the external circuit, electrons migrate from the I-|I2 electrode (anode) to the F-|F2 electrode (cathode)

In the salt bridge, anions migrate from the F-|F2

A voltaic cell is a type of cell in which a spontaneous redox reaction generates an electric current. This current is generated by the migration of electrons from the anode to the cathode in the external circuit, and the migration of anions in the salt bridge maintains electrical neutrality.

A voltaic cell is a type of electrochemical cell where a spontaneous redox reaction generates an electric current. In this particular voltaic cell, the anode reaction is 2I-(aq) -> I2(s) + 2e-, where iodide ions are oxidized to solid iodine (losing electrons). The cathode reaction is F2(g) + 2e- -> 2F-(aq), where gaseous fluorine is reduced to fluoride ions (gaining electrons).

In the external circuit, electrons migrate from the anode (I-|I2 electrode) to the cathode (F-|F2 electrode). This migration of electrons generates the electric current. Lastly, in the salt bridge, anions migrate from the anode compartment (I-|I2) to the cathode compartment (F-|F2), allowing the cell to maintain electrical neutrality throughout the redox reaction.

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Rock candy is formed through a chemical process known as crystallization which requires a supersaturated sugar solution. In the scenario, batch A prepared with hot water is more likely to form rock candy as it can dissolve more sugar creating a supersaturated solution. Batch B prepared at room temperature may not form as much rock candy due to lesser sugar dissolution.

The question relates to the chemical process of crystallization, particularly in the formation of rock candy. When making rock candy, a supersaturated solution of sugar and water is required. This condition is achieved when as much sugar as possible is dissolved in hot water. Once cooled, the oversaturated solution starts to crystallize and forms rock candy. So in the given scenario, batch A is more likely to produce rock candy because it involves the preparation of a supersaturated solution through dissolving sugar in hot water. Batch B, prepared at room temperature, may not dissolve as much sugar as batch A, and thus, less or no rock candy might be formed.

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Batch A, which dissolved sugar in hot water that was then cooled, is more likely to form rock candy as the cooling process can lead to the crystallization of sugar from a supersaturated solution.

The process of making rock candy involves dissolving sugar in water to create a saturated solution from which sugar crystals can form upon cooling or evaporation of the solvent. The solubility of sugar increases with temperature, which means hot water can dissolve more sugar than room temperature water. Therefore, for batch A, dissolving sugar in hot water likely supersaturates the solution, and as it cools to room temperature, excess sugar will crystallize out.

Batch B, on the other hand, dissolves sugar at room temperature, potentially creating a saturated solution, but without the temperature change, it is less likely to form a supersaturated environment and thus may yield less crystallization compared to batch A. In summary, batch A is more likely to form rock candy due to the temperature-dependent solubility of sugar, and the cooling process allows crystals to form from the supersaturated solution.

Answer:

CHECK THE ATTACHMENT TO SEE THE STRUCTURAL DIAGRAM

Explanation:

The spectral data suggests the product is ethylbenzene, which can be synthesized from benzene using ethylchloride and a catalyst like AlCl3 via a Friedel-Crafts alkylation reaction.

The spectral data provided corresponds to the molecule ethylbenzene. The presence of a 13C NMR signal at 142 ppm indicates a carbon directly invested in a aromatic system. The signals at 38 ppm, 24 ppm and 13 ppm are indicative of the ethyl side chain. The IR peaks at 3100 cm-1 and 2900 cm-1 indicate C-H stretching of aromatic and aliphatic hydrogens respectively. The absence of an absorbance near 1700 cm-1 indicates absence of carbonyl group. The lack of a M+2 peak in MS data suggests no halogens are part of the structure. To obtain ethylbenzene from benzene, the reagents used would include ethylchloride and aluminum chloride (AlCl3) in a Friedel-Crafts alkylation reaction.Benzene, ethylchloride and AlCl3 are the key components to this reaction.

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Explanation:

Given that,

The wavelength of high‑frequency radio wave, [tex]\lambda=0.25\ km=250\ m[/tex]

We need to find the energy of exactly one photon of this radio wave radiation. It is given by :

[tex]E=\dfrac{nhc}{\lambda}[/tex]

Here, n = 1

[tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{250}\\\\E=7.95\times 10^{-28}\ J[/tex]

So, the energy of exactly one photon of this radio wave radiation is [tex]7.95\times 10^{-28}\ J[/tex].

The energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].

Given that,

Based on the above information, the calculation is as follows:

We know that

[tex]E = nhc \div \lambda\\\\= (6.63 \times 10^{-34} \times 3 \times 10^8) \div 250[/tex]

= [tex]7.95 \times 10^{-28}J[/tex]

Therefore we can conclude that the energy of exactly one photon of this radio wave radiation is [tex]7.95 \times 10^{-28}J[/tex].

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Answer:

Density= 1.7g/dm3

Explanation:

Applying

P×M= D×R×T

P= 2atm, Mm= 28, D=? R= 0.082, T= 400K

2×28= D×0.082×400

D= (2×28)/(0.082×400)

D= 1.7g/dm3

Answer:

0.1046M NaOH solution

Explanation:

a. KHP is a salt used as primary standard because allows direct standarization of bases solutions. The reaction of KHP with NaOH is:

KHP + NaOH → H₂O + KP⁻ + Na⁺

As you can see, KHP has 1 acid proton that reacts with NaOH.

Molar mass of KHP is 204.22g/mol; 0.5527g of KHP contains:

0.5527g KHP × (1mol / 204.22g) = 2.706x10⁻³moles of KHP. As 1 mole of KHP reacts per mole of NaOH, at equivalence point you must add 2.706x10⁻³moles of NaOH

As you spent 25.87mL of the solution, molarity of the solution is:

2.706x10⁻³moles of NaOH / 0.02587L = 0.1046M NaOH solution

Answer:

NaOH + KHC8H4O4 → NaKC8H4O4 + H2O

Concentration NaOH = 0.105 M

Explanation:

Step 1: Data given

Mass of KHP = 0.5527 grams

Molar mass KHP = 204.22 g/mol

Volume of NaOH = 25.87 mL

Step 2: The balanced equation

NaOH + KHC8H4O4 → NaKC8H4O4 + H2O

Step 3: Calculate moles KHP

Moles KHP = mass KHP / molar mass KHP

Moles KHP = 0.5527 grams / 204.22 g/mol

Moles KHP = 0.002706 moles

Step 4: Calculate moles NaOH

For 1 mol NaOH we need 1 mol KHP to react

For 0.002706 moles KHP we need 0.002706 moles NaOH

Step 5: Calculate concentration NaOH

Concentration = moles / volume

Concentration NaOH = 0.002706 moles / 0.02587 L

Concentration NaOH = 0.105 M

Answer:

1. Cu

2. Cu

3. 2 electrons.

Explanation:

Step 1:

The equation for the reaction is given below:

3Cu(s) + 8HNO3(aq) -> 2NO(g) 3Cu(NO3)2(aq) + 4H2O(l)

Step 2:

Determination of the change of oxidation number of each element present.

For Cu:

Cu = 0 (ground state)

Cu(NO3)2 = 0

Cu + 2( N + 3O) = 0

Cu + 2(5 + (3 x -2)) =0

Cu + 2 (5 - 6) = 0

Cu + 2(-1) = 0

Cu - 2 = 0

Cu = 2

The oxidation number of Cu changed from 0 to +2

For N:

HNO3 = 0

H + N + 3O = 0

1 + N + (3 x - 2) = 0

1 + N - 6 = 0

N = 6 - 1

N = 5

NO = 0

N - 2 = 0

N = 2

The oxidation number of N changed from +5 to +2

The oxidation number of oxygen and hydrogen remains the same.

Note:

1. The oxidation number of Hydrogen is always +1 except in hydride where it is - 1

2. The oxidation number of oxygen is always - 2 except in peroxide where it is - 1

Step 3:

Answers to the questions given above

From the above illustration,

1. Cu is oxidize because its oxidation number increased from 0 to +2 as it loses electron.

2. Cu is the reducing agent because it reduces the oxidation number of N from +5 to +2.

3. The reducing agent i.e Cu transferred 2 electrons to the oxidising agent HNO3 because its oxidation number increase from 0 to +2 as it loses its electrons. This means that Cu transfer 2 electrons.

In the given chemical reaction, the element oxidized and the reducing agent is Copper (Cu). The number of electrons transferred from the reducing agent to the oxidizing agent is six.

In the reaction equation, 3Cu(s) + 8HNO3(aq) --> 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l), the element oxidized is Cu (Copper), as Copper goes from an oxidation state of 0 in Cu(s) to +2 in Cu(NO3)2. Hence, the reducing agent is also Copper (Cu). Oxidation is the process of losing electrons, and in this chemical reaction, each copper atom loses 2 electrons (going from 0 to +2). Since the equation shows the reaction of 3 copper atoms, therefore, the total number of electrons transferred from reducing to oxidizing agent in the equation, as written, is 6.

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Answer : The line-bonds structure of oleic acid (cis‑9‑octadecenoic acid) is shown below.

Explanation :

In line-bonds formula, the organic structures of the compound are represented in such a way that covalent bonds are represented by line and frame the structure by zig-zag straight lines which emits all the hydrogen atom.

The terminals of the line and vertex represents carbon atoms whose valences are satisfied by formation of single bonds with H atom.

The given compound is, (cis‑9‑octadecenoic acid)

In this compound, the parent chain is 18 membered and the carboxylic acid functional group is attached to it.

Answer:

a) True

Explanation:

a) From the definition of the equilibrium. When a reversible reaction is carried out in a closed vessel, a stage is reached when the forward and the backward reactions proceed with the same speed. This stage is known as chemical equilibrium.

Answer:

Methane is produced as a new product.

Explanation:

[tex]CH_{3}MgBr[/tex] acts as a base toward -COOH group in p-methylbenzoic acid.

Hence an acid-base reaction occurs between p-methylbenzoic acid and [tex]CH_{3}MgBr[/tex] to produce methane and p-methylbenzoate.

[tex]H_{3}O^{+}[/tex]addition will convert p-methylbenzoate back to p-methylbenzoic acid.

Hence, methane is produce as a new product.

Reaction sequences are given below.

Answer:

The equilibrium constant at 2000 K is 0.7139

The equilibrium constant at 3000 K is 8.306

ΔH = 122.2 kJ/mol

Explanation:

Step 1: Data given

the standard molar Gibbs energy of formation of X(g) is 5.61 kJ/mol at 2000 K

the standard molar Gibbs energy of formation of X(g) is -52.80 kJ/mol at 3000 K

Step 2: The equation

1/2X2(g)⟶X(g)

Step 3: Determine K at 2000 K

ΔG = -RT ln K

⇒R = 8.314 J/mol *K

⇒T = 2000 K

⇒K is the equilibrium constant

5610 J/mol = -8.314 J/molK * 2000 * ln K

ln K = -0.337

K = e^-0.337

K = 0.7139

The equilibrium constant at 2000 K is 0.7139

Step 4: Determine K at 3000 K

ΔG = -RT ln K

⇒R = 8.314 J/mol *K

⇒T = 3000 K

⇒K is the equilibrium constant

-52800 J/mol = -8.314 J/molK * 3000 * ln K

ln K = 2.117

K = e^2.117

K = 8.306

The equilibrium constant at 3000 K is 8.306

Step 5: Determine the value of ΔH∘rxn

ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)

ln 8.306 /0.713 = -ΔH/8.314 * (1/3000 - 1/2000)

2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)

2.455 = -ΔH/8.314 * (-1.67*10^-4)

-14700= -ΔH/8.314

-ΔH = -122200 J/mol

ΔH = 122.2 kJ/mol

When The constant at 2000 K is 0.7139

Then The constant at 3000 K is 8.306

ΔH = 122.2 kJ/mol

Step 1: Data is given

When the quality molar Gibbs effectiveness of the formation of X(g) is 5.61 kJ/mol at 2000 K

When the quality molar Gibbs effectiveness of the formation of X(g) is -52.80 kJ/mol at 3000 K

Step 2: The equation is:

[tex]1/2X2(g)⟶X(g)[/tex]

Step 3: Then Determine K at 2000 K

ΔG = -RT ln K

[tex]⇒R = 8.314 J/mol *K[/tex]

⇒[tex]T = 2000 K[/tex]

⇒ at that time K is that the constant

Then 5610 J/mol = [tex]-8.314 J/molK * 2000 * ln K[/tex]

ln K is = [tex]-0.337[/tex]

K is =[tex]e^-0.337[/tex]

K is = [tex]0.7139[/tex]

When The constant at 2000 K is 0.7139

Step 4: Then Determine K at 3000 K

ΔG = -RT ln K

⇒[tex]R = 8.314 J/mol *K[/tex]

⇒[tex]T = 3000 K[/tex]

⇒K is that the constant

[tex]-52800 J/mol = -8.314 J/mol K * 3000 * ln K[/tex]

ln [tex]K = 2.117[/tex]

K = e^2.117

K = 8.306

The constant at 3000 K is 8.306

Step 5: at the moment Determine the worth of ΔH∘rxn

ln K2/K1 = -ΔH/r * (1/T2 - 1/T1)

ln 8.306 /0.713 = -Δ[tex]H/8.314 * (1/3000 - 1/2000)[/tex]

2.455 = -ΔH/8.314 * (3.33*10^-4 - 0.0005)

2.455 = -ΔH/8.314 * [tex](-1.67*10^-4)[/tex]

-14700= -ΔH/8.314

-ΔH = [tex]-122200 J/mol[/tex]

Then ΔH = [tex]122.2 kJ/mol[/tex]

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Answer: The pressure that is needed is around 405kPa-755kPa

To summarize into 405PPM-755PPM

Explanation: In which 405 PPM-755 PPM which is about the same amount of pressure that water pressurises a car in water or a better example is that it's the same amount of pressure as if a penny was dropped from the sky towards a person holding a square piece of cardboard in which then the penny would directly go straight through the piece of cardboard.

Answer:

See the attached file for the structure

Explanation:

See the attached file for the explanation

Correct answer:

•Does pressure have an effect on the volume of a gas?

•Which brand of soap is the best for cleaning grease off dishes?

The questions that can be answered by science must be empirical, that is they must be answerable by experiments.

The question that can be answered by science is; Which brand of soap is the best for cleaning grease off dishes?

This question can be answered by using various brands of soap to clean the same type of dish with the same type of grease and comparing the results. The answer to this question is pure empirical.

The other questions listed can not be answered by experiment. Their answer may vary from person to person therefore they are not scientific questions.

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Answer:

The reduced coenzymes generated by the citric acid cycle donate electrons in a series of reactions called the electron transport chain. The energy from the electron transport chain is used for oxidative phosphorylation.

a)The compounds that donate electrons to the electron transport chain are NADH and . FADH2

b) O2 is the final electron acceptor.

c) The final products of the electron transport chain and oxidative phosphorylation are NAD+, H2O, ATP and FAD

Explanation:

Question:

The question is incomplete. See the attached file for the complete question and answer.

Explanation:

Find attached for explanation.

The first two pages is the additional question while the 3rd and last page is the answer .

Answer:

The standard entropy change for the reaction is -197.8 J/mol*K

Explanation:

Step 1: Data given

S°(CO(g)) = 197.7 J/(mol*K)

S°(CO2(g)) = 213.8 J/(mol*K)

S°(NO(g)) = 210.8 J/(mol*K)

S°(N2(g)) = 191.6 J/(mol·K)

Step 2: The balanced equation

2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g)

Step 3: Calculate ΔS°

ΔS° = ∑S°(products) - ∑S°(reactants)

ΔS° = (191.6 + 2*213.8) - (2*210.8+2*197.7)  J/mol*K

ΔS° = 619.2 J/mol*K - 817.0 J/mol *K

ΔS° = -197.8 J/mol* K

The standard entropy change for the reaction is -197.8 J/mol*K

Answer:

Explanation:

Since it first order, we use order rate equation

In ( [tex]\frac{A1}{A0}[/tex]) = -kt where A1 is the final quality = 0.8 (80%), A0 is the initial quality = 1 ( 100%)

also, t half life = [tex]\frac{In2}{k}[/tex] where k is rate constant

k = [tex]\frac{In 2}{45 days}[/tex] = 0.0154

In ( [tex]\frac{0.8}{1}[/tex]) = - 0.0154 t

-0.223 / -0.0154 = t

t = 14.49 approx 14.5 days from the date the yogurt was packaged

Answer:

B) I2(s) is a stronger oxidizing agent than Al3 (aq), and Al(s) is a stronger reducing agent than I-(aq).

Explanation:

An oxidizing agent accepts electrons in a redox reaction and become reduced while a reducing agent looses electrons and become oxidized.

Hence in a redox reaction, oxidizing agents are reduced while reducing agents are oxidized.

Looking at I2 and Al3+, I2 is a better oxidizing agent since it has a reduction potential of +0.54V compared to -1.66V for Al3+.

Given the statements above, the converse must be true, that is; All is a better reducing agent compared to I-

Answer:

The correct answer is option C. The unknown solution definitely has Hg22+ present.

Explanation:

In the analysis of group 1 metal cation, the unknown solution is treated with sufficient quantity of 6 M HCl solution and if group 1 metal cations are present then white precipitate of Agcl, PbCl2 or Hg2Cl2 is formed. The precipitate of PbCl2 is soluble in hot water but the other two remains insoluble after treating with hot water. Precipitate of AgCl disappears upon treatment of NH3 solution but Hg2Cl2 becomes black in the reaction with NH3. The black Colour appears due to the formation of metallic Hg.

Balanced chemical equation of the reation is -

Hg2Cl2 + 2NH3  ---------> HgNH2Cl (white ppt.) + Hg (black ppt.) + NH4Cl

Therefore, from the given information the conclusion which can be drawn is that the unknown solution definitely has Hg22+ present.

Answer:

C-  the unknown solution definitely has Hg22+ present.

Explanation:,

Answer: 112.5moles

Explanation:

P= 0.6atm, V= 4000L, R= 0.082, T= 260K

Applying PV = nRT

0.6×4000= n×0.082×260

Simplify n= (0.6× 4000)/(0.082×260)

n= 112.5moles

Using the Ideal Gas Law, we calculated that the weather balloon near the top of Mt. Rainier, with given conditions, contains approximately 113.1 moles of helium.

The Ideal Gas Law formula is PV = nRT, where:

We can use this formula to calculate the number of moles of helium in the weather balloon.

Given:

Using the Ideal Gas Law:

Plugging in the given values:

Thus, the number of moles of helium in the weather balloon is approximately 113.1 moles.

Complete question:

The student performs a second titration using the 0.10MNaOH(aq) solution again as the titrant, but this time with a 20.mL sample of 0.20MHCl(aq) instead of 0.10MHCl(aq).

1. The box below to the left represents ions in a certain volume of 0.10MHCl(aq). In the box below to the right, draw a representation of ions in the same volume of 0.20MHCl(aq). (Do not include any water molecules in your drawing.)

Answer:

The concentration of H+ and Cl- will be doubled.

Explanation:

See the attached photo for the representation of ions in the same volume.

Answer:

C. adding more powder solute

Explanation:

The concentration of a mixture can be increased by removing solvent.

The concentration of a mixture can be increased in several ways. If we consider a distillery wanting to achieve a higher concentration of alcohol, they need to remove solvent, which in this case is water, from their product. This can be achieved through processes such as distillation, which separates alcohol from water due to their different boiling points. Another approach is the addition of a substance that reacts with the water, effectively removing water content from the mixture.

To increase the concentration of a solution in general, one can also add more solute (the substance being dissolved) into the solution or remove solvent (the substance dissolving the solute). It is important not to confuse the process of dilution, which involves adding solvent and decreases solute concentration, with the process of concentrating a solution, which involves removing solvent and increases solute concentration. Therefore, methods such as evaporation, reverse osmosis, or chemical reaction can be employed to increase the concentration of a solute in a solution.

Answer:

[tex]m_{ice} = 65.336\,g[/tex]

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

[tex]Q_{water} = Q_{ice}[/tex]

[tex](100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)[/tex]

The amount of ice that is melt is:

[tex]m_{ice} = 65.336\,g[/tex]

Final answer:

To determine the new concentrations of NH3 and NH4+ after the addition of HNO3, the moles of HNO3 added are calculated and the amounts of NH3 converted to NH4+ are accounted for. The final concentrations are found to be approximately 0.277 M for NH3 and 0.339 M for NH4+, after considering the change in total volume.

Explanation:

The student is asking to calculate the concentration of buffer components in a solution after the addition of a strong acid. To solve this, we have to determine how much the strong acid (HNO3) will neutralize the NH3 component of the buffer, and how it will change the concentrations of NH3 and NH4+ (the buffer components).

First, calculate the number of moles of HNO3 added:

Since NH3 and NH4+ are in a 1:1 molar ratio in the solution and they react with HNO3 in a 1:1 ratio, the added HNO3 will react with the NH3:

And NH4+ will increase by 0.009 moles (since each mole of NH3 reacts to form one mole of NH4+):

The new volume of the solution would be the initial volume of buffer solution plus the volume of HNO3 added, which totals 288.50 mL or 0.28850 L.

The new concentrations are therefore:

This change in concentrations due to the addition of HNO3 means the buffer will have adjusted to these new concentrations of NH3 and NH4+.

Answer:

These three factors are required for ionization potential or ionization energy.

Explanation:

Ionization potential refers to the amount of energy which is required for the removal of outermost electron of the atom. If the atom size is big so the outermost electron is far from the nucleus and low energy is required for its removal due to lower force of attraction between nucleus and outermost electron. If the nuclear charge is higher, so the electron is tightly held by the nucleus and require more energy for its removal. Nuclear charge means number of protons present in the nucleus.

The statement which are true for a neutral, aqueous solution at 25 °C are:

The statement which is/are true for a neutral, aqueous solution regardless of temperature is;

We must know that at standard temperature, 25°C, a neutral, aqueous solution has it's pH = pOH = 7. Additionally, the hydrogen ion concentration, [H+] is equal to its hydroxide ion concentration, [OH-]

Ultimately, the pH of a solution decreases with increase in temperature and vice versa.

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Answer:funk

hot dog cat

Explanation:

uhughuhuhuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

Answer:

They occur when the sun reaches its highest or lowest point in the sky

Final answer:

Balancing chemical reactions involves trial and error, with the aim of achieving the same number of atoms for each element on both sides of the equation. Atoms that are in only one of the reactants and only one of the products should be done last, and the final coefficients should be the biggest numbers possible.

Explanation: