During the titration, a student pulls out the pH electrode from the titration beaker several times (with about 0.25 mL of solution on it each time) and rinses it off with DI water into a waste container. Will this affect the measured equivalent mass? If so, will the equivalent mass come out higher or lower?

Answer:

Problem 1:

The osmotic pressure, π, can be calculated using the formula:

π = iMRT

where i represents the van't Hoff factor (the number of particles into which a solute dissociates), M represents the molar concentration, R represents the gas constant (0.082 Latm/molK), and T is the temperature in Kelvin.

The osmotic pressure of pure water is 0. As a result, sea water's osmotic pressure must be equal to the pressure necessary to prevent osmosis.

Assuming that sea water is an ideal solution, the total dissolved ion concentration is 1.13 mol/L. Because each dissolved salt molecule dissociates into two ions, the effective particle concentration is 2.26 mol/L.

Filling in the blanks in the formula:

0.918 atm = (2)(2.26 mol/L)(0.082 Latm/molK)(298 K)

Therefore, a pressure of 0.918 atm must be applied to prevent osmotic flow of pure water into sea water.

Problem 2: Glucose has a molar mass of 180.16 g/mol. The solution contains the following number of moles of glucose:

n = 6.65 g / 180.16 g/mol = 0.0369 mol

The molarity of the solution is:

M = n / V = 0.0369 mol / 0.350 L = 0.105 M

Substituting the values into the formula:

π = iMRT = (1)(0.105 M)(0.082 L·atm/mol·K)(308 K) = 2.74 atm

As a result, the osmotic pressure of the solution is 2.74 atm.

Problem 3:

Following the same procedure as in Problem 2, the molarity of the solution is:

M = n / V = 0.02 mol / 0.450 L = 0.044 M

Substituting the values into the formula:

π = iMRT = (1)(0.044 M)(0.082 L·atm/mol·K)(308 K) = 1.14 atm

As a result, the osmotic pressure of the solution is 1.14 atm.

Problem 4:

The molar mass of propanol is 60.10 g/mol. The number of moles of propanol in the solution is:

n = 11.0 g / 60.10 g/mol = 0.183 mol

The molarity of the solution is:

M = n / V = 0.183 mol / 0.850 L = 0.215 M

Substituting the values into the formula:

π = iMRT = (1)(0.215 M)(0.082 L·atm/mol·K)(298 K) = 4.59 atm

Therefore, the osmotic pressure of the solution is 4.59 atm.

Problem 5:

Following the same procedure as in Problem 2, the molarity of the solution is:

M = n / V = 65 g / (180.16 g/mol × 35 L) = 0.104 M

Substituting the values into the formula:

π = iMRT = (1)(0.104 M)(0.082 L·atm/mol·K)(288 K) = 2.06 atm

Therefore, the osmotic pressure of the solution is 2.06 atm.

Physical methods include monitoring temperature, pressure, and color change. Chemical methods include titration and gas analysis.

Monitoring the rate of chemical reactions is important to understand the kinetics of the reaction and optimize the reaction conditions. Physical and chemical methods are used for this purpose.

Physical methods include measuring the change in temperature, volume, and pressure of the reactants and products with time. The rate of reaction can be calculated from the rate of change of these parameters.

Chemical methods include monitoring the concentration of reactants and products with time. This can be done by techniques such as spectroscopy, chromatography, and electrochemistry. These methods are often more accurate and precise than physical methods.

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The order of the resonance structures is C > A > B.

In the order of thermodynamic stability we have A > C >D >B

The compounds have one stereogenic center.

The average of all potential resonance structures for a molecule or ion is represented by the hypothetical molecule known as a resonance hybrid.

When a molecule or ion may be described by two or more Lewis structures that only differ in the arrangement of electrons, resonance occurs. In other words, resonance structures have the same atomic arrangement but have different electron distributions.

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The density, in g/L, of an ideal gas when it is at 1.48 atm and 94.06 °C is 6.67g/L.

The density of an ideal gas can be calculated by dividing the mass of the substance by its volume in litres.

According to this question, the pressure and temperature of the ideal gas is given. The number of moles occupied by the gas can be calculated as follows;

PV = nRT

1.48 × 22.4 = n × 0.0821 × 367.06

33.152 = 30.14n

n = 1.1 moles

mass of gas = 1.1 mol × 145.63g/mol = 160.18g

Density = 160.18g ÷ 22.4L = 6.67g/L

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The process used 4.86 moles of hydrogen gas to generate 3.24 moles of ammonia gas.

According to the balanced chemical equation, the stoichiometry of the reaction shows that 3 moles of hydrogen gas (H₂) react with 1 mole of nitrogen gas (N₂) to produce 2 moles of ammonia gas (NH₃).

So, for every 2 moles of NH₃ produced, we need 3 moles of H₂ consumed. Therefore, to determine the moles of H₂ consumed, set up a proportion:

3 moles H₂ / 2 moles NH₃ = x moles H₂ / 3.24 moles NH₃

where x is the number of moles of H₂ consumed.

Solving for x:

x = (3 moles H₂ / 2 moles NH₃) x (3.24 moles NH₃) = 4.86 moles H₂

Therefore, 4.86 moles of hydrogen gas were consumed in the reaction to produce 3.24 moles of ammonia gas.

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The friend whose statement I agree with most about the cycling of carbon dioxide and oxygen in the ecosystem is Flynn.

I agree with Flynn's statement about the cycling of carbon dioxide and oxygen in the ecosystem.

Animals do indeed take in oxygen and release carbon dioxide during respiration, while plants take in carbon dioxide and release oxygen during photosynthesis. This process of exchanging gases between plants and animals is known as the carbon cycle, which is crucial for the survival of all living things.

The carbon cycle ensures that there is a balance between the amount of carbon dioxide in the atmosphere and the amount that is used up by living organisms. In addition, the cycling of other nutrients such as nitrogen and phosphorus is also important for the functioning of the ecosystem.

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If a beaker contains 15.6 moles of water, then it  represents 9.39 × 10²⁴ molecules of water, 0.317 moles of PbO contains approximately 1.91 × 10²³ formula units of PbO,  6.01 × 10²⁵ atoms of cesium is equivalent to 99.7 moles of cesium, and 3.54 × 10²¹ molecules of sulfur dioxide represents approximately 5.88 × 10⁻³ moles of sulfur dioxide.

If a beaker contains 15.6 moles of water (H₂O), we can calculate the number of molecules using Avogadro's number, which is approximately 6.022 × 10²³ molecules/mol.

Number of molecules = Number of moles × Avogadro's number

Number of molecules = 15.6 moles × 6.022 × 10²³ molecules/mol

Let's plug in the value and calculate;

Number of molecules = 15.6 moles × 6.022 × 10²³ molecules/mol

Number of molecules ≈ 9.39 × 10²⁴ molecules

So, 15.6 moles of water represents approximately 9.39 × 10²⁴ molecules of water.

The number of formula units of PbO (lead(II) oxide) in 0.317 moles of PbO can be calculated using Avogadro's number, which is approximately 6.022 × 10²³ formula units/mol.

Number of formula units = Number of moles × Avogadro's number

Number of formula units = 0.317 moles × 6.022 × 10²³ formula units/mol

Let's plug in the value and calculate;

Number of formula units = 0.317 moles × 6.022 × 10²³ formula units/mol

Number of formula units ≈ 1.91 × 10²³ formula units

The number of moles of cesium (Cs) equivalent to 6.01 × 10²⁵ atoms of cesium can be calculated using Avogadro's number, which is approximately 6.022 × 10²³ atoms/mol.

Number of moles = Number of atoms / Avogadro's number

Number of moles = 6.01 × 10²⁵ atoms / 6.022 × 10²³ atoms/mol

Let's plug in the value and calculate;

Number of moles = 6.01 × 10²⁵ atoms / 6.022 × 10²³ atoms/mol

Number of moles ≈ 99.7 moles

So, 6.01 × 10²⁵ atoms of cesium is equivalent to approximately 99.7 moles of cesium.

The number of moles represented by 3.54 × 10²¹ molecules of sulfur dioxide (SO₂) can be calculated using Avogadro's number, which is approximately 6.022 × 10²³ molecules/mol.

Number of moles = Number of molecules / Avogadro's number

Number of moles = 3.54 × 10²¹ molecules / 6.022 × 10²³ molecules/mol

Let's plug in the values and calculate;

Number of moles = 3.54 × 10²¹ / 6.022 × 10²³

Number of moles ≈ 5.88 × 10⁻³ moles

So, 3.54 × 10²¹ molecules of sulfur dioxide represents approximately 5.88 × 10⁻³ moles of sulfur dioxide.

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There are several physical methods that can be used to monitor the rate of a chemical reaction are; Spectrophotometry, Conductometry, and Turbidity measurement

Spectrophotometry involves measuring the changes in the intensity of light absorbed or transmitted by a solution during a chemical reaction. Spectrophotometers are used to measure the amount of light absorbed or transmitted by a sample at different wavelengths.

Conductometry  involves measuring the changes in electrical conductivity of a solution during a chemical reaction. Conductivity meters are used to measure the electrical conductivity of a solution, which can change as the concentration of ions in the solution changes during a chemical reaction.

Turbidity measurement involves measuring the changes in the clarity or turbidity of a solution during a chemical reaction. Turbidimeters or nephelometers can be used to measure the amount of light scattered by a sample, which can change as particles form or dissolve during a reaction.

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--The given question is incomplete, the complete question is

"What are the physical methods of monitoring the rate of a chemical reaction?"--

The frequency of the electromagnetic wave is 8.1 x 10 ¹⁸ Hz.

The speed of light in a vacuum is given as 3.00 x 10⁸ m/s. The speed of light is also related to the wavelength and frequency of the electromagnetic wave by the equation:

c = λν

where c is the speed of light, λ is the wavelength, and ν is the frequency.

Rearranging the equation to solve for frequency, we get:

ν = c/λ

Substituting the values given in the problem, we get:

ν = (3.00 x 10⁸ m/s) / (3.7 x 10⁻¹¹ m)

ν = 8.1 x 10 ¹⁸ Hz

Therefore, the frequency of the electromagnetic wave is 8.1 x 10 ¹⁸ Hz, and the correct answer is (A) 8.1 x 10 ¹⁸ Hz.

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Since it reflects no change in volume – it is most likely that the data recorded in row B is erroneous.

In order to pinpoint the row with a probable error, we will scrutinize the correlation between temperature and volume.

Typically, whenever there is an increase in temperature, there would be an accompanying escalation in the volume of a given substance.

Let's take a closer look at the values reflected in the table:

From A to B: Temperature rises but the volume remains stagnant.

From B to C: Both temperature and volume show an increment.

Similarly, from C to D, as well as from D to E, we see that for every rise in temperature, there is subsequently more volume illustrated.

Given the dissimilarity between rows A and B – where the former shows a temperature increase while the latter reflects no change in volume – it is most likely that the data recorded in row B is erroneous.

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Which row in the table below is likely to represent an error in the student's data collection?

Temperature (°C) Volume (cm³)

A 2.4 5.8

B 21.5 5.8

C 39.6 6.7

D 55.6 6.9

E 71.2 7.4

Answer:

A student collected the data shown above. Row

✔ B

may represent an error.

Explanation:

edge 2023

The pressure in atm, in a 4.00 L tank with 5.15 moles of nitrogen gas at 69.6˚C is 36.18 atm.

According to the ideal gas law,

PV = nRT

Here,                  

Volume (V) = 4.00 L

Number of moles (n) = 5.15 mol

Temperature (T) = 69.6 ˚C = 69.6+273.15 = 342.75 K

Universal gas constant (R) = 0.082 L atm mol-1 K-1

Pressure (P) =  ?

Therefore, mathematically

 P × 4.00 = 5.15 × 0.082 × 342.75

            P  =   [tex]\frac{(5.15)(0.082)(342.75)}{4.00}[/tex]

            P  =   [tex]\frac{144.74}{4.00}[/tex]

            P  =  36.18 atm

The pressure in atm, in a 4.00 L tank with 5.15 moles of nitrogen gas at 69.6˚C is 36.18 atm.

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There are 5.79 x 10²¹ representative particles in 2.62g of the molecular compound.

Determine the number of moles of the molecular compound.

We can use the formula:

moles = mass / molar mass

where mass is 2.62g and molar mass is 273g/mol.

moles = 2.62g / 273g/mol

moles = 0.00961 mol

Use Avogadro's number to convert from moles to representative particles.

We can use the formula:

representative particles = moles x Avogadro's number

where Avogadro's number is 6.022 x 10²³.

representative particles = 0.00961 mol x 6.022 x 10²³

representative particles = 5.79 x 10²¹

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Both 238U and 232Th are naturally occurring radioactive isotopes found in the Earth's crust, and they undergo radioactive decay by emitting alpha particles. However, the decay chain of 238U includes the isotope 222Rn, which is a gas and can easily be inhaled into the lungs. This presents a greater threat of lung cancer because the alpha particles emitted by 222Rn and its decay products can damage the cells in the lining of the lungs.

Furthermore, 222Rn has a relatively short half-life of 3.8 days, which means that it decays relatively quickly into other radioactive isotopes, including polonium-218 and lead-214, which are also alpha emitters. This continuous decay chain can create a buildup of radioactive particles in the lungs, increasing the risk of lung cancer.

On the other hand, the decay chain of 232Th does not include any gas isotopes, and its decay products tend to be heavy metals rather than alpha emitters. This means that the decay products are less likely to be inhaled into the lungs, and even if they are, they are less likely to cause as much damage as alpha particles.

In summary, the natural underground decay of 238U poses a greater threat of lung cancer than the decay of 232Th because it includes a gas isotope (222Rn) that can easily be inhaled into the lungs, and its decay products are alpha emitters that can cause significant damage to the cells in the lining of the lungs.

0.5 moles of lithium chloride are dissolved in .05 liters of water. 10M  is the molarity of the solution.

The total amount of moles of solute found within a specific number of litres of the solution, or moles per litre of a solution, is known as molar concentration or molarity. Please explain the difference amongst the terms "solute" and "solvent" before we continue.

'Solution' for making it simpler to comprehend the topics that will follow. Solutes are simply substances that exist in solutions because a solution is defined as a homogenous mixture that comprises one or more solutes.

Molarity = moles/volume of solution in liter

              = 0.5/ .05

               = 10M

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0.01 mol/L is the molarity of given solution. 0.002 moles is added to 0.2 L solvent to make desired solution.

The amount of moles of solute found in a specific number of litres of the solution, or moles per litre of a solution, is known as molar concentration or molarity. Solutes are simply substances that can be found in solutions because a solution is defined as a homogenous mixture that comprises one or more solutes.

molar mass =342g /mol

number of moles=mass of solute / molar mass

0.45 /342 =0.002 moles

Volume solution =  50 mL / 1000 =0.2 L

M = n / V

M = 0.002 / 0.2

M = 0.01 mol/L

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There are 0.2107 moles of KBr are dissolved in 60.2 mL of a 3.50 M solution.

The molarity of a substance is defined as the number of moles of solute present in 1 litre of a solution.

According to the given data, the molarity of the solution tells us that there are 3.50 moles of KBr in 1000mL of solution. But we only have 60.2mL of solution, so with a mathematical rule of three we can calculate the amount of moles in 60.2mL:

1000 ml - 3.50 moles

60.2 ml -x = 60.2 ml× 3.50 moles/1000 ml

x= 60.2 ml -0.2107 moles

So, there are 0.2107 moles of KBr.

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The volume in which the weather ballon bursts is 640.21L.

The volume of a gas given the temperature and pressure can be calculated using the combined gas law as follows:

PaVa/Ta = PbVb/Tb

Where;

According to this question, a weather balloon is filled with 28.6 L helium at sea level where the pressure is 1.00 atm at 20.0 °C.

1 × 28.6/293 = 0.034 × Vb/223

0.0976109215 × 223 = 0.034Vb

Vb = 21.767 ÷ 0.034

Vb = 640.21L

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The efficiency of a chemical reaction is measured as a percentage yield, which shows what fraction of the theoretical yield was actually obtained in the laboratory. It is calculated by multiplying by 100% and dividing the theoretical yield (the largest amount of product that can be obtained under perfect conditions) by the actual yield (amount of product obtained in the experiment).

The theoretical yield of CasP₂ from question #6 was calculated to be 0.143 moles.

To calculate the percent yield, we use the formula:

Percent yield = (actual yield/theoretical yield) x 100%

Substituting the given values:

Percent yield = (0.11/0.143) x 100%

Percent yield = 76.92%

Therefore, the percent yield of CasP₂ is 76.92%.

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The amount of volume that is needed to completely react with 4.85 grams of calcium carbonate is 32 mL.

The amount of volume needed to complete a chemical reaction can be calculated using the following formula;

molarity = no of moles ÷ volume

According to this question, hydrochloric acid reacts with calcium carbonate to produce calcium chloride, water and carbondioxide.

2 moles of HCl reacts with 1 mole of calcium carbonate. 4.85g of calcium carbonate is equivalent to 0.0485 moles.

0.0485 moles of calcium carbonate will react with 0.097 moles of HCl.

volume = 0.097 mol ÷ 3M = 32 mL

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It would be easier to communicate information about mass to extraterrestrial civilizations.

Mass and weight are two different concepts that are often used interchangeably in everyday language.

Mass is a measure of the amount of matter in an object and is measured in kilograms (kg) or grams (g). Weight, on the other hand, is a measure of the force exerted on an object due to gravity and is measured in newtons (N).

If we were to communicate with an extraterrestrial civilization, it would be easier to communicate information about mass rather than weight because mass is an intrinsic property of an object that does not depend on gravity.

Therefore, mass would be more universal and easier to understand by extraterrestrial civilizations.

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Under these conditions, the cell potential, Ecell, would be somewhat lower than the usual cell potential, E°cell.

To determine whether the cell potential, Ecell, would be greater than, less than, or equal to the standard cell potential, E°cell, use the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

where:

Ecell = cell potential under the given conditions

E°cell = standard cell potential

R = gas constant (8.314 J/(mol·K))

T = temperature in Kelvin (25°C = 298 K)

n = number of electrons transferred in the overall cell reaction

F = Faraday constant (96,485 C/mol)

Q = reaction quotient

The balanced equation for the cell reaction is:

2Al(s) + 3Sn₂⁺(aq) → 2Al₃⁺(aq) + 3Sn(s)

The cell reaction involves the transfer of 3 electrons, so n = 3.

At standard conditions (1 M concentration for all species and 25°C), the cell potential, E°cell, can be calculated using the standard reduction potentials for the half-reactions:

Al₃⁺(aq) + 3e- → Al(s) E° = -1.66 V

Sn₂⁺(aq) + 2e- → Sn(s) E° = -0.14 V

E°cell = E°(cathode) - E°(anode) = (-0.14 V) - (-1.66 V) = 1.52 V

To calculate Q, use the given concentrations of Al3+ and Sn2+:

Q = ([Al₃⁺]²/[Sn⁺]³) = (1.90 M)² / (0.25 M)³ = 231.2

Now use the Nernst equation to calculate Ecell under the given conditions:

Ecell = E°cell - (RT/nF)ln(Q)

Ecell = 1.52 V - [(8.314 J/(mol·K))(298 K)/(3 mol)(96,485 C/mol)ln(231.2)]

Ecell = 1.52 V - (0.012 V) = 1.508 V

Therefore, the cell potential, Ecell, under these conditions would be slightly less than the standard cell potential, E°cell. This is because the concentration of Al₃⁺ is higher and the concentration of Sn₂⁺is lower than the standard conditions, which shifts the reaction towards the side with lower concentration of Al₃⁺ and higher concentration of Sn₂⁺. This means that the reaction is not occurring under standard conditions and the Nernst equation must be used to account for the non-standard conditions.

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Answer:

If the substance has a half-life of 14 days, then we know that after every 14 days, the amount of the substance remaining will be divided by two.

Since 70 days is five half-lives (70 ÷ 14 = 5), we can find the amount of substance remaining after 70 days by dividing the initial amount by 2 five times.

Amount remaining = 10,000 grams / (2^5) = 10,000 grams / 32

Amount remaining = 312.5 grams

Therefore, after 70 days, you will have 312.5 grams of the substance remaining.

The difference in lengths between the steel and copper bars at 41.0°C is 8.729 mm.

The difference in the lengths of the two bars can be calculated using the linear expansion equation;

ΔL = α × L × ΔT

where; ΔL = Difference in length

α = Coefficient of linear expansion

L = Initial length

ΔT = Change in temperature

Given; Initial length (L) = 1.500 m

Change in temperature (ΔT) = 41.0°C - (-12.0°C) = 53.0°C

The coefficient of linear expansion for steel is typically around 11 x 10⁻⁶ /°C, and for copper, it's around 16 x 10⁻⁶ /°C.

For the steel bar;

α (steel) = 11 x 10⁻⁶ /°C

For the copper bar;

α (copper) = 16 x 10⁻⁶ /°C

Now we can calculate the difference in lengths for both bars.

For the steel bar;

ΔL (steel) = α (steel) × L × ΔT

ΔL (steel) = 11 x 10⁻⁶ /°C × 1.500 m × 53.0°C

For the copper bar;

ΔL (copper) = α (copper) × L × ΔT

ΔL (copper) = 16 x 10⁻⁶ /°C × 1.500 m × 53.0°C

Converting the length difference from meters to millimeters (1 m = 1000 mm):

ΔL (steel) = 11 x 10⁻⁶ /°C × 1.500 m × 53.0°C × 1000 mm/m = 8.729 mm (rounded to three decimal places)

ΔL (copper) = 16 x 10⁻⁶ /°C × 1.500 m × 53.0°C × 1000 mm/m = 12.168 mm (rounded to three decimal places)

Therefore, the difference in the lengths of the two bars will be 8.729 mm.

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Answer:  

What is a double replacement reaction? Double replacement reactions—also called double displacement, exchange, or metathesis reactions—occur when parts of two ionic compounds are exchanged, making two new compounds.

Explanation:

Only comb jellies whose genes mutate to no longer glow will survive and pass on this trait. This describes the most probable change in the comb jelly population over time. Therefore, the correct option is option A.

The term "population" is frequently used to describe the total number of people living in a particular location. To estimate the number of the resident population inside a certain territory, governments conduct censuses. The phrase has particular use in the domains of ecology and genetics and is also used to refer to plants, animals, and microbes.

A population is frequently referred to as a group of organisms in genetics where any two individuals can breed with each other. A breeding group known as a gamodeme is one that may routinely exchange gametes to create children who are normally viable. Only comb jellies whose genes mutate to no longer glow will survive and pass on this trait. This describes the most probable change in the comb jelly population over time.

Therefore, the correct option is option A.

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When we dissolve 123g KBr into 689g of water, the mass percent is 15.1 %.

The mass per cent of a solution is defined as the mass of solute in grams per grams of solution, multiplied by 100 so as to get the mass percentage.

The formula of mass per cent  is expressed as solving for the molar mass and for the mass of each element present in 1 mole of the compound.

The mass of the solution

= mass of solute (KBr) + mass of solvent (water)

= 123 g + 689 g

= 812 g

123 g KBr present in 812 g solution

Let, X be present in  100 g solution

X = 100 g solution x 123 g KBr/812 g solution

   = 15.1 g KBr

   = 15.1 % by mass

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a) The solute in solution A is Sodium (Na)

b) The percent concentration of solution A is 4.72 %

Solute is that component of the solution which gets dissolved therefore here sodium is the solute and water is the solvent. Solute is also the minor component of a solution, the component which is present in less amount.

Percent concentration is simply the grams of solute present in 100 grams of a solution.

To calculate the percent concentration, the formula given below could be used

% concentration = [tex]\rm \dfrac{ Mass \ of \ solute}{Mass \ of \ solution} \times 100[/tex]

                           = [tex]\frac{460}{97.4}[/tex]

                           = 4.72 %

Therefore,

a) The solute in solution A is Sodium (Na)

b) The percent concentration of solution A is 4.72 %

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