Calculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 200.0 mL of 0.12 M NH3. The Kb for NH3 is 1.8 × 10-5.

Answer:

A. Increasing the temperature will favor forward reaction and more CaCo3 formed.

B. More CaCo3 will be formed.

C. CaCo3 will decrease and more react ants formed.

D. Less CaCo3 will be formed.

E. Iridium is a catalyst so there is no effect

Explanation:

A. Temperature will increase because it's an endothermic reaction.

B. Adding Cao will favor forward reaction and more CaCo3 formed.

C. Removing methane, more react ants are formed and CaCo3 decreases.

D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.

Answer:

A. Ka = [CO2] / [C] [O2]^1/2

B. Kb = [CO2] / [CO] [O2]^1/2

Explanation:

Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:

A. Determination of the expression for equilibrium constant Ka.

This is illustrated below:

C(s) + 1/2 O2(g) <==> CO(g)

Ka = [CO2] / [C] [O2]^1/2

B. Determination of the expression for equilibrium constant Kb.

This is illustrated below:

CO(g) + 1/2 O2(g) <==> CO2(g)

Kb = [CO2] / [CO] [O2]^1/2

Answer:

1. C

Carbon atoms

2. C

CH3CH2OH

3. C

extended structure

AND 4. B

One grey sphere, four white spheres, one red sphere

TOOK THE QUIZ. YOUR WELCOME :))

The term which describe the repeated arrangement of same molecules is extended structure. Thus option D is correct.    

Extended structure can be defined as a structure in which the subunits are arranged in a repeating pattern and occur in a consistent ratio.

Sodium chloride and diamond are some of the example of extended structure.

The expanded or extended  structure departs from the bilayer configuration, with well-separated hydrophilic and hydrophobic layers made up of propagating chains along the calixarene cavity axis are some characteristics.  

Thus, the term which describe the repeated arrangement of same molecules is extended structure. Thus option D is correct.  

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Your question is incomplete but probably your complete question was

Which term describes the repeated arrangement of the molecule?

A. Bonds

B. Atoms

C. Molecular Model

D. Extended Structure

.....................

Cellulose is insoluble in water which is made up of glucose because it possesses high inter and intramolecular hydrogen bonding between the hydroxyl groups of the neighboring chains.

Due to being insoluble in nature, it serves as the fundamental component of the cell membrane in the plant. This is why the cell wall of plant cells are made of cellulose.

The chains of the cellulose are strongly bonded to each other. So, it is very difficult for the molecules of water to rupture or destruct these bonds between the chains. It is a hydrogen bond cross-linked polymer and more complex than glucose.

Therefore, due to high inter and intramolecular hydrogen bonding between the hydroxyl groups of the neighboring chains, cellulose is insoluble in water.

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Answer:

Explanation:

Answer:

1, 1, 1, 1  

Explanation:

potassium hydrogen sulfate + potassium hydroxide ⟶ potassium sulfate + water(l)

                 KHSO₄                   +               KOH              ⟶            K₂SO₄   + H₂O

1. Put a 1 in front of the most complicated-looking formula (K₂SO₄?):

KHSO₄  + KOH ⟶ 1K₂SO₄ + H₂O

2. Balance S:

We have fixed 1 S on the right. We need 1 S on the left. Put a 1 in front of KHSO₄ to fix it.

1KHSO₄  + KOH ⟶ 1K₂SO₄ + H₂O

3. Balance K:

We have fixed 2 K on the right and 1 K on the left. We need 1 more K on the left. Put a 1 in front of KOH.

1KHSO₄  + 1KOH ⟶ 1K₂SO₄ + H₂O

4. Balance O

We have fixed 4 O on the right and 5 O on the left. We need 1 more O on the right. Put a 1 in front of H₂O.

1KHSO₄  + 1KOH ⟶ 1K₂SO₄ + 1H₂O

Every formula has a coefficient. The equation should be balanced.

5. Check that atoms balance:

[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{H} & 2 & 2\\\text{S} & 1 & 1\\\text{O}&5&5\\\end{array}[/tex]

It checks.

The coefficients are 1, 1, 1, 1.

Answer:

ΔT = 2.28°C

Explanation:

Heat, H = 75.4J

Number of moles = 0.25 mol

Specific heat capacity, c = 0.861 J/g·°C

Change in temperature, ΔT = ?

These quantities are related by the following equation;

H = mc ΔT

Mass, m = Number of moles * Molar mass

m = 0.25mol * 153.82 g/mol

m = 38.455g

S back to the equation;

H = mc ΔT

Substituting the values;

75.4 = 38.455 * 0.861 * ΔT

ΔT = 75.4 /  33.11

ΔT = 2.28°C

The change in temperature is 2.28 °C

First, we will determine the mass of CCl₄ absorbed

From the given information,

Number of moles of CCl₄ absorbed = 0.25 mol

Using the formula

Mass = Number of moles × Molar mass

Molar mass of CCl₄ = 153.82 g/mol

∴ Mass of CCl₄ absorbed = 0.25 × 153.82

Mass of CCl₄ absorbed = 38.455 g

Now, using the formula

Q = mcΔT

Where Q is the quantity of heat

m is the mass

c is the specific heat of substance

and ΔT is the change in temperature

From the given information

Q = 75.4 J

c = 0.861 J/g.°C

Putting the parameters into the formula, we get  

75.4 = 38.455 × 0.861 ×ΔT

75.4 = 33.109755 × ΔT

∴ ΔT = 75.4 ÷ 33.109755

ΔT = 2.28 °C

Hence, the change in temperature is 2.28 °C

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Answer:

[tex]M_{base}=0.709M[/tex]

Explanation:

Hello,

In this case, since the reaction between potassium hydroxide and nitric acid is:

[tex]KOH+HNO_3\rightarrow KNO_3+H_2O[/tex]

We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:

[tex]n_{acid}=n_{base}[/tex]

That in terms of molarities and volumes is:

[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]

Thus, solving the molarity of the base (KOH), we obtain:

[tex]M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M[/tex]

Regards.

Answer:

The smallest whole-number coefficient for OH⁻ is 2

Explanation:

Step 1: The equation redox reaction is divided into two half equations

Reduction half equation: MnO₄⁻  ----> MnO₂

Oxidation half-equation: I⁻  ---> IO₃⁻

Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;

MnO₄⁻ +  2H₂O ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O

Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation

MnO₄⁻ +  2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction

2MnO₄⁻ +  4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

Step 5 : addition of the two half equations to yield a net ionic equation

2MnO₄⁻  + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻

The smallest whole number coefficient for OH⁻ is 2

A redox reaction is divided into two half equations which are shown below:

Reduction half equation: MnO₄⁻  ----> MnO₂

Oxidation half-equation: I⁻  ---> IO₃⁻

Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;

MnO₄⁻ +  2H₂O ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O

The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation

MnO₄⁻ +  2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.

2MnO₄⁻ +  4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

The two half equations are then added and written together to form a net ionic equation

2MnO₄⁻  + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻

The smallest whole-number coefficient for OH⁻ is therefore 2.

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Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]

3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.

2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]

1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.

3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]

1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.

4)  [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]

1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.

Answer:

B

Explanation:

because it is a trigonal planar compound

Three bonding pairs around a central atom results in a trigonal planar compound. Option B is correct.

In a trigonal planar arrangement, the central atom is surrounded by three bonding pairs of electrons, forming a flat, triangular shape. The bond angles between the bonding pairs are approximately 120 degrees.

This molecular geometry is observed when a molecule has a central atom with three bonded pairs and no lone pairs. Examples of compounds with  trigonal planar geometry include boron trifluoride (BF₃) as well as formaldehyde (H₂CO).

The other options are not correct for a molecule with three bonding pairs;

Octahedral refers to a molecular geometry with six bonding pairs around a central atom.

Tetrahedral corresponds to a molecular geometry with four bonding pairs around a central atom.

Linear represents a molecular geometry with two bonding pairs around a central atom.

Hence, B. is the correct option.

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Answer:

The correct option is C.

Note the full question and structure of the moleculesis found in the attachment below.

Explanation:

Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.

The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.

During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.

From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.

The correct option therefore, is C

Answer:

The difference between a rock and a boulder can be explained in terms of size and detachment. A rock is defined by geologists as an aggregate of minerals. A boulder is a type of rock, specifically a large detached one. All boulders are rocks, but not all rocks are boulders.

Answer:

(B). it's metallic bonding

Answer: 6.31×10⁻³ moles Fe

Explanation:

To calculate moles when given atoms, we need to use Avogadro's number.

Avogadro's number: 6.022×10²³ atoms/mol

[tex](3.8*10^2^1 atoms)*\frac{mol}{6.022*10^2^3 atoms} =6.31*10^-^3 mols[/tex]

The atoms cancel out, and we are left with moles. There are 6.31×10⁻³ moles Fe.

Answer:

The atomic is 6, the mass number is 12

Explanation:

The element being described is Carbon.

The number of protons in the nucleus of an atom is called its atomic number, Z.

The number of protons pluss the number of neutronsin the nucleus of an atom gives the atom's mass number, A.

Answer:

14.297 g

Explanation:

From the question;

1 mo of the compound requires 1320.0 kJ

From the molar mass;

1 ml of the compound weighs 30.55g

How many grams requires 617.30kJ?

1 ml = 1320

x mol = 617.30

x = 617.30 / 1320

x = 0.468 mol

But 1 mol = 30.55

0.468 mol = x

x = 14.297 g

Answer:

B) hydroxide concentration

Explanation:

Hello,

In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.

After that, due to the fact that the pH is related with the pOH as shown below:

pH=14-pOH

We can directly compute the pH.

Best regards.

Answer:

D. 2:5

Explanation:

It has 5 valency electrons

[tex].[/tex]

In the given question, the electronic configuration of nitrogen is 2,5. The correct answer is option D.

The electronic configuration of an atom describes the arrangement of its electrons in different energy levels or orbitals.

Nitrogen has an atomic number of 7, which means it has 7 electrons.

Therefore, the electronic configuration of nitrogen is 2,5, which means it has 2 electrons in its first energy level and 5 electrons in its second energy level. Option D is the correct answer.

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Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

1 cm³ = 1 mL

mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g

mass of freshwater + cup = 734.265 + 25 = 759.265 g

Therefore,  mass of equal volume of seawater = 759.265 g

Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

Therefore, amount of salt in 1 L seawater = 34 g

A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]  = 0.10 M

The chemical equation for this reaction is :

[tex]\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}[/tex]

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = [tex]\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}[/tex] = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  [tex]\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}[/tex]

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log [tex]\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}[/tex]

The pH of this solution = 4.74 + log [tex]\mathtt{\dfrac{0.135}{0.065}}[/tex]

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places

Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = 4.74

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

Answer:

0.015 M

Explanation:

For a first order reaction;

ln[A] =ln[A]o - kt

[A] = final concentration

[A]o =initial concentration

k= rate constant

t= time taken

ln[A] =ln[A]o - kt

ln[A] = ln(0.0500) - 6.7 x 10-4 (30 × 60)

ln[A] = -2.9957 - 1.206

ln[A] = -4.202

e^ln[A] = e^(-4.202)

A= 0.015 M

Answer:

Each molecule of O2 is made up of 2 oxygen atoms. So 1 mole of O2 molecules is made up of 2 moles of oxygen atoms. Therefore 1 mole of oxygen gas contains 2 moles of oxygen atoms. And 0.4 moles of oxygen gas contains 0.8 moles of oxygen atoms.

There are 0.8 moles of oxygen atoms in 0.4 moles of oxygen gas.

Oxygen gas (O₂) consists of two oxygen atoms bonded together. Therefore, to determine the number of moles of oxygen atoms present in a given amount of oxygen gas, we can simply multiply the number of moles of oxygen gas by the number of oxygen atoms per molecule, which is 2.

Given that we have 0.4 moles of oxygen gas, we can calculate the number of moles of oxygen atoms as follows:

Number of moles of oxygen atoms = Number of moles of oxygen gas × Number of oxygen atoms per molecule

= 0.4 moles × 2

= 0.8 moles

Therefore, there are 0.8 moles of oxygen atoms present in 0.4 moles of oxygen gas.

This calculation is based on the stoichiometry of oxygen gas, which indicates that each molecule of O₂ contains two oxygen atoms. By considering the mole ratio between oxygen gas and oxygen atoms, we can determine the number of moles of oxygen atoms in a given quantity of oxygen gas.

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Answer:

The rms speed of its molecules becomes. (T) has become four times. Therefore, v_(rms) will become two times,...

Answer:

Exceed the buffer capacity and Raise the pH by several units

Explanation:

Options are:

Raise the pH slightly

Lower the pH slightly

Raise the pH by several units

Lower the pH by several units

Not change the pH

Exceed the buffer capacity

The hypochlorous acid, HClO, is in equilibrium with Hypochlorite ion (From potassium hypochlorite, ClO⁻) producing a buffer. Using H-H equation, pH of initial buffer is:

pH = pKa + log [ClO⁻] / [HClO]

pKa for hypochlorous acid is 7.53

pH = 7.53 + log [0.581M] / [0.436M]

pH = 7.65

Barium hydroxide reacts with HClO producing more ClO⁻, thus:

Ba(OH)₂ + 2HClO →  2ClO⁻ + 2H₂O

As 0.479 moles of Barium hdroxide are added. For a complete reaction you require 0.479mol * 2 = 0.958 moles of HClO

As you have just 0.436 moles (Volume = 1L),

The addition will:

The Ba(OH)₂ that reacts is:

0.436 moles HClO * (1mole (Ba(OH)₂ / 2 mol HClO) = 0.218 moles Ba(OH)₂ and will remain:

0.479 mol - 0.218 mol = 0.261 moles Ba(OH)₂

As 1 mole of Ba(OH)₂ contains 2 moles of OH⁻, moles of OH⁻ and molarity is:

0.261 moles* 2 = 0.522 moles OH⁻ = [OH⁻]

pOH = -log [OH⁻]

pOH = 0.28

And pH = 14 - pOH:

pH = 13.72

Thus, after the addition the pH change from 7.65 to 13.62:

Answer:

The process will be spontaneous above 702 K.

Explanation:

Step 1: Given data

Step 2: Calculate the temperature range in which the process will be spontaneous

The reaction will be spontaneous when the standard Gibbs free energy (ΔG°) is negative. We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T × ΔS°

When ΔG° < 0,

ΔH° - T × ΔS° < 0

ΔH° < T × ΔS°

T > ΔH°/ΔS°

T > (308,000 J/mol)/(439 J/mol.K)

T > 702 K

The process will be spontaneous above 702 K.

Answer:

848 torr  

Explanation:

The only variables are the pressure and the volume, so we can use Boyle's Law.

p₁V₁ = p₂V₂

Data:

p₁ = 565 torr; V₁ = 24.0 L

p₂ = ?;            V₂ =  16.0 L

Calculations:

[tex]\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{565 torr} \times \text{24.0 L} & = & p_{2} \times \text{16.0 L}\\\text{13 560 torr} & = & 16.0p_{2}\\p_{2} & = & \dfrac{\text{13 560 torr}}{16.0}\\\\& = &\textbf{848 torr}\\\end{array}\\\text{The final pressure of the gas is $\large \boxed{\textbf{848 torr}}$}[/tex]

Answer:

[tex][H^+]=0.000123M[/tex]

[tex]pH=3.91[/tex]

Explanation:

Hello,

In this case, dissociation reaction for acetic acid is:

[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]

For which the equilibrium expression is:

[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]

Which in terms of the reaction extent [tex]x[/tex] could be written as:

[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]

Thus, solving by using a solver or quadratic equation we obtain:

[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]

And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:

[tex][H^+]=0.000123M[/tex]

Now, the pH is computed as follows:

[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]

Best regards.

Answer:

a. the Kb of NO₂⁻ to find the hydroxide concentration.

Explanation:

When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.

NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)

Na⁺ comes from NaOH (strong base) so it doesn't react with water.

NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.

NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)

This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to  find the hydroxide concentration.

Answer:

The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"

Explanation:

[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]

In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor  and each other suspend

[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]

[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]

The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.

The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.

The cell potential has been changed when there has been a change in the potential of the half cells.

The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.

With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.

Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

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