In nature, the element X consists of two naturally occurring isotopes. 107X with abundance 53.84% and isotopic mass 106.9051 amu and 109X with isotopic mass 108.9048 amu. Use the given information to calculate the atomic mass of the element X to an accuracy of .001% (Report your answer like this yyy.yyyy)

Answer: carbonyl group C=O

Explanation:

Formaldehyde is an organic compound, it is the simplest form of Aldehydes. It formula is CH2O and has a carbonyl functional group, C=O. The general formula for adehydes is R-COH. The carbon atom is bonded to oxygen with a double bond and one of the two remaining bonds is occupied by hydrogen, and the other by an alkyl group.

One of the properties of adehydes is their solubility in water. The lower members (up to 4 carbons) of aldehydes are soluble in water due to H-bonding. Ofcourse the the higher members are not soluble in water because their hydrophobic long chains.

Aldehydes contain carbonyl group, therefore they undergo reactions like nucleophilic addition reactions, oxidation, reduction, halogenation.

Answer:

[tex]\beta=63.85^{\circ}[/tex] from the vertical OR [tex]16.15^{\circ}[/tex] from horizontal.

Explanation:

Given:

Thrust of launching the rocket, [tex]F=5\times 10^6\ N[/tex]

angle of launch from the horizontal, [tex]\theta=37^{\circ}[/tex]

mass of the rocket, [tex]m=200000\ kg[/tex]

Now the direction of acceleration due to the thrust force is in the direction of force:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{5000000}{200000}[/tex]

[tex]a=25\ m.s^{-2}[/tex]

And the acceleration due to gravity is always directed towards the center of the earth i.e. vertically downwards.

Since acceleration is a vector quantity we, approach accordingly:

[tex]\tan\beta=\frac{a\cos\theta}{g}[/tex]

[tex]\tan\beta=\frac{25\cos37^{\circ}}{9.8}[/tex]

[tex]\beta=63.85^{\circ}[/tex] from the vertical OR [tex]16.15^{\circ}[/tex] from horizontal.

Answer:

The amount of sucrose that must be added is 1.66 moles

Explanation:

Colligative property of lowering vapor pressure has this formula:

Vapor pressure of pure solvent (P°) - Vapor pressure of solution = P° . Xm

We have both vapor pressure (pure solvent and solution9, so let's determine the ΔP

ΔP = 92.6 Torr - 72 Torr = 20.6 Torr

Let's add the data in the formula

20.6 Torr = 92.6 Torr . Xm

Xm = Mole fraction of solute → (mol of solute/ mol of solute + mol of solvent)

Mol of solvent = 5.83 mol (data from the problem)

Therefore Xm = 20.6 Torr / 92.6 Torr → 0.222

Let's find out the moles of solute (our unknown value)

0.22 = moles of solute / moles of solute + 5.83 moles of solvent

0.222 (moles of solute + 5.83 moles of solvent) = moles of solute

0.222 moles of solute + 1.29 moles of solvent = moles of solute

1.29 moles of solvent = moles of solute - 0.222 moles of solute

1.29 moles = 0.778 moles of solute

1.29 / 0.778 = moles of solute → 1.66 moles

Answer:

The answer to this is the concentration of glucose (C6H12O6) in mg/dL = 95.48 mg/dL

Explanation:

To solve this we list out the known variables thus

Measured concentration of  glucose  (C6H12O6) = 5.3mM per liter

The molar mass of glucose  = 180.156 g/mol

From the above, it is seen that one mole of glucose contains 180.156 grams of  C6H12O6 therefore 5.3 mM which is 5.3 × 10⁻³ moles contains

5.3 × 10⁻³ moles × 180.156 g/mol = 0.9548 grams of glucose

Also 1 d L = 0.1 L  or 1 L = 10 dL and 1 mg = 1000 g, hence

thus 0.9548 grams per liter is equivalent to 1000/10 × 0.9548 milligrams per dL or 95.48 mg/dL

Answer:

option d

Explanation:

Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.

Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.

Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’

Calcium carbonate

Explanation:

The chemical most commonly used to counteract the effects of acid precipitation on aquatic ecosystems is calcium carbonate.

Acid rain or acid deposition or acid precipitation is any form of precipitation with an elevated level of hydrogen ion concentration in them.

To nullify this acidic precipitation in aquatic ecosystem, we need to use an environmentally friendly alkaline agent.

The most desired is calcium carbonate. The carbonate neutralizes the acid by producing carbon dioxide, water and calcium salts.

Answer:

True

Explanation:

This are acts or actions that concur with situational expectations.

Answer:

                      ΔT  =  20.06 °C

Explanation:

The equation used for this problem is as follow,

                                    Q  =  m Cp ΔT   ----- (1)

Where;

           Q  =  Heat  =  1.17 kJ = 1170 J

           m  =  mass  =  24.1 g

           Cp  =  Specific Heat Capacity  =  2.42 J.g⁻¹.°C⁻¹

           ΔT  =  Change in Temperature  =  ??

Solving eq. 1 for ΔT,

                                ΔT  =  Q / m Cp

Putting values,

                                ΔT  =  1170 J / 24.1 g × 2.42 J.g⁻¹.°C⁻¹

                                ΔT  =  20.06 °C

The net ionic equation for the reaction of strontium chloride and potassium sulfate, forming strontium sulfate solid, is Sr2+(aq) + SO42-(aq) → SrSO4(s)

When an aqueous solution of strontium chloride is mixed with an aqueous solution of potassium sulfate, strontium sulfate precipitates out and potassium and chloride ions remain in the solution. The balanced net ionic equation for this reaction is as follows:

Sr2+(aq) + SO42-(aq) → SrSO4(s)

This equation represents the change where strontium ions from strontium chloride and sulfate ions from potassium sulfate are combined to form strontium sulfate solid. The charges are balanced with the positive 2 charge of the strontium ion balancing the negative 2 charge of the sulfate ion.

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Answer:

10mol

Explanation:

3H2 + N2 -> 2NH3

Stoichiometry is a tool that chemists can use to find the amount of substance present in any part of a reaction. The arrow (->) suggests that the reaction goes to completion (100%), so assume that left side = right side.

3H2

15 mol

You can divide the amount of moles by the coefficient to find the number of moles when you have a coefficient of 1. This number can then be used to find the value of moles for the rest of the products/reactants:

15/3=5mol

NH3 has a coefficient of 2, so we have to multiply the value we got (5mol) by 2. This results in having 10mol of ammonia as the end result.

Final answer:

The production of ammonia from hydrogen and excess nitrogen follows a 3:2 mole ratio, according to the balanced equation N2(g) + 3H2(g) → 2NH3(g). From 15 moles of hydrogen, 10 moles of ammonia are produced.

Explanation:

The question asks about the production of ammonia (NH3) from hydrogen (H2) in the presence of excess nitrogen (N2) based on the chemical reaction provided. According to the stoichiometry of the balanced equation, N2(g) + 3H2(g) → 2NH3(g), there is a 3:2 mole ratio between hydrogen and ammonia. Therefore, for every 3 moles of hydrogen, 2 moles of ammonia are produced.

To calculate the amount of ammonia produced from 15 mol of hydrogen, we use the mole ratio from the balanced equation. Since 3 moles of hydrogen produce 2 moles of ammonia, the amount of ammonia produced from 15 moles of hydrogen can be found using cross-multiplication:

(2 mol NH3) / (3 mol H2) = (x mol NH3) / (15 mol H2)

x = (15 mol H2 × 2 mol NH3) / 3 mol H2

x = 10 mol NH3

The answer is that 10 moles of ammonia can be produced from 15 moles of hydrogen reacting with excess nitrogen.

Answer:

The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point

Explanation:

To solve the above question we have the given variable as follows

ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole

However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.

The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles

Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ

Answer:

[tex]-5.226\times 10^{-7} C[/tex] is the charge on each sphere.

Explanation:

Coulomb's law is given as ;

[tex]F=K\times \frac{q_1\times q_2}{r^2}[/tex]

[tex]q_1,q_2[/tex] = Charges on both charges

r = distance between the charges

K = Coulomb constant =[tex]9\times 10^{9} N m^2/C^2[/tex]

We have ;

Charge of ion =[tex]q_1=-q[/tex]

Charge of electron =[tex]q_2=-q[/tex]

[tex]r=26.55 cm =0.2655 m[/tex]

Force between the charges  at r distance will be :  F

F = 0.03500 N

[tex]0.03500 N=9\times 10^{9} N m^2/C^2\times \frac{(-q)\times (-q)}{(0.2655 m)^2}[/tex]

[tex]q=5.226\times 10^{-7} C[/tex]

[tex]-5.226\times 10^{-7} C[/tex] is the charge on each sphere.

Answer: The correct statement is, It behaves as the electron donor.

Explanation:

According to the Lewis concept:

A Lewis-acid is defined as a substance that accepts electron pairs.

A Lewis-base is defined as a substance which donates electron pairs.

For example : Acid + Base ⇄ Acid-base adduct

[tex]H^++NH_3\rightleftharpoons NH_4^+[/tex]

As per question, the characteristic of a Lewis base is that it behaves as the electron donor.

Hence, the correct statement is, It behaves as the electron donor.

Answer:

A

Explanation:

It behaves as the electron donor

Answer: A. Perform a gravity filtration with a stemless funnel.

Explanation: Gravity filtration with a stemless funnel prevent blocking up the funnel. Undissolved sample may come out in the stem when the solution cools down thus will be blocking the funnel. Using stemless funnel prevent this problem.

Answer:

1. 8.7moles of H2

2. 2.25moles of O2

Explanation:

1. 2NH3 —> N2 + 3H2

From the equation,

2moles of NH3 produce 3 moles of H2.

Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e

Xmol of H2 = (5.8x3)/2 = 8.7moles

2. C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

5moles of O2 produced 4moles of H2O.

Therefore, Xmol of O2 will produce 1.8mol of H2O i.e

Xmol of O2 = (5x1.8)/4 = 2.25moles

Answer:

molecules

Explanation:

Answer:

The rate at which [tex]P_4[/tex] is being produced is 0.0228 M/s.

The rate at which [tex]PH_3[/tex] is being consumed is 0.0912 M/s.

Explanation:

[tex]4PH_3\rightarrow P_4(g)+6H_2(g)[/tex]

Rate of the reaction : R

[tex]R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

The rate at which hydrogen is being formed = [tex]\frac{d[H_2]}{dt}=0.137 M/s[/tex]

[tex]R=\frac{1}{6}\frac{d[H_2]}{dt}[/tex]

[tex]R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s[/tex]

The rate at which [tex]P_4[/tex] is being produced:

[tex]R=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

[tex]0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]

The rate at which [tex]PH_3[/tex] is being consumed :

[tex]R=\frac{-1}{4}\frac{d[PH_3]}{dt}[/tex]

[tex]0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}[/tex]

[tex]\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s[/tex]

The rate at which P4 is being produced is 0.034 M/s and the rate at which PH3 is being consumed is 0.2055 M/s.

The given reaction is 4PH3(g) → P4(g) + 6H2(g). We are given the rate at which molecular hydrogen is being formed, which is 0.137 M/s. To find the rate at which P4 is being produced, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of PH3 consumed, 1 mole of P4 is produced. Therefore, the rate at which P4 is being produced is 0.137/4 or 0.034 M/s.

Similarly, to find the rate at which PH3 is being consumed, we can use the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of PH3 consumed, 6 moles of H2 is produced. Therefore, the rate at which PH3 is being consumed is (6/4) * 0.137 or 0.2055 M/s.

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Answer:

Explanation:

If our entire Solar System were the size of a quarter, the planets and the sun is now a tiny speck of dust. The flat disc of the coin can represent the orbits of the planets.

Using this scale, the diameter of our Milky Way galaxy will be about the size of North America.

The nearest star, other than our own Sun, is about four light years away. That means it takes four years for its light to reach us. Since light travels at a speed of 3.0 x 10^8 meters per second, each light year is such a great distance. Proxima Centauri Milky way would be another quarter, two soccer fields away.

A much further star, Deneb is actually 1,800 light years away, the nearest star would be about 24,000 miles away.

Answer:

the correct answer is d

Explanation:

The HA bar on the right must be converted completely to H3O+ and A-. Strong acids completely dissociate in solution. Complete dissociation would mean that there is no HA bar left on the right of the arrow.

To represent a strong acid in the graph, the 'HA' bar has to be almost non-existent while the H3O+ and A- bars significantly increase reflecting the fact that strong acids completely disassociate in water. The correct answer is 'The HA bar on the right must be converted completely to H3O+ and A-'.

The correct answer to the given question is option D).

In the context of this question, the graph represents a moderately weak acid and we're asked to determine what changes would occur in the graph for a relatively strong acid.

Here, the 'HA' would represent the weak acid that partially disassociates into H3O+ (hydronium ions) and A- (the conjugate base). One characteristic of a strong acid is that it completely disassociates in water.

Therefore, to represent a strong acid, the 'HA' bar on the right would need to be much lower or even non-existent (to indicate complete disassociation). In turn, the H3O+ and A- bars on the right would need to increase significantly/acquire all the 'HA' bar's original height to represent the products of the strong acid's complete disassociation.

So, the correct answer would be (D) 'The HA bar on the right must be converted completely to H3O+ and A-'.

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Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀([tex]1 - \frac{z}{\sqrt{z^{2} + R^{2} } }[/tex])

Substituting the values into the equation, it becomes

E = σ/ε₀([tex]1 - \frac{z}{\sqrt{z^{2} + R^{2} } }[/tex]) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²([tex]1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }[/tex]) = 7.12 × 10⁵([tex]1 - \frac{0.12}{0.1216}[/tex]) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Answer: The average potential energy of the PIB is 0 irrespective of the wave function.

Explanation:

⟨H⟩=⟨KE⟩+⟨V⟩

the nn quantum number

⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )

the average kinetic energy of the wavefunction is dependent on

⟨V⟩=∫sin(kx)0sin(kx)dx=0

The average potential energy of the PIB is 0 irrespective of the wave function.

⟨H⟩=⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )

Answer:

The complete question is:

Question: What disorder is indicated by these findings? A client comes to the emergency department with status asthmaticus. His respiratory rate is 48 breaths/minute, and he is wheezing. An arterial blood gas analysis reveals a pH of 7.52, a partial pressure of arterial carbon dioxide (PaCO2) of 30 mm Hg, PaO2 of 70 mm Hg, and bicarbonate (HCO3−) of 26 mEq/L.

A. Metabolic acidosis

B. Respiratory acidosis

C. Metabolic alkalosis

D. Respiratory alkalosis

Answer: The correct answer is:

D. Respiratory alkalosis

Explanation:

In Respiratory alkalosis the Partial Pressure of Arterial Carbondioxide (PaCO2) become decreased (i.e. less than 35 mm Hg) and the pH of blood become increased (i.e. more than 7.45). Alveolar hyperventilation causes respiratory alkalosis.

Alveolar hyperventilation occurs when alveolar ventilation is increased than the arterial carbondioxide tension and carbondioxide production.

Alveolar ventilation is the gaseous exchange between alveoli and the external environment.

Whereas, in metabolic acidosis, bicarbonate (HCO3) become decreased (i.e. less than 22 mEq/l and the pH of blood become decreased (i.e. less than 7.35); in respiratory acidosis,  the pH of blood also become decreased (i.e. less than 7.35) and the PaCO2 become increased (i.e. more than 45 mm Hg); and in metabolic alkalosis,  the bicarbonate (HCO3) become increased (i.e. more than 26 mEq/l and the pH become increased (i.e. more than 7.45).

Phosphoric acid is a triprotic acid that can donate three protons in solution, forming three anions. The pH of a 0.100 M solution is approximately 1.0, assuming it only ionizes once. The concentrations of the formed species are estimated to be highest for H2PO4⁻ and much lower for HPO4²⁻ and PO4³⁻.

Phosphoric acid, H3PO4 is a triprotic acid, meaning it can donate three hydrogen ions in a solution. This results in the formation of three different species:  H2PO4⁻, HPO4²⁻, and PO4³⁻.

The estimated pH of a 0.100 M phosphoric acid solution will depend on the degree of dissociation, but for the first ionization, we can approximate it using the expression pH=-log[H⁺], where [H⁺] is the hydronium ion concentration. Given that a 0.100 M phosphoric acid solution ionizes mostly once, we have [H⁺]≈0.100 M, leading to an estimated pH around 1.0.

Next, the concentrations of the species in equilibrium can be calculated without exact Kb or Ka values as long as we make the approximation that dissociation after the first hydrogen ion is minimal in a dilute solution like 0.100 M. In this case, we will assume [H2PO4⁻]≈0.100 M and [HPO4²⁻] and [PO4³⁻] will be much less than [H2PO4⁻].

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Answer : The pressure of the helium gas inside the container would be, 32.1 atm

Explanation :

To calculate the pressure of the gas we are using ideal gas equation as:

[tex]PV=nRT[/tex]

where,

P = Pressure of [tex]He[/tex] gas = ?

V = Volume of [tex]He[/tex] gas = 7.5 L

n = number of moles [tex]He[/tex] = 10 mole

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of [tex]He[/tex] gas = [tex]20^oC=273+20=293K[/tex]

Putting values in above equation, we get:

[tex]P\times 7.5L=10mole\times (0.0821L.atm/mol.K)\times 293K[/tex]

[tex]P=32.1atm[/tex]

Thus, the pressure of the helium gas inside the container would be, 32.1 atm

Answer:

65.4 is the mass for 1.9×10²⁴ atomsof Pb

Explanation:

1mol of atoms of Pb has → NA (6.02×10²³ atoms) and weighs → 207.2 g

Therefore 1.9×10²⁴ atomsof Pb may weigh (1.9×10²⁴ . 207.2) / NA = 65.4 g

Final answer:

Approximately 34.97 kilocalories of heat energy were absorbed by the water in the velomobile seats.

Explanation:

In this scenario, we need to calculate the amount of heat energy absorbed by the water in the velomobile seats. To do this, we can use the formula:

Where:

Let's calculate the heat energy absorbed:

Converting this to kilocalories:

Therefore, approximately 34.97 kilocalories of heat energy were absorbed by the water in the velomobile seats.

Answer:

weaker and longer

Explanation:

Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger

thus the carbon-carbon double bond in ethene is weaker and longer than the carbon-carbon triple bond in ethyne

Answer:

The answer to this question is 45.33 moles of aluminum sulfate is produced when 125 moles of aluminum hydroxide and 136 moles of sulfuric acid react

Explanation:

To solve this, we write out the chemical equation of he reaction thus

Al(OH)3(s) + 3 H2SO4(aq) -----> Al2 (SO4)3(aq) + 6 H2O(l)

here it is seen that one moles of aluminum hydroxide reacts with three moles of sulfuric to produce one mole of aluminum sulfate and six moles of water

hence

136 moles of sulfuric acid reacts with 136/3 or 45.33 moles of aluminum hydroxide to produce 136/3 or 45.33 moles of aluminum sulfate and 2× 136 moles of water

Hence the amount in moles of aluminum sulfate produced is 45.33 moles

Final answer:

Using the balanced chemical equation 3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O, and knowing that aluminum hydroxide is the limiting reactant, 125 moles of aluminum hydroxide will produce 41.67 moles of aluminum sulfate.

Explanation:

The question asks how many moles of aluminum sulfate will be produced when reacting 125 moles of aluminum hydroxide with 136 moles of sulfuric acid. To answer this, we need the balanced chemical equation:

3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O

The stoichiometry of the reaction shows that 3 moles of aluminum hydroxide react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate. Since there are more moles of sulfuric acid present, aluminum hydroxide is the limiting reactant. Therefore, we can calculate the moles of aluminum sulfate produced by dividing the moles of aluminum hydroxide by 3, which gives us:
125 moles Al(OH)3 ÷ 3 = 41.67 moles Al2(SO4)3 (rounded to two decimal places as per the significant figures in the provided moles of reactants).

Answer:

Percent composition of the solution is 26 % of sucrose and 74 % of water

Explanation:

Percent composition is the mass of solute, either of solvent in 100 g of solution.

Mass of solution = Mass of solvent + Mass of solute

Mass of solute = 35 g

Mass of solvent = 100 g

As we know, water density = 1g/mL

So 1g/mL . 100 mL = 100 g

35 g + 100 g = 135 g → Mass of solution

(Mass of solute / Mass of solution) . 100 =

(35 g / 135 g) . 100 = 26 %

(Mass of solvent / Mass of solution) . 100 =

(100 g / 135 g) . 100 = 74 %

To calculate the percent composition of sucrose in the solution, divide the mass of sucrose (35 grams) by the total mass of the solution (sucrose plus water, which is 135 grams) and multiply by 100%. The solution has a percent composition of approximately 25.93% sucrose.

The question involves calculating the percent composition of a solution by mass. If a solution contains 35 grams of sucrose (C12H22O11) in 100 mL of water (noting that the density of water is roughly 1 g/mL, so we have 100 grams of water), the total mass of the solution is the sum of the mass of the solute (sucrose) and the solvent (water), which is 35 g + 100 g = 135 g. To find the percent by mass of sucrose in the solution, we use the formula:

Percent by mass of sucrose = (Mass of sucrose / Total mass of solution) × 100%

Inserting the values we have:

Percent by mass of sucrose = (35 g / 135 g) × 100% ≈ 25.93%

Therefore, the percent composition of sucrose in the solution is approximately 25.93%.

Answer:

Explanation:

You must find how many grams of a 14.0% solution contains 35.0 g of thiophene (solute) and then evaluate if the amount available (1.0 kg) is enough.

Formula:

Substitute and solve for the mass of solution:

        [tex]14.0 =(35.0g/x)\times 100\\\\14.0/100=35.0g/x\\\\x=35.0g\times100/14\\\\x=250g[/tex]

Hence, the student should use 250g of the 14.0% w/w solution of thiophene in ethanol. Since, 1.0 kg is 1,000g there is enough available.

Final answer:

To acquire 35.0 grams of thiophene from a 14.0% w/w thiophene solution, the student should measure out 250 grams of the solution. The 1.0 kg of available solution is more than sufficient for the student's needs.

Explanation:

To calculate the mass of the thiophene solution that the student needs, we first need to understand the percentage concentration of the solution. A 14.0% w/w solution of thiophene in ethanol means that there are 14 grams of thiophene for every 100 grams of solution. To find out how many grams of solution contain the required 35.0 grams of thiophene, we use the following formula:

Mass of thiophene = (Percentage/100) × Total mass of solution

Therefore, we can rearrange this to solve for the total mass of solution:

Total mass of solution = Mass of thiophene / (Percentage/100)

Total mass of solution = 35.0 g / (14.0/100) = 250 g

The student should use 250 grams of the solution to obtain 35.0 grams of thiophene. Since the available solution is 1.0 kg, which is 1000 grams, there is enough solution for the experiment.