A 7.41 mass % aqueous solution of sodium chloride has a density of 1.14 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.

Answer:

Explanation:

The triple point of carbon dioxide is 5.11 atmosphere at -56.6 degree celsius . At pressure greater than 5.11 , solid carbon dioxide liquefies ,  as it is warmed. At pressure lesser than 5.11 atmosphere , it will go into gaseous state without liquefying . Excessive pressure helps liquification process.

So maximum pressure required is 5.11 atmosphere. Beyond this pressure , solid CO2 will liquify.

Final answer:

Dry ice (solid CO₂) sublimes without melting at pressures below its triple point of 5.11 atm, transitioning directly from a solid to a gas at a temperature of -78.5°C, notably at atmospheric pressure of 1 atm.

Explanation:

The maximum pressure at which solid CO₂ (dry ice) can be converted into CO₂ gas without melting is at pressures below the substance's triple point. The triple point of CO₂ is at -56.6°C and 5.11 atm, which is the condition where the solid, liquid, and gaseous phases of CO₂ coexist in equilibrium. At pressures below 5.11 atm, solid CO₂ does not melt but instead undergoes sublimation, transitioning directly from a solid to a gas.

Specifically, at standard atmospheric pressure of 1 atm, dry ice sublimes at a temperature of -78.5°C. Thus, solid CO₂ will sublime at any pressure below its triple point without melting. Hence, the answer to the student's question is that solid CO₂ can be converted into gas without melting at pressures below 5.11 atm, and this will especially be observed at the standard atmospheric pressure where sublimation occurs directly.

Answer:

pKa of the acid HA with given equilibrium concentrations is 6.8

Explanation:

The dissolution reaction is:

HA ⇔ H⁺ + A⁻

So at equilibrium, Ka is calculated as below

Ka = [H⁺] x [A⁻] / [HA] = 2.00 x 10⁻⁴ x 2.00 x 10⁻⁴ / 0.260

    = 15.38 x 10⁻⁸

Hence, by definition,

pKa = -log(Ka) = - log(15.38 x 10⁻⁸) = 6.813

Answer:

213.89 lb of CO2.

Explanation:

Equation for the reaction:

C8H18 + 25/2O2(g) --> 8CO2(g) + 9H2O(g)

Given:

Volume of gasoline = 12 gallon

Converting gallon to ml,

12 gallon * 3.785 l/1 gallon * 1000 ml/1 l

= 45420 ml

Density of the gasoline = 0.692 g/ml

Mass = density * volume

= 45420 * 0.692

= 31430 g

Molar mass of octane = (8*12) + (18*1)

= 114 g/mol.

Number of moles = mass/molar mass

= 31430/114

= 275.702 mol.

From the above equation, 1 mole of octane was completed burnt to give off 8 moles of CO2.

By stoichiometry,

Number of moles of CO2 = 275.702 * 8

= 2205.614 mol of CO2.

Molar mass of CO2 = 12 + (2*16)

= 44 g/mol

Mass of CO2 = number of moles * molar mass

= 2206.614 * 44

= 97047.02 g

Converting g to pound,

= 97047.02 g *1 kg/1000 g * 2.204 lb/1kg

= 213.89 lb of CO2.

Answer:

Option (B)

Explanation:

In the field of mineralogy, cleavage is usually defined as a surface along which a mineral has the tendency to break. These surfaces are the plane of weaknesses that are formed due to deformation as well as metamorphism.

Fluorite is a mineral that has a hardness of 4, according to the Mohs hardness scale. It is comprised of four cleavage directions. This means that it has four sets of perfect cleavage, and this mineral often breaks into small pieces that are octahedron in shape.

Thus, the correct answer is option (B).

Answer:

Option a . 0.154L

Explanation:

P₁ . V₁ = P₂ . V₂

when we have constant temperature and constant moles for a certain gas.

At sea level, pressure is 1 atm so:

0.5 L . 1atm = V₂ . 3.25 atm

(0.5L . 1atm) / 3.25 atm = 0.154 L

"0.154 L" is the volume of the balloon.

Given:

Pressure,

Volume,

As we know,

→ [tex]P_1. V_1 = P_2 .V_2[/tex]

or,

→      [tex]V_2 = \frac{P_1. V_1}{P_1}[/tex]

By substituting the values, we get

            [tex]= \frac{0.5\times 1}{3.25}[/tex]

            [tex]= \frac{0.5}{3.25}[/tex]

            [tex]= 0.154 \ L[/tex]

Thus the above answer i.e., "option a" is correct.

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Answer:

331.2 grams of [tex]NO_2[/tex] are theoretically produced .

Explanation:

[tex]S + 6HNO_3\rightarrow H_2SO_4 + 6NO_2 + 2H_2O[/tex]

Moles of sulfur = 1.20 mol

Moles nitric acid = 9.90 moles

According to reaction ,1 mole of S reacts with 6 moles of sulfuric acid.

Then 1.20 moles of S will react with :

[tex]\frac{6}{1}\times 1.20 mol=7.2 moles[/tex] of nitric acid

This means that S is in limiting amount and nitric acid is in excessive amount.

So, amount of [tex]NO_2[/tex] gas will depend upon amount of S.

According to reaction, 1 m,ole of S gives 6 moles of [tex]NO_2[/tex] gas .

Then 1.20 moles of S will give:

[tex]\frac{6}{1}\times 1.20 mol=7.2 moles[/tex] of [tex]NO_2[/tex]

Mass of 7.2 moles of [tex]NO_2[/tex]

7.2 g × 46 g/mol = 331.2 g

331.2 grams of [tex]NO_2[/tex] are theoretically produced .

The question involves a stoichiometry problem on the production of Nitrogen dioxide (NO2) from Sulfur (S) and Nitric acid (HNO3). The theoretical yield of NO2 is constrained by the amount of Sulfur, the limiting reactant in this case. With given quantities, according to stoichiometric calculations, 331.27 grams of NO2 are theoretically produced.

The subject of this question is stoichiometry, a concept in chemistry which is the basis of predicting how much product can be made in a chemical reaction based on the quantity of reactants.

The reaction in the question is: S + 6HNO3 → H2SO4 + 6NO2 + 2H2O, according to which 1 mole of S will react with 6 moles of HNO3 to produce 6 moles of NO2. If we start with 1.20 moles of S, the reaction theoretically produces 6*1.20 = 7.20 moles of NO2.

However, when we consider the moles of HNO3, 9.90 moles of it can theoretically produce 9.90 moles of NO2 as per the balanced equation. However, this is more than what can be produced by the available S. Hence, S becomes the limiting reactant here. Therefore, the theoretical yield of NO2 is constrained by the amount of S and is 7.20 moles. When converting from moles to grams, using the molar mass of NO2 which is approximately 46.01 g/mol, we find that 7.20 moles of NO2 equals 331.27 grams.

Thus, the amount of NO2 that can be theoretically produced from these amounts of reactants is 331.27 grams.

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Answer:

In explanation

Explanation:

a. Selenium, atomic number 34

Argon is 18, adding the other electrons gives 34.

Other elements include: oxygen O , Sulphur S , Tellurium Te and Polonium Po

b. Hafnium, element 72

Xe is 54, adding the other electrons makes the total 72.

Other elements; Titanium Ti, Zirconium Zr, Rutherfordium Rf

c. Manganese, element 25

Other group members are Technetium Tc , Rhenium Re and Bohrium Bo

Answer:

A. 1.79kg

Explanation:

The value of the mass of a particular substance could be obtained if there is adequate information about the volume of that particular substance and the density of that substance.

Mathematically expressed, the mass of a substance is the product of the density and the volume .

In this particular question, the density is 0.01932kg/cm3 while the volume is 92.5cm3

The mass is thus = 92.5 * 0.01932 = 1.7871kg which is approximately 1.79kg

The mass of a 92.5 cm3 sample of gold is approximately 1.79 kg.

The mass of an object can be calculated by multiplying its density by its volume. In this case, the density of gold is given as 0.01932 kg/cm3, and the volume of the gold sample is 92.5 cm3. To find the mass, we can use the formula:

mass = density * volume

Substituting the values, we have:

mass = 0.01932 kg/cm3 * 92.5 cm3

Calculating this, we find that the mass of the gold sample is approximately 1.79 kg.

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A) The higher level that the electron reached from ground state is; n = 5

B) The intermediate level that the electron reached from ground state is; n = 3

C) The wavelength of the second photon emitted is; λ = 103 nm

A) We are given;

Wavelength of photon absorbed by ground state H atoms; λ_g = 94.91 nm = 94.91 × 10⁻⁹ m

Formula to get the higher level is Rydberg's formula;

1/λ = R(1/n₁² - 1/n₂²)

where;

R is rydberg constant = 1.097 × 10⁷ m⁻¹

Thus;

1/(94.91 × 10⁻⁹) = 1.097 × 10⁷(1/1² - 1/n₂²)

0.9605 = 1 - 1/n₂²

1/n₂² = 1 - 0.9605

1/n₂² = 0.0395

n₂ = √(1/0.0395)

n₂ ≈ 5

B) We want to find the intermediate level where wavelength = 1281 nm = 1281 × 10⁻⁹ m

Thus;

1/(1281 × 10⁻⁹) = 1.097 × 10⁷(1/n₂² - 1/5²)

0.0712 = 1/n₂² - ¹/₂₅

1/n₂² = 0.0712 + ¹/₂₅

1/n₂² = 0.1112

n₂ = √(1/0.1112)

n₂ ≈ 3

C) Formula for energy of photon is;

E = hc/λ

where;

h is Planck's constant = 6.626 × 10⁻³⁴ m².kg/s

c is speed of light = 3 × 10⁸ m/s

Thus;

E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(1281 × 10⁻⁹)

E = 1.552 × 10⁻¹⁹ J

The energy at ground state is;

E = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(94.91 × 10⁻⁹)

E = 20.944 × 10⁻¹⁹ J

Thus;

Energy of second photon = (20.944 × 10⁻¹⁹) - (1.552 × 10⁻¹⁹)

Energy of second photon = 19.352 × 10⁻¹⁹ J

Wavelength of second photon emitted is;

λ = hc/E

λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/19.352 × 10⁻¹⁹

λ = 103 nm

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The electron moves to the third energy level after the initial photon absorption, then drops to the fourth energy level releasing a photon of 1281 nm wavelength, and finally back to the ground state with a recorded emission of 97.08 nm wavelength.

To solve this, we can use the Rydberg formula for hydrogen: 1/λ = R (1/n1^2 - 1/n2^2), where λ is the wavelength of the light, R is the Rydberg constant (1.096776 x 10^7 m^-1), n1 is the lower energy level, and n2 is the higher energy level.

(a) The photon absorbed takes the electron to a higher energy level, so we use the wavelength 94.91 nm, and n1 = 1 (ground state). This calculates to n2 = 2.32, meaning the electron moves to roughly the third energy level as energy levels are only in whole numbers.

(b) The first emitted photon has wavelength 1281 nm, which brings the electron to an intermediate level. By substituting the values, we calculate n1 = 3 and get n2 = 4. So, the intermediate energy level is 4.

(c) For the second photon emitted, we know the electron goes from n = 4 to the ground state n = 1. Rearranging the Rydberg formula, we find λ = 97.08 nm, which is the wavelength of the second photon emitted.

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Completed question:

A sample of steam with a mass of 0.535 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.40 g of water at 5.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C)

Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer:

73.9°C

Explanation:

The steam is already at the water boiling point (100°C at 1 atm), so it will first lose heat to be condensed, by the equation:

Q1 = -n*ΔH∘vap*1000

Where n is the number of moles, ΔH∘vap is the enthalpy of evaporation, and the minus sign indicates that the heat is being lost. The equation is multiplied by 1000 because ΔH∘vap is in kJ, and the result must be in J.

Then, when it is in a liquid state, it will change heat with the cold water presented. The hot one will lose heat (Q2) and the cold one will gain heat (Q3) as the equation:

Q2 = mh*c*ΔTh

Q3 = mc*c*ΔTc

Where mh is the mass of the hot water, ΔTh is the variation of the temperature of the hot water (final - initial), mc is the mass of the cold water, and ΔTc is the variation of the temperature of the cold water.

At equilibrium they both will have the same final temperature (T), and, because the system doesn't change heat with the surroundings, the sum of all those heats must be 0.

n = mass/molar mass

Molar mass of water: 18g/mol

n = 0.535/18 = 0.0297 mol

Q1 + Q2 + Q3 = 0

-0.0297*40.7*1000 + 0.535*4.18*(T - 100) + 4.40*4.18*(T - 5.0) = 0

-1208.79 + 2.2363T - 223.63 + 18.392T - 91.96 = 0

20.6283T = 1524.38

T = 73.9°C

The final temperature of the mixture is mathematically given as

T = 73.9°C

Question Parameter(s):

Generally, the equation for the heat  is mathematically given as

Q1 = -n*dH∘vap*1000

Q2 = mh*c*dTh

Q3 = mc*c*dTc

Therefore

Q1 + Q2 + Q3 = 0

-0.0297*40.7*1000 + 0.535*4.18*(T - 100) + 4.40*4.18*(T - 5.0) = 0

20.6283T = 1524.38

T = 73.9°C

In conclusion, The temperature is

T = 73.9°C

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Complete question:

A sample of steam with a mass of 0.535 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.40 g of water at 5.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer:

Kyanite (Al2SiO5) - silicate

Ilmenite (FeTiO3) - Oxides

Rhodochrosite (MnCO3) - carbonate

Celestite (SrSO4) - sulphate

Chalcocite (Cu2S) -  sulphide

Explanation:

Minerals are classified according to their chemical composition. For example those that hve the CO32- ion are called carbonates and those with the SO42- ion are called sulphates while the ones with S2- ion are called sulphides

Answer:

Half life = 79.67 sec

Explanation:

Given that:

k = [tex]8.7\times 10^{-3}\ s^{-1}[/tex]

The expression for half life is shown below as:-

[tex]t_{1/2}=\frac{\ln2}{k}[/tex]

Where, k is rate constant

So,  

[tex]t_{1/2}=\frac{\ln2}{8.7\times 10^{-3}\ s^{-1}}[/tex]

[tex]t_{1/2}=114.94252\ln \left(2\right)\ sec=79.67\ sec[/tex]

Half life = 79.67 sec

Answer:

(a) n₂ = 2

(b) n₂ = 5

(c) n₂ = 4

(d) n₂ = 3

(e) n₂ = 6

Explanation:

The Rydberg equation give us the wavelength of the transition between energy levels according to the formula:

1/λ = Rh x ( 1/n₁² - 1/n₂² )

where n₁ is the final state and n₂ is the initial state.

The strategy here, since we are given the wavelength, is to solve for λ, and then by substituting for n₂ combinations find which ones match our question.

λ = 1 / [ Rh x  ( 1/n₁² - 1/n₂² ) ]

Lets express Rh in 1/Angstrom

1.097 x 10 ⁷ / [m x ( 1 m/ 10¹⁰ A) ] = 0.011 / Å

⇒ λ  = 1 / [0.011 A x  ( 1/n₁² - 1/n₂² )] = 911.6 Å / ( 1/n₁² - 1/n₂² )

For n₁ = 2  n₂ = 3, 4, 5,.......

λ  ( n₂ = 3 ) = 911.6 A / ( 1/2² - 1/3² ) = 6563.5 Å

λ  ( n₂ = 4 ) = 911.6 A / ( 1/2² - 1/4² ) = 4861.9  Å

λ  ( n₂ = 5 ) = 911.6 A / ( 1/2² - 1/5² ) = 4341.0  Å

So we have matched three of the transitions

Now for n₁ = 1    n₂ = 2, 3, 4....

λ  ( n₂ = 2 ) = 911.6 A / ( 1/1² - 1/2² ) = 1215.5 Å

For  n₁ = 3          n₂ = 4, 5, 6....

λ  ( n₂ = 4 ) = 911.6 A / ( 1/3² - 1/4² ) = 18752.9 Å

λ ( n₂ = 5 ) =  911.6 A / ( 1/3² - 1/5² ) = 12819.4  Å

λ ( n₂ = 6 ) =  911.6 A / ( 1/3² - 1/6² ) = 10939.2 Å

a. For [tex]\lambda=1212.7\;\rm \r{A}[/tex], initial transition level is 2.

b. For  [tex]\lambda=4340.5\;\rm \r{A}[/tex], initial transition level is 5.

c. For  [tex]\lambda=4861.3\;\rm \r{A}[/tex], initial level of transition is 4.

d. For [tex]\lambda=6562.8\;\rm \r{A}[/tex], initial level of transition is 3.

e. For  [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level of transition has been 6.

The wavelength from the transition has been given by the Rydberg equation as:

[tex]\dfrac{1}{\lambda}=\text Rh\dfrac{1}{N_1}-\dfrac{1}{N_2}[/tex]

Where, wavelength of the radiation, [tex]\lambda[/tex]

The initial transition level, [tex]N_1[/tex]

The final transition level, [tex]N_2[/tex]

The constant [tex]Rh=1.097\;\times\;10^7[/tex], or in Armstrong it can be given as, [tex]0.011/\r{\rm A}[/tex]

The wavelength can be given as:

[tex]\lambda=\dfrac{1}{0.011\;\r{\rm A}\;\times\;(\frac{1}{N_1^2} -\frac{1}{N_2^2}) }[/tex]

[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{N_1^2}-\frac{1}{N_2^2} }[/tex]

a. The initial level ([tex]N_2[/tex]) of transition has been given for [tex]\lambda=1212.7\;\rm \r{A}[/tex], and [tex]N_2=2[/tex] has been given, with substituting [tex]N_1=1[/tex]

[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{1^2}-\frac{1}{2^2} }\\\lambda=1215.5\;\r{A}[/tex]

Thus, for [tex]\lambda=1212.7\;\rm \r{A}[/tex], initial transition level is 2.

b. For, [tex]\lambda=4340.5\;\rm \r{A}[/tex], and [tex]N_2=2[/tex], [tex]N_1=5[/tex]

[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{5^2} }\\\lambda=4341.0\;\r{A}[/tex]

Thus, for [tex]\lambda=4340.5\;\rm \r{A}[/tex], initial transition level is 5.

c. For, [tex]\lambda=4861.3\;\rm \r{A}[/tex], the final transition has been, [tex]N_2=2[/tex], the initial level has been substituted as [tex]N_1=4[/tex]:

[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{4^2} }\\\lambda=4861.3\;\r{A}[/tex]

Thus, for [tex]\lambda=4861.3\;\rm \r{A}[/tex], initial level of transition is 4.

d. For [tex]\lambda=6562.8\;\rm \r{A}[/tex], the final level has been 2, the initial level has been substituted as, [tex]N_1=3[/tex]

[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{2^2}-\frac{1}{3^2} }\\\lambda=6562.8\;\r{A}[/tex]

Thus, for [tex]\lambda=6562.8\;\rm \r{A}[/tex], initial level of transition is 3.

e. For [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level has been substituted as, [tex]N_1=6[/tex]

[tex]\rm \lambda=\dfrac{911.6\;\r{A}}{\frac{1}{3^2}-\frac{1}{6^2} }\\\lambda=10,938\;\r{A}[/tex]

Thus, for [tex]\lambda=10,938\;\rm \r{A}[/tex], the initial level of transition has been 6.

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Answer :  The molecular geometry of the molecule is, tetrahedral.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, [tex]SiH_4[/tex]

As we know that silicon has '4' valence electrons and hydrogen has '1' valence electron.

Therefore, the total number of valence electrons in [tex]SiH_4[/tex] = 4 + 4(1)  = 8

According to Lewis-dot structure, there are 8 number of bonding electrons and 0 number of non-bonding electrons.

Now we have to determine the hybridization of the given molecules.

Formula used  :

[tex]\text{Number of electron pair}=\frac{1}{2}[V+N-C+A][/tex]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

The given molecule is, [tex]SiH_4[/tex]

[tex]\text{Number of electrons}=\frac{1}{2}\times [4+4]=4[/tex]

The number of electron pair are 4 that means the hybridization will be [tex]sp^3[/tex] and the electronic geometry and the molecular geometry of the molecule will be tetrahedral.

Hence, the molecular geometry of the molecule is, tetrahedral.

Silicon tetrahydride has a tetrahedral molecular geometry with the silicon atom being sp³ hybridized and bond angles approximately 109.5°.

The structure of silicon tetrahydride, also known as silane, can be represented according to Lewis theory by a central silicon atom surrounded by four hydrogen atoms, each bonded to the silicon with a single bond. The Lewis structure reflects that all the valence electrons of silicon are used to form the bonds with hydrogen atoms, leaving no lone pairs on the silicon. The associated molecular geometry of silicon tetrahydride is tetrahedral, as silicon is sp³ hybridized, with the four regions of electron density (the four Si-H bonds) arranged in a tetrahedral fashion. This geometry results in bond angles that are approximately 109.5°.

Answer: The empirical formula for the given compound is [tex]CuCl_3[/tex]

Explanation:

We are given:

Mass of copper chloride in 1 L or 1000 mL of solution = 42.62 grams

Taking Trial A:

Volume of solution = 49.6 mL

Applying unitary method:

In 1000 mL of solution, the mass of copper chloride present is 42.62 grams

So, in 49.6 mL of solution, the mass of copper chloride will be = [tex]\frac{42.62}{1000}\times 49.6=2.114g[/tex]

We are given:

Mass of filter paper = 0.908 g

Mass of filter paper + copper = 1.694 g

Mass of copper = [1.694 - 0.908] g = 0.786 g

Mass of chlorine in the sample = [2.114 - 0.786]g = 1.328 g

To formulate the empirical formula, we need to follow some steps:

Moles of Copper =[tex]\frac{\text{Given mass of Copper}}{\text{Molar mass of Copper}}=\frac{0.786g}{63.5g/mole}=0.0124moles[/tex]

Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{1.328g}{35.5g/mole}=0.0374moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0124 moles.

For Copper = [tex]\frac{0.0124}{0.0124}=1[/tex]

For Chlorine = [tex]\frac{0.0374}{0.0124}=3.02\approx 3[/tex]

The ratio of Cu : Cl = 1 : 3

Hence, the empirical formula for the given compound is [tex]CuCl_3[/tex]

Based on the experimental data, the empirical formula of the copper chloride is CuCl₂.

To determine the empirical formula of the copper chloride, we'll use the provided experimental data for three trials.

First, calculate the mass of copper produced in each trial:

Next, calculate the moles of copper produced using its molar mass (Cu = 63.55 g/mol):

Now, find the moles of copper chloride (assuming CuCl₂):

1. [tex]\text{Trial A:} & \quad \text{Volume of solution} = 49.6 \, \text{mL}, \\[/tex]

2. [tex]\text{Trial B:} & \quad \text{Volume of solution} = 48.3 \, \text{mL}, \\[/tex]

3. [tex]\text{Trial C:} & \quad \text{Volume of solution} = 42.2 \, \text{mL}, \\[/tex]

Lastly, the moles of chlorine (as Cl2) can be determined as the difference between the moles of Cu and total moles of CuCl₂:

Averaging these values, the ratio of moles of Cu to Cl is approximately 1:2, suggesting an empirical formula of CuCl₂.

Complete Question: -

In a student experiment, the empirical formula of a copper halide was found by adding aluminum metal to an aqueous solution of the halide, displacing copper metal. The copper metal was filtered, washed with distilled water, dried, and weighed. Three separate determinations were performed. The copper halide solution contained 42.62 g of copper chloride per liter. The student recorded the following experimental data:

Trial A:

Trial B:

Trial C:

Write the empirical formula of copper chloride based on the experimental data.

Answer: pH of the solution is 12

Explanation:

NaOH —> Na+ + OH-

NaOH = 0.01M

Na+ = 0.01M

OH- = 0.01M

pOH = —log [OH-] = —log [0.01]

pOH = 2

pH + pOH = 14

pH = 14 — pOH = 14 — 2 = 12

pH = 12

Explanation:

1. 2,3-dimethylhexane

2. 2,4-dimethylhexane

3. 2,5-dimethylhexane

4. 3,4-dimethylhexane

Below are the structures of the isomers.

Answer : The electric potential on proton is, 27.1 V

Explanation :

To calculate the electric potential we are using the formula.

[tex]V=\frac{1}{4\pi \epsilon_0}\frac{q}{r}[/tex]

where,

V = electric potential

q = charge on proton = [tex]1.6\times 10^{-19}C[/tex]

r = radius = [tex]0.530\times 10^{-10}m[/tex]

[tex]\frac{1}{4\pi \epsilon_0}=8.99\times 10^9[/tex]

Now put all the given values in the above formula, we get:

[tex]V=(8.99\times 10^9)\times \frac{1.6\times 10^{-19}}{0.530\times 10^{-10}}[/tex]

[tex]V=27.1V[/tex]

Thus, the electric potential on proton is, 27.1 V

Answer:

Aragonite is most likely be formed under high pressure.

Explanation:

Relating the densities of Aragonite and Calcite which both has the same chemical composition, it shows that the density of Aragonite is more than Calcite which says to some large extent that there is every possibility that Aragonite is most likely to be formed under high pressure.

Answer:

4.7 x 10⁻²³

Explanation:

The strategy here is to use de Broglie relation to answer this question:

mv = h/λ

where m is the mass, h is Planck´s constant, and λ is the wavelength:

v = h/ mλ

h = 6.626 x 10⁻³⁴ J·s

m = 142 g = 0.142 kg  ( we are working in the metric system )

λ = 100 pm = 100 pm  x  ( 1 x 10⁻¹² m/pm ) = 1 x 10⁻¹⁰ m

⇒ v =  6.626 x 10⁻³⁴ J·s / ( 0.142 kg x 1 x 10⁻¹⁰ m ) =  4.7 x 10⁻²³ m/s

This calculation shows why we do not talk about quantum effects at the macro level, notice the extreme low velocity that the baseball will have to have a wavelength equal to that of an x-ray photon.

A baseball with a mass of 142 grams would need to travel at a velocity of about 4.67 x 10^7 m/s in order to have a de Broglie wavelength equal to that of an x-ray photon with a wavelength of 100 pm.

To calculate the required velocity for the baseball to have a de Broglie wavelength equal to an x-ray photon, we use the de Broglie equation given as lambda = h / (m * v), where lambda is the de Broglie wavelength, h is Planck's constant (6.626 x 10^-34 kg m²/s), m is the mass of the particle, and v is the velocity. Given the wavelength lambda = 100 pm = 100 x 10^-12 m, and the mass of the baseball 142 g = 0.142 kg, we can rearrange the equation for velocity to get v = h / (m * lambda). Using the values, we get a velocity of approximately 4.67 x 10^7 m/s.

The seemingly large velocity can be attributed to the fact that de Broglie wavelengths are typically observable at small scales (in the quantum realm). For a macroscopic object like baseball, the velocities required for corresponding de Broglie wavelengths become unfeasibly high.

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Answer:   It is important to wet the filter paper in the Buchner funnel first with cold re crystallization solvent before the re crystallization mixture being filtered to minimize gaps around the edges of the filter paper which can prevent mechanical impurities from passing through. This gives better filtration where most impurities can be filtered. Furthermore, it provides good vacuum.

Answer:

The new partial pressures after equilibrium is reestablished:

[tex]PCl_3,p_1'=6.798 Torr[/tex]

[tex]Cl_2,p_2'=26.398 Torr[/tex]

[tex]PCl_5,p_3'=223.402 Torr[/tex]

Explanation:

[tex]PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g) [/tex]

At equilibrium before adding chlorine gas:

Partial pressure of the [tex]PCl_3=p_1=13.2 Torr[/tex]

Partial pressure of the [tex]Cl_2=p_2=13.2 Torr[/tex]

Partial pressure of the [tex]PCl_5=p_3=217.0 Torr[/tex]

The expression of an equilibrium constant is given by :

[tex]K_p=\frac{p_1}{p_1\times p_2}[/tex]

[tex]=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245[/tex]

At equilibrium after adding chlorine gas:

Partial pressure of the [tex]PCl_3=p_1'=13.2 Torr[/tex]

Partial pressure of the [tex]Cl_2=p_2'=?[/tex]

Partial pressure of the [tex]PCl_5=p_3'=217.0 Torr[/tex]

Total pressure of the system = P = 263.0 Torr

[tex]P=p_1'+p_2'+p_3'[/tex]

[tex]263.0Torr=13.2 Torr+p_2'+217.0 Torr[/tex]

[tex]p_2'=32.8 Torr[/tex]

[tex]PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g) [/tex]

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

[tex]K_p=\frac{p_3'}{p_1'\times p_2'}[/tex]

[tex]1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}[/tex]

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

[tex]p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr[/tex]

[tex]p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr[/tex]

[tex]p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr[/tex]

Answer: 1g of formaldehyde

Explanation:

Molar Mass of benzene (C6H6) = (12x6) + (6x1) = 78g/mol

1mole(78g) of benzene contains 6.02x10^23 molecules.

1g of benzene with contain = (6.02x10^23) /78 = 7.72x10^21 molecules

Molar Mass of formaldehyde (CH2O) = 12 +2+16 = 30g/mol

1mole (30g) of formaldehyde contains 6.02x10^23 molecules.

1g of formaldehyde will contain = (6.02x10^23) /30 = 2.01x10^22 molecules

Molar Mass of TNT(C7H5N3O6) = (12x7)+(5x1)+(14x3)+(16x6) = 227g/mol

1mole(227g) of TNT contains 6.02x10^23 molecules.

1g of TNT will contain = (6.02x10^23)/227 = 2.65x10^21 molecules

Molar Mass of naphthalene (C10H8) = (12x10) +8 =128g/mol

1mole(128g) of naphthalene contains 6.02x10^23 molecules.

1g of naphthalene will contain = (6.02x10^23)/128 = 4.7x10^21 molecules

Molar Mass of glucose(C6H12O6) = (12x6) + 12 +(16x6) = 180g/mol

1mole(180g) of glucose contains 6.02x10^23 molecules.

1g of glucose will contain = (6.02x10^23)/180 = 3.34x10^21 molecules

Therefore, 1g of each of the compound contains the following:

Benzene = 7.72x10^21 molecules

Formaldehyde = 2.01x10^22 molecules

TNT = 2.65x10^21 molecules

Naphthalene = 4.7x10^21 molecules

Glucose = 3.34x10^21 molecules

From the above, we see clearly that formaldehyde has the largest number of molecules

Final answer:

The compound with the largest number of molecules is formaldehyde with 2.00 x 10^22 molecules.

Explanation:

The substance with the largest number of molecules can be determined by calculating the number of moles of each substance and then using Avogadro's number to convert moles to molecules. The formula to calculate the number of moles is moles = mass / molar mass. By using this formula:

For benzene: moles = 1g / 78.11g/mol = 0.0128 moles

For formaldehyde: moles = 1g / 30.03g/mol = 0.0333 moles

For TNT: moles = 1g / 227.13g/mol = 0.0044 moles

For naphthalene: moles = 1g / 128.2g/mol = 0.0078 moles

For glucose: moles = 1g / 180.16g/mol = 0.0055 moles

Then, we can use Avogadro's number (6.022 x 10^23 molecules/mol) to convert moles to molecules:

For benzene: molecules = 0.0128 mol x 6.022 x 10^23 molecules/mol = 7.72 x 10^21 molecules

For formaldehyde: molecules = 0.0333 mol x 6.022 x 10^23 molecules/mol = 2.00 x 10^22 molecules

For TNT: molecules = 0.0044 mol x 6.022 x 10^23 molecules/mol = 2.65 x 10^21 molecules

For naphthalene: molecules = 0.0078 mol x 6.022 x 10^23 molecules/mol = 4.68 x 10^21 molecules

For glucose: molecules = 0.0055 mol x 6.022 x 10^23 molecules/mol = 3.31 x 10^21 molecules

Therefore, the compound with the largest number of molecules is formaldehyde with 2.00 x 10^22 molecules.

Answer:

0.01 mol of H+.

Explanation:

Equation of reaction:

Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)

Molar concentration = number of moles/volume

= 1*0.01

= 0.01 mol.

Dissociation:

2HCl --> 2H+ + 2Cl-

By stoichiometry,

If 2 mole of HCl dissociates to guve 2 moles of H+

Number of moles of H+ = 0.01 mol.

Explanation:

Formula to calculate work done by motor is as follows.

         Work done by motor = [tex]mass \times g \times height[/tex]

where,   g = gravitational constant = 10 [tex]m/s^{2}[/tex]

Therefore, work done by motor is as follows.

 Work done by motor = [tex]1.00 kg \times 10 m/s^{2} \times 10.0 m[/tex]  

                                  = 100.0 J

Now, heat lost by water will be calculated as follows.

            q = [tex]mC \times \Delta T[/tex]

                = [tex]g \times 4.184 J/g^{o}C \times 0.024^{o}C[/tex]

                = 10.0 J

Hence, heat gained by motor = heat lost by water

As, heat gained by motor = 10.0 J

So, change in energy = heat gained - work done

Therefore, change in energy will be calculated as follows.  

   Change in energy = heat gained - work done

                                  = (10.0 J) - (100.0 J)

                                   = -90.0 J

Thus, we can conclude that change in the energy of the battery contents is -90.0 J.

Answer :

(a) The charge and full ground-state electron configuration of the monatomic ion is, (+1) and [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]

(b) The charge and full ground-state electron configuration of the monatomic ion is, (-3) and [tex]1s^22s^22p^6[/tex]

(c) The charge and full ground-state electron configuration of the monatomic ion is, (-1) and [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]

Explanation :

For the neutral atom, the number of protons and electrons are equal. But, they are unequal when the atoms present in the form of ions or the atom has some charges.

When an unequal number of electrons and protons then it leads to the formation of ionic species.

Ion : An ion is formed when an atom looses or gains electron.

When an atom looses electrons, it will form a positive ion known as cation.

When an atom gains electrons, it will form a negative ion known as anion.

(a) The given element is, Rb (Rubidium)

As we know that the rubidium element belongs to group 1 and the atomic number is, 37

The ground-state electron configuration of Rb is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1[/tex]

This element will easily loose 1 electron and form [tex]Rb^+[/tex] ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Rb ion is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]

(b) The given element is, N (Nitrogen)

As we know that the nitrogen element belongs to group 15 and the atomic number is, 7

The ground-state electron configuration of N is:

[tex]1s^22s^22p^3[/tex]

This element will easily gain 3 electrons and form [tex]N^{3-}[/tex] ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of N ion is:

[tex]1s^22s^22p^6[/tex]

(c) The given element is, Br (Bromine)

As we know that the bromine element belongs to group 17 and the atomic number is, 35

The ground-state electron configuration of Rb is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^5[/tex]

This element will easily gain 1 electron and form [tex]Br^-[/tex] ion  which attain stable noble gas electronic configuration.

The full ground-state electron configuration of Br ion is:

[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]

The most likely monatomic ions formed by Rb, N, and Br are Rb+, N3-, and Br-, respectively.

(a) The most likely monatomic ion formed by Rb is Rb+. The ground-state electron configuration of Rb is [Kr]5s1, and when it loses one electron to form the ion, its configuration becomes [Kr].

(b) The most likely monatomic ion formed by N is N3-. The ground-state electron configuration of N is 1s22s22p3, and when it gains three electrons to form the ion, its configuration becomes 1s22s22p6.

(c) The most likely monatomic ion formed by Br is Br-. The ground-state electron configuration of Br is [Ar]4s23d104p5, and when it gains one electron to form the ion, its configuration becomes [Ar]4s23d104p6.

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Final answer:

The quantum numbers n, l, m_l, and m_s describe the energy level, shape of the orbital, orbital orientation, and spin of an electron, respectively. No two electrons can have the same set of four quantum numbers due to the Pauli exclusion principle, allowing only two with opposite spins in the same orbital.

Explanation:

Quantum Numbers and Electron Configuration

Electrons in atoms have quantized energies, and their states are described by four quantum numbers. These quantum numbers include:

According to the Pauli exclusion principle, no two electrons in the same atom can have identical sets of these quantum numbers. This means that within a single orbital defined by n, l, and ml, only two electrons can exist and they must have opposite spins (ms).

Final answer:

The four quantum numbers with their allowed values are the principal quantum number (n) with positive integers, the azimuthal quantum number (l) with values from 0 to n-1, the magnetic quantum number ([tex]m_l[/tex]) with values between -l and +l, and the spin quantum number ([tex]m_s[/tex]) with values of (+1/2) or (-1/2).

Explanation:

Allowed Values for Quantum Numbers

Each electron in an atom is described by four quantum numbers: the principal quantum number (n), the azimuthal quantum number (l), the magnetic quantum number ([tex]m_l[/tex]), and the spin quantum number ([tex]m_s[/tex]). The allowed values for these quantum numbers are:

These quantum numbers are based on the solutions to the Schrödinger Equation for atoms and are fundamental to understanding electron configurations in atoms.

Answer:

CuSO4.5H2O —> CuSO4 + 5H2O

Explanation:

This reaction shows how hydrate copper sulphate losses its water of crystallization to become anhydrous copper sulphate

CuSO4.5H2O —> CuSO4 + 5H2O

Answer: The decreasing order of atomic size:

For a:  Pb > Sn > Ge

For b:  Sr > Sn > Te

For c:  Na > F > Ne

For d:  Mg > Na > Be

Explanation:

Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom.

As moving from top to bottom in a group, there is an addition of shell around the nucleus and the outermost shell gets far away from the nucleus and hence, the distance between the nucleus and outermost shell increases. Thus, increasing the atomic radii of the atom.

As moving from left to right in a period, more and more electrons get added up in the same shell and the attraction between the last electron and nucleus increases, which results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom on moving towards right of the periodic table.

For the given options:

Germanium (Ge) is present in Group 14 and period 4 of the periodic table.

Lead (Pb) is present in Group 14 and period 6 of the periodic table.

Tin (Sn) is present in Group 14 and period 5 of the periodic table.

So, the decreasing order of atomic size is:

Pb > Sn > Ge

Tin (Sn) is present in Group 14 and period 5 of the periodic table.

Tellurium (Te) is present in Group 16 and period 5 of the periodic table.

Strontium (Sr) is present in Group 2 and period 5 of the periodic table.

So, the decreasing order of atomic size is:

Sr > Sn > Te

Fluorine (F) is present in Group 17 and period 2 of the periodic table.

Neon (Ne) is present in Group 18 and period 2 of the periodic table.

Sodium (Na) is present in Group 1 and period 3 of the periodic table.

So, the decreasing order of atomic size is:

Na > F > Ne

Beryllium (Be) is present in Group 2 and period 2 of the periodic table.

Magnesium (Mg) is present in Group 2 and period 3 of the periodic table.

Sodium (Na) is present in Group 1 and period 3 of the periodic table.

So, the decreasing order of atomic size is:

Mg > Na > Be

The decreasing order of atomic size are the following:

a:  Pb > Sn > Ge

b:  Sr > Sn > Te

c:  Na > F > Ne

d:  Mg > Na > Be

Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom. As move from top to bottom in a group,  addition of shell occurs and the atomic radii of the atom increases.

While on the other hand. as we move from left to right, atomic size of an atom decreases due to addition of nuclear charge in the nucleus.

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(a) ΔH∘rxn: (1) -1471.1 kJ/mol, (2) 3491.9 kJ/mol, (3) -480 kJ/mol. (b) Complete burn: -1259.6 kJ. (c) Incomplete burn: 387.8 kJ, Soot: -53.3 kJ.

Let's work through the given questions step by step.

(a) Calculating ΔH∘rxn for each response:

The responses you gave are combustion responses of paraffin wax ([tex]C_2_1H_4_4[/tex]). To discover the ΔH∘rxn for each response, we'll utilize the given ΔH∘f for [tex]C_2_1H_4_4[/tex] and the standard enthalpies of arrangement for the items.

Response (1): Total combustion

[tex]C_2-1H_4_4 + 32 O_2 = 21CO_2 + 22H_2O[/tex]

ΔH∘rxn =( (ΣΔH∘f(products)) - (ΣΔH∘f(reactants)))

ΔH∘rxn = ((21) × (ΔH∘f[tex](CO_2[/tex])) + ((22) × (ΔH∘f([tex]H_2O[/tex])) -(ΔH∘f([tex]C_2_1H_4_4[/tex]))

ΔH∘f(CO2) =( -393.5 kJ/mol and (ΔH∘f(H2O)) = -285.8 kJ/mol:

ΔH∘rxn(1) = ((21) × (-393.5 kJ/mol) + (22) × (-285.8 kJ/mol)) - -476 kJ/mol

ΔH∘rxn(1) = (-8233.5 kJ/mol + 6288.4 kJ/mol) - (-476 kJ/mol)

ΔH∘rxn(1) = (-1471.1 kJ/mol)

Response (2): Complete combustion

[tex]C_2_1H_4_4 + 22O_2 → 21CO + 22H_2O[/tex]

Using the same approach:

(ΔH∘rxn(2)) = ((21) × ΔH∘f(CO)) + (22) × ΔH∘f([tex]H_2O[/tex])] - (ΔH∘f([tex]C_2_1H_4_4[/tex]))

Given ΔH∘f(CO) = -110.5 kJ/mol:

ΔH∘rxn(2) = [21 × (-110.5 kJ/mol) + 22 × (-285.8 kJ/mol)] - (-476 kJ/mol)

ΔH∘rxn(2) = -2320.5 kJ/mol + 6288.4 kJ/mol - (-476 kJ/mol)

ΔH∘rxn(2) = 3491.9 kJ/mol

Response (3): Formation of soot

[tex]C_2_1H_4_4 + 11O_2[/tex] → 21 C (graphite) +[tex]22H_2O[/tex]

ΔH∘rxn(3) = [21 × ΔH∘f(C) + 22 × ΔH∘f([tex]H_2O[/tex])] - ΔH∘f([tex]C_2_1H_4_4[/tex])

Given ΔH∘f(C) = kJ/mol (since graphite is the standard reference state):

ΔH∘rxn(3) = [21 × kJ/mol + 22 × (-285.8 kJ/mol)] - (-476 kJ/mol)

ΔH∘rxn(3) = -6292.6 kJ/mol + 6288.4 kJ/mol - (-476 kJ/mol)

ΔH∘rxn(3) = -480 kJ/mol

(b) Calculating q for complete combustion:

To calculate q (warm discharged) when the candle burns totally, ready to utilize the condition:

q = n × ΔH∘rxn

Given the mass of the candle is 254 g and the molar mass of [tex]C_2_1H_4_4[/tex] is around 296.66 g/mol:

n = 254 g / 296.66 g/mol = 0.856 mol

q = 0.856 mol × (-1471.1 kJ/mol) = -1259.6 kJ

So, when the candle burns totally, it discharges around 1259.6 kJ of warm.

(c) Calculating q for fragmented combustion and sediment arrangement:

For inadequate combustion, the response is the same as response (2), and for sediment arrangement, the reaction is the same as response (3). We have to calculate the warm discharged for each of these cases independently.

To begin with, calculate the mass of the candle included in each response:

Deficient combustion: 8.00% by mass

Sediment arrangement: 5.00% by mass

Add up to mass burned = 8.00% + 5.00% = 13.00% = 0.13 (decimal frame)

Presently, for fragmented combustion:

n_incomplete = 0.13 ×254 g / 296.66 g/mol = 0.111 mol

q_incomplete = 0.111 mol × 3491.9 kJ/mol = 387.8 kJ

For sediment arrangement:

n_soot = 0.13 × 254 g / 296.66 g/mol = 0.111 mol

q_soot = 0.111 mol × (-480 kJ/mol) = -53.3 kJ

So, when 8.00% of the candle burns not completely, it discharges roughly 387.8 kJ of heat, and when 5.00% experiences sediment arrangement, it assimilates roughly 53.3 kJ of warm.

If it's not too much trouble note that the negative sign for q_soot demonstrates that warm is ingested, which is anticipated for an endothermic prepare-like sediment arrangement.

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The given problems ask to calculate the enthalpy change of the full and partial combustion of a paraffin candle to CO2, CO and elemental carbon (soot), and the heat released in these combustion reactions. The calculated ΔHrxn of each reaction is used with the given or calculated mass to get the heat released.

For the question asked, first we look at the combustion reactions of a candle wax paraffin formula C21H44.

By using the standard enthalpy of formation (ΔHf) values, we can calculate ΔHrxn for each reaction. By mass of candle burnt, we can find q by using the formula q = mass x specific heat x ΔT; q can also be calculated from the calculated ΔHrxn multiplied by the moles involved.

For the final part (c), if 8.00% by mass of the candle burns incompletely and another 5.00% undergoes soot formation, q for each partial scenario can be calculated from the relevant ΔHrxn multiplied by the moles involved (mass burnt / molar mass), and adding these two values together.

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