1. You dissolve 123g KBr into 689g of water. Calculate the mass percent.

The best inference Tara can draw about the commercial is the fact that it is making pseudoscientific claims based on opinions rather than scientific evidence, according to the message at the bottom of the screen.

The claims claimed in the commercial are not supported by scientific data and may be exaggerated or inaccurate because the pill has not been examined by physicians and has not been put through rigorous testing. Therefore, before making any judgements regarding the use of the product, it is crucial to exercise caution and skepticism towards the assertions made in such advertising and to look for trustworthy sources of information.

Therefore, the correct option is D.

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Your question is incomplete, most probably the complete question is:

Tara sees a commercial for a brand of pills that promise to boost energy and help people perform better in work and school. As people in the commercial talk about how much better they feel when taking the pill, Tara notices a message on the bottom of the screen that reads, "This pill is not approved by doctors and has not been tested in controlled experiments." What is the best conclusion that Tara should make about the commercial?

Mg + CO[tex]_2[/tex]→ MgO + CO is an oxidation-reduction reaction. Therefore, the correct option is option E among all the given options.

Redox reactions involve oxidation-reduction chemical processes in which the oxidation states of the reactants change. Redox is a shortened version of reduction-oxidation. Two distinct processes—a reduction process or an oxidation process—can be used to describe all redox reactions.

In redox and oxidation-reduction processes, the oxidation or reduction reactions usually take place concurrently. In a chemical reaction, the material that is being reduced is referred to as the reducing agent, and the substance that is being oxidised is the oxidising agent. Mg + CO[tex]_2[/tex]→ MgO + CO is an oxidation-reduction reaction.

Therefore, the correct option is option E.

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The substance E₁ is made of lead (Pb), and electrode E₂ is made of palladium (Pd) and the species in solution S₁ is Pb(NO₃)₂(aq), while the species in solution S₂ is Pd(NO₃)₂(aq).

The galvanic cell given in the question can be represented as;

Pb(s) | Pb(NO₃)₂(aq) || Pd(NO₃)₂(aq) | Pd(s)

where '|' represents a phase boundary, and '||' represents a salt bridge or porous disk separating the two half-cells.

The anode is the electrode where oxidation occurs, while the cathode is the electrode where reduction occurs.

To write the balanced equation for the half-reaction that occurs at the cathode, we need to identify the reduction half-reaction. Looking at the cell notation, we can see that Pd(NO₃)₂(aq) is being reduced to Pd(s), so the reduction half-reaction can be written as;

Pd²⁺(aq) + 2 e⁻ → Pd(s)

To write the balanced equation for the half-reaction that occurs at the anode, we need to identify the oxidation half-reaction. The anode is where the Pb(s) is oxidized to Pb²⁺(aq). The balanced oxidation half-reaction is;

Pb(s) → Pb²⁺(aq) + 2 e⁻

The species E₁ and E₂ refer to the electrodes at the anode and cathode, respectively. The electrode E₁ is made of lead (Pb), and electrode E₂ is made of palladium (Pd).

Therefore, the species in solution S₁ is Pb(NO₃)₂(aq), while the species in solution S₂ is Pd(NO₃)₂(aq).

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The solubility of MgF₂ in water is 0.00635 moles per liter, if Ksp for the compound is 6.4 x 10⁻⁹.

The solubility of a compound in water is determined by its solubility product constant (Ksp), which is a measure of the extent to which the compound dissociates into its constituent ions in water.

For the given compound, magnesium fluoride (MgF₂), the Ksp is 6.4 x 10⁻⁹.

MgF₂ dissociates in water according to the following equation;

MgF₂(s) ↔ Mg²⁺(aq) + 2F⁻(aq)

The Ksp expression for MgF₂ is then;

Ksp = [Mg²⁺] × [F⁻]²

where [Mg²⁺] represents the concentration of Mg²⁺ ions in solution, and [F⁻] represents the concentration of F⁻ ions in solution.

Since MgF₂ dissociates into one Mg²⁺ ion and two F⁻ ions, the stoichiometry of the reaction is 1:2. This means that for every mole of MgF₂ that dissolves, one mole of Mg²⁺ ions and two moles of F⁻ ions are formed.

Let's assume that the solubility of MgF₂ in water is "x" moles per liter. Therefore, the concentration of Mg²⁺ ions and F⁻ ions in solution will also be "x" moles per liter.

Substituting these values into the Ksp expression, we get;

Ksp = [Mg²⁺] × [F⁻]²

6.4 x 10⁻⁹ = x × (2x)²

6.4 x 10⁻⁹ = 4x³

Now, we can solve for "x";

4x³ = 6.4 x 10⁻⁹

x³ = (6.4 x 10⁻⁹ / 4

x³ = 1.6 x 10⁻⁹

x = (1.6 x 10⁻⁹(1/3)

x ≈ 0.00635

Therefore, the solubility of MgF₂ in water is approximately 0.00635 moles per liter.

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The percent yield of the reaction is 45.44%

Percentage yield is basically,

 [tex]\rm Percent\ yield\ = \frac{actual\ yield}{theoretical\ yield} \times 100[/tex]

2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O

To calculate the theoretical yield, consider the stoichiometry of the reaction.

2 moles of NaOH reacts with 1 mole of H₂SO₄ to give 1 mole of Na₂SO₄ and 2 moles of H₂O. The mole ratio of H₂SO₄ and Na₂SO₄ is 1:1

For this mole ratio to be useful, convert the given concentration of H₂SO₄ into moles.  

     [tex]\rm Molarity\ =\ \frac{No.\ of\ moles}{Volume\ of\ solution (L)}[/tex]

            [tex]\rm 3.10\ =\ \frac{No.\ of\ moles}{0.125}[/tex]

[tex]\rm No.\ of\ moles\ =\ 3.10 \times 0.125\[/tex]

                     [tex]=\ 0.38[/tex]

Since, mole ratio of H₂SO₄ and Na₂SO₄ is 1:1

Amount of Na₂SO₄ formed would be also 0.38 mol

Convert this amount in moles to amount in grams

[tex]\rm No.\ of\ moles\ =\ \frac{Mass\ formed\ }{Molecular\ mass}[/tex]

[tex]\rm Mass\ formed\ =\ No.\ of\ moles\times molecular\ mass[/tex]

                     [tex]\rm =\ 0.38\times 142.04[/tex]

                     [tex]\rm =\ 53.97\ grams[/tex]

Theorical yield of Na₂SO₄ is 53.97 grams

Therefore,   [tex]\rm Percent\ yield\ = \frac{actual\ yield}{theoretical\ yield} \times 100[/tex]

                                          [tex]\rm =\ \frac{24.5}{53.97}\times 100[/tex]

                                          =  45.44%

The percent yield of the reaction is 45.44%

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The concentration of iodide ions in a saturated solution of lead (II) iodide is 3.03 x 10⁻³ M. The solubility product constant of PbI₂ is 1.4 x 10-8.

The solubility product constant, which is denoted by Ksp​ is defined as the equilibrium constant for a solid substance which  dissolves in an aqueous solution.

when PbI₂ dissolves, it dissociates as follows

PbI₂ --> Pb²⁺ + 2I⁻

Molar solubility is the number of moles of salt that can be dissolved in 1 L of solution

If molar solubility of PbI₂ is x , then molar solubility of Pb²⁺ is x and I⁻ is 2x

ksp is solubility product constant.

ksp = [Pb²⁺][I⁻]²

ksp = [x][2x]²

ksp = 4x³

4x³ = 1.4 x 10⁻⁸

x³ = 0.35 x 10⁻⁸

x = 1.51 x 10⁻³ M

since molar solubility of I⁻ is 2x ,

then molar solubility of I⁻ or concentration of iodide ions is 3.03 x 10⁻³ M

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Algal blooms and high concentrations of nitrogen in water are effects of water pollution. Overgrazing and the use of chemical fertilizers cause water pollution.

Algal blooms are an effect of water pollution. They occur when there is an excessive amount of nutrients, particularly nitrogen and phosphorus, in the water due to pollution. The overgrowth of algae depletes the oxygen levels in the water, which can harm fish and other aquatic animals.

Overgrazing can cause water pollution by increasing the sedimentation rate of waterways. This sedimentation can carry nutrients, bacteria, and other pollutants into the water, which can degrade water quality and cause harm to aquatic life.

The use of chemical fertilizers to enhance production is a cause of water pollution. When fertilizer is overused, it can leach into waterways and cause nutrient pollution, which can lead to algal blooms and other forms of water pollution.

High concentrations of nitrogen in water are often an effect of water pollution. This can be caused by the overuse of fertilizers or the discharge of untreated sewage into waterways. High nitrogen levels can cause algal blooms, which can lead to oxygen depletion and harm aquatic life.

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Answer:

Cause:

: : use of chemical fertilizers to enhance production

: : overgrazing

Effect:

: : high concentration of nitrogen in water

: : algal blooms

Hope this helps ;)

The heat change or enthalpy change, ΔH°rxn, for the slow reaction of zinc with water is +417.7 kJ/mol.

The heat change or enthalpy change, ΔH°rxn, for the reaction of zinc (Zn) with water (H₂O) can be calculated using the standard enthalpies of formation for each species involved in the reaction.

Balanced chemical equation for the reaction is;

Zn(s) + 2H₂O(l) → Zn²⁺(aq) + H₂(g)

The standard enthalpy change for the reaction, ΔH°rxn, can be calculated as the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants, each multiplied by their respective stoichiometric coefficients;

ΔH°rxn = Σ(nΔH°f, products) - Σ(mΔH°f, reactants)

where n and m are the stoichiometric coefficients of the products and reactants, respectively, and ΔH°f is the standard enthalpy of formation.

Assuming standard conditions (25°C and 1 atm), the standard enthalpies of formation for Zn²⁺(aq) and H₂(g) are typically tabulated values. Let's assume their values to be ΔH°f(Zn²⁺(aq)) = -153.9 kJ/mol and ΔH°f(H₂(g)) = 0 kJ/mol, respectively.

The standard enthalpy of formation of water (H₂O) is -285.8 kJ/mol.

Put the values into the equation, we get;

ΔH°rxn = [ΔH°f(Zn²⁺(aq)) + ΔH°f(H₂(g)] - [ΔH°f(Zn(s)) + 2ΔH°f(H₂O(l))]

ΔH°rxn = [-153.9 + 0] - [0 + 2(-285.8)]

ΔH°rxn = -153.9 + 571.6

ΔH°rxn = 417.7 kJ/mol

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The mass of iron(III) oxide produced is 115.29 grams.

To determine the mass of iron(III) oxide produced, we can use the stoichiometry of the balanced chemical equation provided.

The balanced equation shows that 2 moles of iron (Fe) react with 3 moles of water vapor (H₂O) to produce 1 mole of iron(III) oxide ( Fe₂O₃).

Given that we have 6.00 L of water vapor at 50.2 °C and 0.121 atm, we can use the ideal gas law to calculate the moles of water vapor;

PV = nRT, where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature in Kelvin.

First, let's convert the temperature from Celsius to Kelvin;

T = 50.2 + 273.15 = 323.35 K

Now, we can plug in the values and solve for n;

0.121 atm × 6.00 L = n × 0.0821 atm L/mol K × 323.35 K

n = (0.121 atm × 6.00 L) / (0.0821 atm L/mol K × 323.35 K)

n = 0.361 mol

According to the stoichiometry of the balanced equation, the mole ratio between Fe and  Fe₂O₃ is 2:1. Therefore, 0.361 mol of water vapor will react with 2 × 0.361 mol = 0.722 mol of Fe.

Now, we can calculate the molar mass of  Fe₂O₃;

Fe₂O₃ = 2 × atomic mass of Fe + 3 × atomic mass of O

Fe₂O₃ = 2 × 55.85 g/mol + 3 × 16.00 g/mol

Fe₂O₃ = 159.70 g/mol

Finally, we can calculate the mass of  Fe₂O₃ produced;

Mass of  Fe₂O₃ = moles of  Fe₂O₃ × molar mass of  Fe₂O₃

Mass of  Fe₂O₃ = 0.722 mol × 159.70 g/mol

Mass of Fe₂O₃ = 115.29 g

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The concentration of N₂O₄ is 0.514 M.

The equilibrium constant (K) is a ratio of the concentrations of products to reactants at equilibrium. In this problem, the equilibrium constant (K) is given as 0.028 for the reaction N₂O₄ ← → 2NO₂.

To find the concentration of N₂O₄, we can use the expression for K and the initial concentration of NO₂. Since there is no initial concentration of N₂O₄ given, we assume that it is x M.

Using the equilibrium constant (K) and the initial concentration of one of the species.

[N₂O₄] = √(K/[NO₂])

[N₂O₄] = √(0.028/0.042)

[N₂O₄] = 0.514 M

Therefore, the concentration of N₂O₄ is 0.514 M.

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The ranking of the elements by increasing ionization energy is; Cs < Sn < W < Xe.

Xenon (Xe) has the highest ionization energy among the given elements. It is a noble gas in Group 18 of the periodic table and has a full complement of valence electrons, making it very stable and difficult to remove an electron.

Tungsten (W) has higher ionization energy compared to tin. It is a transition metal in Group 6 of the periodic table and has even more valence electrons, which are held more tightly by the increased effective nuclear charge, resulting in higher ionization energy.

Tin (Sn) has higher ionization energy compared to cesium. It is a metal in Group 14 of the periodic table and has more valence electrons, which are closer to the nucleus, requiring more energy to remove.

Cesium (Cs) has the lowest ionization energy among the given elements. It is in the alkali metal group (Group 1) of the periodic table and has only one valence electron, which is relatively far from the nucleus, making it easier to remove.

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To identify the limiting reactant, excess reactant, and theoretical yield, we first need to determine the amount of each reactant in moles.

Using the molar masses of Ca and P:

Number of moles of Ca = 17.0 g / 40.08 g/mol = 0.424 mol

Number of moles of P = 18.0 g / 30.97 g/mol = 0.581 mol

Next, we need to determine the stoichiometric ratio of the reactants. From the balanced chemical equation, we see that the ratio of Ca to P is 3:2.

3 Ca + 2 P → Ca3P2

To use the stoichiometric ratio to determine the limiting reactant, we need to compare the actual ratio of the reactants to the stoichiometric ratio.

Actual ratio of Ca to P = (0.424 mol Ca) / (0.581 mol P) ≈ 0.73

Stoichiometric ratio of Ca to P = 3/2 = 1.5

Since the actual ratio is greater than the stoichiometric ratio, Ca is the excess reactant and P is the limiting reactant.

To find the theoretical yield of Ca3P2, we need to use the stoichiometric ratio to determine how many moles of Ca3P2 can be produced from the limiting reactant (P).

From the balanced chemical equation, we see that 2 moles of P react with 3 moles of Ca to produce 1 mole of Ca3P2.

So, the number of moles of Ca3P2 that can be produced from 0.581 mol of P is:

(0.581 mol P) × (1 mol Ca3P2 / 2 mol P) = 0.2905 mol Ca3P2

Therefore, the theoretical yield of Ca3P2 is 0.2905 mol.

The correct ions that have the quoted electron configuration would be [tex]Ag^{3+[/tex], [tex]Gd^{3+[/tex], [tex]Cr^{6+[/tex], [tex]Ni^{2+[/tex], and [tex]Pt^{2+[/tex].

The electron configuration (n = 3, 4, 5, ...) corresponds to the elements in the d-block and f-block of the periodic table. Therefore, the ions with this electron configuration are:

P3+ (phosphorus ion) does not have an electron configuration of (n = 3, 4, 5, ...) as it is a p-block element with the electron configuration [Ne] 3s^2 3p^3.

Therefore, the correct ions that have this electron configuration are [tex]Ag^{3+[/tex], [tex]Gd^{3+[/tex], [tex]Cr^{6+[/tex], [tex]Ni^{2+[/tex], and [tex]Pt^{2+[/tex].

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When the moving car brakes to the stop the kinetic energy of car will be converted to the heat energy.

The mechanical brake will be applies to the friction force and it convert the kinetic energy of the car into the thermal energy that which then dissipates on atmosphere. The process of the braking will follow the principle of the conservation of the energy.

The conservation of the energy is the principle, that is expressed in its the most general form, and it is the first law of the thermodynamics. The first law of thermodynamics explains that "the energy of the universe remains the same."

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This question is incomplete, the complete question is :

A car in motion has kinetic energy. A moving car is suddenly stopped. Why does the car stop? Where did the energy go?

The correct rate law of the reaction from the experimental data is k[NO]^2 [O2]. Option D

The rate of reaction is usually expressed in terms of the amount of reactant consumed or product formed per unit time, and is typically measured in units of moles per liter per second (mol/L/s) or similar units.

We have that;

For NO;

3.4 * 10^-5/8.4 * 10^-6 = 2 * 10^-4/ 2 * 10^-4

4 = 2^n

n = 2

For O2;

8.4 * 10^-6/2.8 * 10^-6 = 3 * 10^-4/ 1 * 10^-4

3 = 3^n

n = 1

Thus the rate law of the reaction is;

Rate = k[NO]^2 [O2]

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The vapor pressure of the SiCl₄ in the mmHg at the 33.0 °C is the 312 mmHg.

The Clausius - Clapeyron equation is as :

ln ( P₂ / P₁ ) = ΔHvap / R ( 1 / T₁ - 1 / T₂ )

P₂ = P₁eˣ

Where,

The temperature, T₁  = 5.4 °C = 278.55 K

The temperature, T₂  = 33.0 °C= 306 K

The pressure, P₁ = 100 mmHg

ΔHvap is the heat of the vaporization = 30.2 kJ /mol = 30200 J/mol

The gas constant, R = 8.314 J / mol K

x = ΔHvap / R ( 1 / T₁ - 1 / T₂ )

x = 30200 / 8.314 ( 1/ 278.55 - 1/ 306 )

x = 1.05

P₂ = 100 [tex]e^{1.05}[/tex]

P₂ = 312 mmHg

The vapor pressure of the  SiCl₄ is 312 mmHg.

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Answer: The pressure in the container is 1949.5 torr when the volume is 7.20 liters and the temperature is 71.0 °C

Explanation: To solve this problem, we can use the combined gas law, which relates the initial and final conditions of pressure (P), volume (V), and temperature (T) for an ideal gas:

(P1 × V1) / T1 = (P2 × V2) / T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.

We can plug in the given values:

P1 = 740.5 torr

V1 = 10.8 L

T1 = 23.0 °C + 273.15 = 296.15 K

V2 = 7.20 L

T2 = 71.0 °C + 273.15 = 344.15 K

(P1 × V1) / T1 = (P2 × V2) / T2

(740.5 torr × 10.8 L) / 296.15 K = (P2 × 7.20 L) / 344.15 K

Solving for P2:

P2 = (740.5 torr × 10.8 L × 344.15 K) / (296.15 K × 7.20 L)

P2 = 1949.5 torr

Therefore, the pressure in the container is 1949.5 torr when the volume is 7.20 liters and the temperature is 71.0 °C.

The half-reaction is; H₂ → 2H⁺ + 2e⁻, the oxidizing agent is chlorine , and the reducing agent is hydrogen. The oxidation half-reaction is where a species loses electrons, the reduction half-reaction is where a species gains electrons, the oxidizing agent is the species that causes another species to be oxidized by accepting electrons itself, and the reducing agent is the species that causes another species to be reduced by donating electrons itself.

The oxidation half-reaction is the half-reaction where a species loses electrons. In this reaction, hydrogen (H₂) is oxidized to form hydrogen ions (H⁺). The half-reaction can be written as;

H₂ → 2H⁺ + 2e⁻

The reduction half-reaction is the half-reaction where a species gains electrons. In this reaction, chlorine (Cl₂) is reduced to form chloride ions (Cl⁻). The half-reaction can be written as follows;

Cl₂ + 2e⁻ → 2Cl⁻

The oxidizing agent is the species that causes another species to be oxidized by accepting electrons itself. In this reaction, chlorine (Cl₂) is the oxidizing agent because it accepts electrons from hydrogen (H₂), causing it to be oxidized.

The reducing agent is the species that causes another species to be reduced by donating electrons itself. In this reaction, hydrogen (H₂) is the reducing agent because it donates electrons to chlorine (Cl₂), causing it to be reduced.

Therefore, chlorine is the oxidizing agent, and hydrogen is the reducing agent.

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To calculate the apparent bond energy of a carbon-sulfur bond in COS, we need to use the bond energies of the individual bonds that make up the molecule. The apparent bond energy of a carbon-sulfur bond in COS is approximately 1092 kJ/mol.

The apparent bond energy is the difference between the energy required to break the bonds in the reactants and the energy released when new bonds form in the products.

The enthalpy change of two (2) moles of COS is:

ΔH = –3.2 × 10⁻²⁴ × 2  × 6.02 × 10²³ = -3.85 kJ / mol

ΔH = ∑bond energy of reactants - ∑bond energy of products

2×C=O + 2 × C=S -[2 × C=O] + [2×C=S)

-3.85 = 2 × C=S -(2 × 552)

2 × C=S = 1104 - 3.85 = 1100.15

C=S =  1100.15 /2 = 550.08 kJ / mol

So apparent energy is 550.08 kJ / mol.

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A Tutor is responsible for helping students to learn and understand the new concepts and complete assignments. They prepare lessons by studying lesson plans, reviewing textbooks in detail.

The role of a teacher is to teach a certain subject to the students according to the school syllabus whereas tutors respond to a child's individual needs by filling gaps. It is one who gives private instruction.

A Tutor for chemistry can be a person with good subject knowledge of chemistry. Generally the person who have taken MSC in chemistry or Mphil can be the apt tutor. They should got these degrees from a recognized university.

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The composition is three number of atoms in each sample.

The number of samples that was given include the following:

Water (H2O) : This sample is made up of three number of atoms but contains only two elements of oxygen and hydrogen in a ratio of 1:2 respectively.

Carbondioxide (CO2) : This sample is made up of three number of atoms but contains only two elements of carbon and oxygen in a ratio of 1:2.

Sugar (CHO) : This sample is made up of three number of atoms but contains three elements of carbon, oxygen and hydrogen in a ratio of 1:1:1.

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If you have an 8.53 mg sample of phosphorus oxychloride and it contains 1.72 mg of phosphorus, 20.6%  is the percent composition of the phosphorus in the compound.

The ratio of each element's amount to the sum of all the individual components present in the compound, multiplied by 100, is what determines any compound's percentage composition. Here, we quantify the amount of the constituents in the solution in terms of grammes.

Any compound's percent composition expresses its makeup in terms of all the components that are present. The chemical analysis reveals the relevance of this composition calculation.

percent composition = (1.72/8.53 )×  100

                                    = 20.6%

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The products formed in the acid-base reaction between NaOH and HCl are NaCl (sodium chloride) and H₂O (water).

NaOH + HCl → NaCl + H₂O

In this reaction, NaOH is a strong base, and HCl is a strong acid. A strong base is one that completely dissociates in water to form hydroxide ions (OH⁻) while a strong acid is one that completely dissociates in water to form hydrogen ions (H⁺).

NaOH + H₂O → Na⁺ + OH⁻

HCl + H₂O → H⁺ + Cl⁻

When NaOH is added to HCl, the hydroxide ions (OH⁻) from the NaOH react with the hydrogen ions (H⁺) from the HCl to form water (H2O). The remaining ions, Na⁺ and Cl⁻, combine to form sodium chloride (NaCl), which is a salt.

Na⁺ + Cl⁻ → NaCl

The reactants in this reaction are a strong base (NaOH) and a strong acid (HCl), and the products are a salt (NaCl) and water (H₂O). This reaction is an example of neutralization, which is a reaction between an acid and a base that produces a salt and water. The Na⁺ ion is the conjugate acid of the strong base NaOH, while the Cl⁻ ion is the conjugate base of the strong acid HCl

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Answer: An onion cell has a cell wall.

Explanation:

One of the main differences in cell structure between an onion cell and a human cheek cell is that an onion cell has a cell wall, while a human cheek cell does not. The cell wall of the onion cell provides it with additional support and protection, which is not required by animal cells, such as the human cheek cell.

The student pulling out the pH electrode from the titration beaker and rinsing it off with DI water into a waste container several times during the titration will not significantly affect the measured equivalent mass.

This is because the equivalent mass of a substance is determined by the stoichiometry of the reaction, which is not influenced by the pH electrode or the rinsing process. However, it is important to note that if the student is rinsing the electrode with a significant amount of water, it could dilute the solution and affect the accuracy of the titration. Therefore, it is recommended to use a minimal amount of water during the rinsing process to minimize any potential dilution effect.

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The ionization constant of propanoic acid is 0.7181

The ionization constant (Ka) of the monoprotic acid can be calculated using the equation:

Ka = ([H⁺][A⁻])/[HA]

where [H⁺] is the concentration of hydrogen ions, [A⁻] is the concentration of the conjugate base, and [HA] is the initial concentration of the acid.

We know that the acid is 0.815 percent ionized, which means that only 0.815 percent of the initial concentration of the acid has ionized into hydrogen ions and the conjugate base.

Therefore, the concentration of hydrogen ions and the conjugate base can be calculated using the following equations:

[H⁺] = 0.815 x 0.200 M = 0.163 M

[A⁻] = 0.815 x 0.200 M = 0.163 M

The initial concentration of the acid ([HA]) can be calculated by subtracting the concentration of hydrogen ions and the conjugate base from the initial concentration of the solution:

[HA] = 0.200 M - 0.163M= 0.037 M

Substituting these values into the equation for Ka, we get:

Ka = (0.163 M)² / 0.037M = 0.7181

Therefore, the ionization constant (Ka) of propanoic acid is 0.7181

To know more about ionization constant here :

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