How many grams of KBr are dissolved in 72.7 mL of a 0.998 M solution?


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



You dissolve 3 moles KBr into 46 moles of water. Calculate the mass percent

Answer:

If you can't measure the length, breadth, or height of an item, you can utilise displacement or guesswork instead.

For example, you may calculate the volume of an item by measuring the displacement of water while it is submerged in it. You may then estimate its dimensions by assuming a certain form or by considering additional metrics such as weight or density.

Another technique is to estimate the object's dimensions using a known object of comparable size or visual cues such as a ruler or grid in an image. These procedures, however, may be less accurate and more prone to mistake.

Photosynthesis is a chemical reaction in which energy from sunlight is used to convert carbon dioxide and water into glucose and oxygen. This process occurs in the chloroplasts of plant cells and is essential for the survival of plants, as well as for many other organisms that depend on plants for food.

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 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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If you have 220 mL of a 3.6M solution of calcium hydroxide, 7.92g is the mass in gram of calcium hydroxide would be leftover if all the water was evaporated.

A body's mass is an inherent quality. Prior to the discoveries of the atom as well as particle physics, it was widely considered to be tied to the amount of matter within a physical body.

It was discovered that, despite having the same quantity of matter in theory, different atoms and elementary particles have varied masses. There are various conceptions of mass in contemporary physics that are theoretically different but physically equivalent.

Molarity = moles/ volume

volume = 220 / 100

            = 2.2L

3.6× 2.2 = moles

mass= 7.92g

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The lattice energy of MX is 459.2 kJ/mol.

The lattice energy (ΔH° lattice) of an ionic compound is the energy released when one mole of the solid is formed from its constituent gaseous ions under standard conditions. The lattice energy is calculated using the Born-Haber cycle, which involves several steps including atomization, ionization, dissociation, and sublimation energies.

The lattice energy is related to the Coulombic attraction between the oppositely charged ions in the solid. To calculate the lattice energy for MX, we can use the following equation:

ΔH° lattice = ΔH° sub + ΔH° ion + ΔH° diss + ΔH° formation

where ΔH° sub is the sublimation energy of M(s), ΔH° ion is the first ionization energy of M(g), ΔH° diss is the dissociation energy of X2(g), and ΔH° formation is the enthalpy of formation of MX(s).

Using the given data, we can calculate each of these values and substitute them into the equation to obtain the lattice energy. The final answer should be in units of kJ/mol.

ΔH° sub (M) = 107.3 kJ/mol

ΔH° ion (M) = 577.5 kJ/mol

ΔH° diss (X2) = 242 kJ/mol

ΔH° formation (MX) = -467.6 kJ/mol

ΔH° lattice = 107.3 + 577.5 + 242 + (-467.6) = 459.2 kJ/mol

As a result, MX has a lattice energy of 459.2 kJ/mol.

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The table that can depict the compound is given below

Sodium +1 Na+ Chloride -1 Cl-

Potassium +1 K+ Oxide -2 O2-

Calcium +2 Ca2+ Nitride -3 N3-

Magnesium +2 Mg2+ Sulfide -2 S2-

Ionic compounds are the compounds made up of ions that form charged particles when an atom gains or loses electrons. A cation is an ion charged positively; an anion is an ion charged negatively.

In this table, the first column lists the names of the ions, the second column shows their corresponding charges, and the third column displays their ion formulas.

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The volume occupied by the 0.75 mole of the gas at 295 K and a pressure of 1.5 atm is 12.11 L

From the question given above, the following data were obtained:

Ideal gas equation is written as shown below:

PV = nRT

When we input the given parameters, the volume occupied is obatined as:

1.5 × V = 0.75 × 0.0821 × 295

Divide both sides by 4.31

V = (0.75 × 0.0821 × 295) / 1.5

V = 12.11 L

Thus, from the above calculation, we can say that the volume occupied by the gas is 12.11 L

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During a scientific investigation on magnets you conduct research. The one that is not a reliable source of information for research is A cartoon with a superhero who has magnetic powers and the correct option is option 2.

In the Scientific investigation, primary sources are documents that provide full description of the original research. For example, a primary source would be a journal article where scientists describe their research on the human immune system.

A secondary source would be an article commenting or analyzing the scientists' research on the human immune system.

Thus, the ideal selection is option 2.

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The acid ionization steady of HCIO₂ (chlorous corrosive) can be spoken to by the taking after condition:

HCIO₂ + H₂O ⇌ H₃O⁺ + CIO₂⁻

The harmony steady expression for this response is:

Ka = [H₃O⁺][CIO₂⁻] / [HCIO₂]

The esteem of the acid ionization consistent (Ka) for HCIO₂ is roughly

1.1 x 10^-2 at 25°C.

It is critical to note that the esteem of Ka can change with temperature, and it is additionally subordinate on the dissolvable utilized.

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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 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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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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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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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  product of combining Aluminum & Hydroxide is option  C. AI(OH)3

Aluminum goes through a double displacement reaction when it reacts with hydroxide. The chemical element aluminum in its cationic form, Al3+, bonds with a set of three anionic hydroxide particles, OH-, resulting in the formation of a chemical compound identified as aluminum hydroxide (AI(OH)3).

The other options are inaccurate as AI(OH) is not a steady substance, and neither AIOH3 nor AI3OH are appropriate formulas for the compound aluminum hydroxide.

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The needed quantity of heat is also 11,627.2 cal.

Water is the sample that would take more heat to raise the temperature by 10.0°C.

7. To calculate the amount of heat required to increase the temperature of the iron block from 25°C to 1535°C, use the formula:

q = m x c x ΔT

where q is the amount of heat, m is the mass of the iron block, c is the specific heat of iron, and ΔT is the change in temperature.

Substituting the values given:

q = (75.0 g) x (0.449 J/g°C) x (1535°C - 25°C)

= 48,571.8 J

Therefore, the amount of heat required to increase the temperature of the iron block is 48,571.8 J.

To convert this to calories, we can use the conversion factor 1 cal = 4.184 J:

q = 48,571.8 J x (1 cal / 4.184 J)

= 11,627.2 cal

Therefore, the amount of heat required is also 11,627.2 cal.

8. To determine which sample would require more heat to raise the temperature by 10.0°C, use the formula:

q = m x c x ΔT

where q is the amount of heat, m is the mass of the sample, c is the specific heat of the sample, and ΔT is the change in temperature.

Substituting the values for water:

q(water) = (10.0 g) x (4.184 J/g°C) x (10.0°C)

= 418.4 J

Substituting the values for ethanol:

q(ethanol) = (10.0 g) x (2.44 J/g°C) x (10.0°C)

= 244.0 J

Therefore, the sample that would require more heat to raise the temperature by 10.0°C is water, since it has a higher specific heat than ethanol. Water can absorb more heat per unit mass than ethanol, which makes it more effective at storing and releasing heat.

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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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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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10.0 g of ethanol would require more heat to raise the temperature by 10.0°C Because it has a higher boiling point.

The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure of the liquid’s environment. At this temperature, the liquid is converted into a vapour.

The normal boiling point is high for liquids with strong intermolecular attractions and low for liquids with weak intermolecular attractions.

10.0 g of ethanol would require more heat to raise the temperature by 10.0°C because it has a higher boiling point.

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Claim 1: Yes, there probably will be a lunar eclipse of the moon of Kepler-47c during a year.

Claim 2: No, there probably won't be a lunar eclipse of the moon of Kepler-47c during a year.

Kepler-47c is a moon orbiting Kepler-47 AB, a binary star system. This implies that the moon's light varies on a regular basis as it circles the two stars. These changes in light might result in a lunar eclipse.

Lunar eclipses occur when the moon passes through the Earth's shadow. While Kepler-47c circles two stars, it is still subject to Earth's gravitational pull and may pass into its shadow.

Lunar eclipses occur on a regular basis, with two to four eclipses occurring on average per year. Given the frequency of lunar eclipses, Kepler-47c is likely to see at least one lunar eclipse every year.

Evidence to support this claim 2:

Lunar eclipses occur when the moon passes through the Earth's shadow. While Kepler-47c is in the Earth's gravitational field, it is not on the same orbital plane as the Earth.

While Kepler-47c circles two stars, its orbit is not known to be stable. Variations in Kepler-47c's orbit may reduce the likelihood of a lunar eclipse.

While lunar eclipses are common, they are not uniformly distributed throughout the year. The locations and motions of the Earth, moon, and sun govern the timing and frequency of lunar eclipses.

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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

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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.

Answer:

specific heat, the quantity of heat required to raise the temperature of one gram of a substance by one Celsius degree. The units of specific heat are usually calories or joules per gram per Celsius degrees.

Explanation:

the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

[tex][Zn^+^2] [Sn^+^2][/tex] is the most plausible source of this deviation. This is because neither the reaction quotient Q nor the conventional equilibrium constant K are equivalent under the present conditions of the reaction. Standard cell potential (E°), real cell potential (E), gas constant (R), temperature (T), Faraday constant (F), and reaction quotient (Q) are all related by the following formula:

E = E° - (RT/nF) ln Q

Q must be greater than K because E is greater than E°. The reaction quotient Q can be calculated by:

Q = [Sn]/[Zn + 2]

The measured voltage will be greater than the normal cell potential if [Zn+2] [Sn+2] and Q > K.

So, the correct option is C.

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

Sn2+ (aq) + Zn (s) → Sn (s) + Zn+2 (aq)

A galvanic cell was constructed using tin and zinc half cells. The voltage produced was measured to be 0.66 V at 25°C. However, the standard cell potential for this reaction is 0.62 V. Which of the following statements is the most likely source of this deviation?

(A) The salt bridge contained Na2SO4.

(B) SO42 was the anion at the anode and Cl was the anion at the cathode. (C) [Zn+2] < [Sn+2]

(D) The anode has a larger mass than the cathode.

(E) The volume of the Sn2 solution is greater than the volume of the Zn+2 solution.

Déterminer le réactif limitant (LR) en comparant les quantités de Ca et de P données :

17 litres de Ca = 17 000 mL de Ca (1 litre = 1000 mL), donc 17 000 mL / 40 g/mL = 425 g de Ca

2,93 x 10 atomes de P / 6,02 x 10^23 atomes/mol = 4,87 x 10^-24 mol de P

Masse molaire de Ca3P2 = 182,2 g/mol

Pour calculer le LR, nous devons convertir les quantités de Ca et de P en moles, en utilisant les coefficients stœchiométriques de l'équation :

Moles de Ca = 425 g / 40 g/mol = 10,6 mol

Moles de P = 4,87 x 10^-24 mol

Pour déterminer la quantité de CasP théorique qui peut être produite, nous devons d'abord calculer la quantité de Ca qui réagit avec le P :

Moles de P nécessaires = 2/3 x 10,6 mol = 7,1 mol

Moles de Ca nécessaires = 3/2 x 7,1 mol = 10,6 mol (car le rapport stœchiométrique est de 3 Ca pour 2 P)

Masse de Ca3P2 produite = 10,6 mol x 182,2 g/mol = 1932 g

Le rendement théorique (TY) de la réaction est de 1932 g de Ca3P2.

Pour calculer le rendement en pourcentage de la réaction réelle, il faut connaître la quantité réelle de Ca3P2 produite.

Si 3,44 litres de Ca3P2 ont été produits, cela équivaut à 3,44 x 10^6 mL x 1,82 g/mL = 6,26 kg.

Le rendement réel (RY) est donc :

RY = (masse réelle de Ca3P2 produite / masse théorique de Ca3P2 produite) x 100%

RY = (6,26 kg / 1,932 kg) x 100% = 323,7%

Le rendement réel est supérieur à 100%, ce qui indique que des erreurs ont été commises lors de la mesure de la quantité de produit ou de réactif, ou qu'il y a eu des pertes pendant la manipulation ou le processus de purification.