A sample weighing 3.110 g is a mixture of Fe 2 O 3 (molar mass = 159.69 g/mol) and Al 2 O 3 (molar mass = 101.96 g/mol). When heat and a stream of H 2 gas is applied to the sample, the Fe 2 O 3 reacts to form metallic Fe and H 2 O ( g ) . The Al 2 O 3 does not react. If the sample residue (the solid species remaining after the reaction) weighs 2.387 g, what is the mass fraction of Fe 2 O 3 in the original sample?

Answer:

15,4g of ice

Explanation:

The dissolution of HNO₃ increases the temperature of the solution thus:

Q = -C×m×ΔT (1)

Where Q is heat produced:

Q = -33 kJ/mol×(0,0130L×14,0M) = -6,00kJ

C is molar heat capacity of solution: 80,8 J/mol°C

m are moles of solution ≈ moles of water = 100mL≡100g×(1mol/18g) = 5,56 moles

And ΔT is final temperature - Initial temperature (X-25°C)

Replacing in (1):

-6000J = -80,8J/mol°C×5,56mol×(X-25°C)

13,4 = X-25°C

X = 38,4°C

Knowing that you want to return the temperature of the system to 25°C, the ice must to absorb 6000J of energy (In the fussion process and increasing each temperature until equilibrium) produced in the dissolution of HNO₃:

6000J = ΔH°fus×X + C×X×ΔT

-Where X are moles of ice-

Replacing:

6000J = 6010J/mol×X + 75,3J/mol°C×X×(38,4°C-25°C)

6000J = 6010J×X + 1006J×X

0,855 = moles of X

In grams:

0,855 moles×(18g/1mol)= 15,4g of ice

I hope it helps!

Answer:

a) 39/19 K : stable nuclide, 40/19 K  : radioactive nuclide.

b) 209B: stable nuclide, 208Bi : radioactive nuclide

c) nickel-58 : stable nuclide, nickel-65 : radioactive nuclide.

Explanation:

As per the rule, nuclides having odd number of neutrons are generally not stable and therefore, are radioactive.

Mass number (A) = Atomic number (Z) + No. of neutrons (N)

Or, N = A - Z

a)

39/19 K and 40/19 K

Calculate no. of neutrons in 39/19 K as follows:

atomic no. = 19, mass no. 39

N = 39 - 19

   = 20 (even no.)

Calculate no. of neutrons in 40/19 K as follows:

atomic no. = 19, mass no. 40

N = 40 - 19

   = 21 (odd no.)

Therefore, 39/19 K is a stable nuclide and 40/19 K is a radioactive

nuclide.

b)

209Bi and 208Bi

Calculate no. of neutrons in 209Bi as follows:

atomic no. = 83, mass no. 209

N = 209 - 83

   = 126 (even no.)

Calculate no. of neutrons in 208Bi as follows:

atomic no. = 83, mass no. 208

N = 208 - 83

   = 125 (odd no.)

Therefore, 209Bi is a stable nuclide and 208Bi is a radioactive nuclide.

c)

nickel-58 and nickel-65

Calculate no. of neutrons in nickel-58 as follows:

atomic no. = 28, mass no. 58

N = 58 - 28

   = 30 (even no.)

Calculate no. of neutrons in nickel as follows:

atomic no. = 28, mass no. 65

N = 65 - 28

   = 37 (odd no.)

Therefore,nickel-58 is a stable nuclide and nickel-65 is a radioactive nuclide.

To determine which nuclide is radioactive and which is stable, we need to analyze the composition of their nuclei. The number of neutrons in a nuclide plays a crucial role in determining its stability. Based on this, we can predict whether a nuclide is radioactive or stable.

In order to predict which nuclide is radioactive and which is stable, we need to determine the composition of their nuclei. Radioactive nuclides have an unstable nucleus that undergoes radioactive decay, while stable nuclides have a more balanced composition.

Therefore, the pairs are:
a. 39/19K (stable) and 40/19K (radioactive)
b. 209Bi (stable) and 208Bi (radioactive)
c. 58Ni (stable) and 65Ni (radioactive)

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Answer:

c. (CH3CH2CH2)2CuLi

Explanation:

(CH3CH2CH2)2CuLi

This reagent is called Lithium di(n propyl)cuprate used to covert benzoyl chloride to phenyl propyl ketone. this reagent is called Gilman Reagent and The reaction occur with nucleophilic substitution of an alkyl group for the leaving group (chloride), forming one new carbon-carbon bond.

Hence the correct answer is C that is   c. (CH3CH2CH2)2CuLi

Answer:

i=1.62 .

Explanation:

Let, i be the Van't Hoff Factor.

Moles of benzamide,=[tex]\dfrac{Mass}{molar\ mass}=\dfrac{70.4}{121.14}=0.58\ mol.[/tex]

Molality of solution, m=[tex]\dfrac{moles  }{mass\ of\ solvent}=\dfrac{0.58}{0.85}=0.68\ molal.[/tex]

Now, we know

Depression in freezing point, [tex]\Delta T=i\times K_f\times m[/tex]  .....1

It is given that,

[tex]\Delta T=2.7^o C\\i=1 ( since\ it\ is\ non\ dissociable\ solutes)\\K_f  ( freezing\ constant)\\[/tex]

Putting all these values we get,

[tex]K_f=3.949\ C/m.[/tex]

Now, moles of ammonium chloride=[tex]\dfrac{70.4}{53.49}=1.316\ mol.[/tex]

molality =[tex]\dfrac{1.316}{0.85}=1.54 molal.\\\Delta T=9.9 .[/tex]

Putting all these values in eqn 1.

We get,

i=1.62 .

Hence, this is the required solution.

Answer:B

Explanation:

The solubility product is the product of the concentration of all the ions present as found in the rate equation. The solubility product is the equilibrium expression that shows the extent to which a substance dissolves in water. The full solution is shown in the image attached.

Answer:

4.0E^-18

which makes the answer A

Answer:

A) - 0.26 V

Explanation:

Here Ni undergoes oxidation by loss of electrons, thus act as anode. Standard hydrogen electrode undergoes reduction by gain of electrons and thus act as cathode.

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

Given that:- [tex]E^0=0.26\ V[/tex]

Also, [tex]E^0_{[H^{+}/H_2]}=0\ V[/tex]

So,

[tex]E^0=E^0_{[H^{+}/H_2]}-E^0_{[Ni^{2+}/Ni]}[/tex]

[tex]0.26\ V=0\ V- E^0_{[Ni^{2+}/Ni]}[/tex]

[tex]E^0_{[Ni^{2+}/Ni]}=-0.26\ V[/tex]

The standard reduction potential for the Ni2+/Ni half-cell is - 0.26 V.

The standard hydrogen electrode is a reference electrode that was arbitrarily assigned an electrode potential of 0.00V. We know that metals that are above hydrogen in the electrochemical series will always have a negative electrode potential.

As such, the standard reduction potential for the Ni2+/Ni half-cell is - 0.26 V.

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Answer:

-1.37 kJ/mol

Explanation:

The expression for the calculation of the enthalpy of dissolution of [tex[NH_4NO_3[/tex] is shown below as:-

[tex]\Delta H=m\times C\times \Delta T[/tex]

Where,  

[tex]\Delta H[/tex]  is the enthalpy of dissolution of [tex[NH_4NO_3[/tex]

m is the mass

C is the specific heat capacity

[tex]\Delta T[/tex]  is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 5.60 g

Specific heat = 4.18 J/g°C

[tex]\Delta T=17.9-22.0\ ^0C=-4.1\ ^0C[/tex]

So,  

[tex]\Delta H=-1.25\times 4.18\times 3.9\ J=-95.9728\ J[/tex]

Negative sign signifies loss of heat.  

Also, 1 J = 0.001 kJ

So,  

[tex]\Delta H=-0.096\ kJ[/tex]

Also,

Molar mass of [tex[NH_4NO_3[/tex] = 80.043 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{5.60\ g}{80.043 \ g/mol}[/tex]

[tex]Moles= 0.06996\ mol[/tex]

Thus, [tex]\Delta H=-\frac{0.096}{0.06996}\ kJ/mol=-1.37\ kJ/mol[/tex]

Answer:

10 g %  w/v

Explanation:

Weight/volume percent concentration means the grams of solute, in 100 mL of solution.

If we have 1 g of B3 in 10mL of solution, the rule of three for the weight volume percent will be:

10mL ___ 1 g

100 mL ____ 10 g

Answer:

The particles in a gas are comparatively very far from each other. Further an ideal gas has not interactions between the particles i.e repulsion. This explains why a gas that is compressed into a liquid collapses into a volume that is small by a factor of 1000. The particles get much closer to each other to form a liquid and they do not repel each other

Answer:

1. The theoretical yield is 0.2mols

2. No, aldol condensation reaction does not proceeds via an E1 mechanism.

Answer: [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa[/tex], NaOH are all ionic compounds.

Explanation:

It is known that ionic compounds are the compounds in which one atom transfer its valence electrons to another atom. Hence, during this transfer partial opposite charges tend to develop on the combing atoms due to which strong force of attraction exists between the atoms.

An ionic bond is always formed between a metal and a non-metal.

For example, [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa[/tex], NaOH are all ionic compounds.

On the other hand, a compound formed due to sharing of electrons between the combining atoms is known as a covalent compound. Generally, a non-metal with same or different non-metal tends to form a covalent bond.

For example, [tex]POCl_{3}[/tex], [tex]SOCl_{2}[/tex] etc are all covalent compounds.

Thus, we can conclude that [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa, NaOH are all ionic compounds.

Answer:

The ice will not melt, and the temperature will remain at 0°C.

Explanation:

The reaction of the sodium in water is exothermic because heat is being released. In an isolated system, the change in heat must be 0, so the released heat must be absorbed by the ice.

The molar mass of Na is 23 g/mol, so the number of moles that reacted was:

n = 0.250 g/ 23g/mol

n = 0.011 mol

By the reaction:

2 moles ------- -368 kJ

0.011 mol ----- x

By a simple direct three rule:

2x = -4.048

x = -2.024 kJ/mol

So the ice will absorbs 2.024 kJ/mol, which is less than the necessary to melt it (6.02 kJ/mol). Then, the ice will not melt.

The temperature of a pure substance didn't change until all of it has changed of phase, so the temperature must remain at 0°C.

Answer:

[tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]

Explanation:

The given cell notation is:-

[tex]Fe_{(s)}|Fe^{3+}_{(aq)}||Cl_2_{(g)}|Cl^-_{(aq)}|Pt[/tex]

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

[tex]X\rightarrow X^{n+}+ne^-[/tex]

The oxidation half reaction of the cell is:-

[tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]

Answer:

0.116

Explanation:

The molar mass of ethanol is 46.07 g/mol. The moles corresponding to 25.0 g are:

25.0 g × (1 mol/46.07 g) = 0.543 mol

The molar mass of water is 18.02 g/mol. The moles corresponding to 75.0 g are:

75.0 g × (1 mol/18.02 g) = 4.16 mol

The total number of moles is:

total moles = mol ethanol + mol water = 0.543 mol + 4. 16 mol = 4.70 mol

The mole fraction of ethanol is:

X(Ethanol) = mol ethanol / total moles = 0.543 mol / 4.70 mol = 0.116

Explanation:

In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes =  1 : 1 : 2

let the :

Number of lattice point = 1x.

Number of octahedral points = 1x

Number of tetrahedral  points = 2x

If anions occupy the HCP lattice points and cations occupy half of the octahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the octahedral points, B =  1x

The formula of the compound will be = [tex]A_{1x}B_{1x}=AB[/tex]

If anions occupy the HCP lattice points and cations occupy all of  the octahedral and the tetrahedral holes.

Number of anions occupying the HCP lattice points, A= 1x

Number of cations occupying the octahedral points, B =  x

Number of cations occupying the tetrahedral points, B =  2x

total number of cations = x + 2x = 3x

The formula of the compound will be = [tex]A_{1x}B_{3x}=AB_3[/tex]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: [tex]m_{(AcO)_2Cu} = 0.972 g[/tex]

Volume of the sodium chromate solution: [tex]V_{Na_2CrO_4} = 150.0 mL[/tex]

Molarity of the sodium chromate solution: [tex]c_{Na_2CrO_4} = 0.0400 M[/tex]

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

[tex](CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)[/tex]

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

[tex]n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol[/tex]

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

[tex]n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol[/tex]

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

[tex]Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)[/tex]

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

[tex]n_{(AcO)_2Cu} = 0.0053515 mol[/tex]

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

[tex]n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol[/tex]

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

[tex]c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M[/tex]

The final molarity of the acetate anion in the solution, after copper(II) acetate is dissolved in a sodium chromate solution, is 0.1 M.

The acetate anion, CH3 CO₂¯, found in copper(II) acetate, is the conjugate base of acetic acid. When copper(II) acetate is dissolved in a solution of sodium chromate, it yields a solution of inert cations and weak base anions, resulting in a basic solution. Given the initial acetate concentration, if the newly formed acetate ion gives a final acetate concentration of (1.0 × 10−²) + (0.01 × 10−²) = 1.01 × 10-² mol NaCH3 CO₂, this would mean there are both 9.9 × 10-3 mol/L of copper(II) acetate and 1.01 × 10-2 mol/L of sodium chromate.

To compute the molarity of acetate anion, you can divide the moles of the acetate by the volume of the solution, yielding a final molarity of the acetate anion in the solution. Hence the molarity of acetate anion in the solution would be 1.01 x 10^-2 mol /0.101 L = 0.1 M.

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Answer:

2 mole H2O = 18.02 g will not work.

Explanation:

Let us check the options one by one:

1 mole of O2 weighs 32 g.

⇒Hence option A is correct.

1 mole of H2O weighs 18 g.

⇒Hence option B is wrong.

From the balanced equation we can say :

1 mole of CH4 reacts with 2 moles of O2 to give 1 mole of CO2 and 2 moles of H2O.

⇒Hence option C and D are correct.

Answer:

Option B: 2 mole H2O = 18.02 g is not correct.

Explanation:

Step 1: The balanced equation

CH4 + 2O2 → CO2 + 2H2O

Step 2: Data given

Molar mass of O2 = 32 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of CH4 = 16.04 g/mol

Step 3: Calculate number of mol

For 1 mol of CH4 we need 2 moles of O2 to produce 1 mol of CO2 and 2 moles of H2O

This means option C is correct: For 2 moles O2 consumed, we'll get 1 mol of CO2 produced.

Also Option D is corect: For 1 mole of CH4 we get 2 moles of H2O

Mass = moles * molar mass

Mass of 1 mole O2 = 1 mol * 32.00 g/mol = 32.00 grams

Mass of 2 moles H2O = 2 mol * 18.02 g/mol = 36.04 grams  (This means option B is not correct.)

Option B: 2 mole H2O = 18.02 g is not correct.

Answer:

B) 16.4

Explanation:

Given that:-

Current, I = 15.0 A

Time, t = 60.0 minutes

Also, 1 minute = 60 seconds

So, t = [tex]60\times 60[/tex] s = 3600 s

F is Faraday constant = 96485 C

Atomic weight of Nickel = 58.69 g/mol

Also, Charge on Ni in [tex]NiCl_2[/tex] = 2

So, equivalent weight of Ni , E = [tex]\frac{58.69}{2}\ g/mol[/tex] = 29.34 g/mol

Thus, according to the Faraday's Law:-

[tex]W=\frac{EIt}{96485}[/tex]

Where, W is the mass of the metal deposited.

So,

[tex]W=\frac{29.34\times 15\times 3600}{96485}\ g=16.4\ g[/tex]

Weight of Ni metal plated out = 16.4 g

Final answer:

To determine the mass of nickel metal plated out, use Faraday's law of electrolysis and the given values of current and time. The answer is 16.4 grams.

Explanation:

To determine the number of grams of nickel metal plated out, we need to use Faraday's law of electrolysis. The formula is:

Mass (g) = (Current (A) * Time (s)) / (Molar mass (g/mol) * Faraday's constant (C/mol))

Given that the current is 15.0 A and the time is 60.0 minutes (or 3600 seconds), we can use the molar mass of nickel (58.69 g/mol) and Faraday's constant (96,500 C/mol) to calculate the mass:

Mass (g) = (15.0 A * 3600 s) / (58.69 g/mol * 96,500 C/mol) = 16.4 g

Therefore, the answer is Choice B) 16.4 g.

Answer:

(a) [tex]2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}[/tex]

(b) 0.726 g

Explanation:

(a)

The balanced chemical equation is shown below as:-

[tex]2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}[/tex]

(b)

Given that:

Temperature = 90 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (90 + 273.15) K = 363.15 K

V = 50.0 mL = 0.05 L ( 1 mL = 0.001 L )

Pressure = 1 atm

Using ideal gas equation as:

[tex]PV=nRT[/tex]

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

1 atm × 0.05 L = n ×0.0821 L atm/ K mol  × 363.15 K

⇒n = 0.001677 mol

Thus, moles of [tex]O_2[/tex] = 0.001677 mol

According to the reaction shown above,

1 mole of [tex]O_2[/tex] is produced when 2 moles of mercury(II) oxide are reacted

0.001677 mole of [tex]O_2[/tex] is produced when [tex]2\times 0.001677[/tex] moles of mercury(II) oxide are reacted

Moles of mercury(II) oxide = 0.003354 moles

Molar mass of mercury(II) oxide = 216.59 g/mol

Mass of mercury(II) oxide = [tex]Moles\times Molar\ mass[/tex] =  [tex]0.003354\times 216.59\ g=0.726\ g[/tex]

0.726 g is the mass of mercury(II) oxide that must have reacted.

The balanced chemical equation for the decomposition of mercury(II) oxide into mercury and dioxygen is: 2HgO(s) → 2Hg(l) + O2(g). The mass of mercury(II) oxide that reacted to produce 50.0 ml of dioxygen gas at 90°C and 1 atm pressure is approximately 0.97 g.

1. The balanced chemical equation for the reaction is: 2HgO(s) → 2Hg(l) + O2(g).

2. First, we need to use the ideal gas law to calculate the number of moles of dioxygen gas produced. At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 L. Therefore, the number of moles of dioxygen is 50.0 ml / 22.4L/mol = 0.00223 moles. Next, we use the balanced chemical equation, which tells us that one mole of HgO produces a half mole of dioxygen gas. So, the amount of HgO reacted is 2 x 0.00223 moles = 0.00446 moles. Finally, we use the molar mass of HgO (216.6 g/mol) to calculate the mass of HgO reacted: 0.00446 moles x 216.6 g/mol = 0.97 g (to three significant digits).

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Answer:

The effective nuclear charge for a valence electron in oxygen atom: [tex]Z_{eff} = 4.55[/tex]

Explanation:

Effective nuclear charge  [tex][Z_{eff}][/tex] is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.

It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s).  

[tex]\Rightarrow Z_{eff} = Z - s[/tex]

For an oxygen atom-

Electron configuration: (1s²) (2s² 2p⁴)

The atomic number (actual nuclear charge): Z = 8

The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:

⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45

Therefore, the effective nuclear charge for a valence electron in oxygen atom is:

[tex]Z_{eff} = Z - s = 8 - 3.45 = 4.55[/tex]

Therefore, the effective nuclear charge for a valence electron in oxygen atom: [tex]Z_{eff} = 4.55[/tex]

To calculate the effective nuclear charge (Zeff) for a valence electron in an oxygen atom, which has an atomic number of 8, we use Slater's rules to determine the shielding constant (S) and subtract it from Z. The result is Zeff = 8 - (2 * 0.85 + 5 * 0.35) = 4.55.

The effective nuclear charge (Zeff) for an electron can be determined using Slater's rules which consider the actual nuclear charge (Z) and the shielding constant (S). To calculate Zeff for a valence electron in an oxygen atom with an atomic number of 8:

The calculated Zeff for a valence electron in an oxygen atom is 4.55.

Answer:

Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L

Explanation:

STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.

According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .

Molar mass of Lithium Oxide = 29.8 g/mol

[tex]Li_{2}O[/tex] = 2(6.9) + 15.99 = 29.8

Mass of 1 mole [tex]Li_{2}O[/tex] = 29.8 g

1 mole [tex]Li_{2}O[/tex] occupies, Volume = 22.4 L

29.8 g [tex]Li_{2}O[/tex] occupies, V = 22.4 L

1 g [tex]Li_{2}O[/tex] occupies ,V = [tex]\frac{22.4}{29.8}[/tex]

1 g [tex]Li_{2}O[/tex] occupies ,V = 0.7516 L

10 g tex]Li_{2}O[/tex] occupies ,V = [tex]0.7516 \times10[/tex] L

V = 7.52 L

So, volume occupied by Lithium Oxide At STP is 7.52 L

Answer:

The following answer are based only on emf values

a).Most anodic element ,Titanium - Ti

b).Metal That will corrode easily is : Titanium - Ti

c). Most reactive metal is Titanium-Ti

d).Al can't be used for making implants because it undergo corrosion easily and do not form strong passive oxide layer.

Ti forms very strong passive layer which prevents further corrosion.

Au is least reactive and does not corrode.

Note : You have to give answer according to E values

Explanation:

Here the oxidation potential data is given (since elements are getting oxidised)

[tex]\Delta^{0} E[/tex]  : Standard Reduction Potential : It is tendency of the element to gain electron and get reduced.

Oxidation Potential is opposite to reduction potential

Standard Oxidation Potential :It is tendency of the element to loose electron and get oxidised.

It is measured in Volts/V

The main application [tex]\Delta^{0} E[/tex] values is in electrochemical cell to predict the products of the reaction. or  predict whether the reaction is spontaneous or not.

Since [tex]Ti\rightarrow Ti^{+}[/tex] has maximum value of [tex]\Delta^{0}E[/tex]  = 2.00V , so it get oxidised as compared to other elements.

From the , E values Ti show maximum tendency of oxidation hence it also get corroded easily.

Final answer:

Titanium is the most anodic and most reactive metal given its high standard electrode potential of +2.00V, which also makes it the most prone to corrosion. Titanium and Gold are suitable for biomedical implants due to their corrosion resistance and biocompatibility, while Aluminum is not used due to potential toxicity.

Explanation:

To address the questions related to electrochemical potentials and the reactivity of metals, we must consider both the given standard electrode potentials and practical implications such as corrosion resistance in the case of biomedically important metals.

Most Anodic and Most Corrosive

The most anodic metal would be the one with the lowest reduction potential, which correlates with the highest oxidation potential. Since oxidation is the loss of electrons, the metal with the highest potential to lose electrons (most positive ΔEº) will be the most anodic and the most susceptible to corrosion. In this list, Titanium (Ti) has the highest standard electrode potential at +2.00 V, indicating it would be the most anodic and would corrode the most in an environment where it can react.

Most Reactive Metal

The reactivity of a metal in chemical terms is often reflected by how easily it can lose electrons (oxidize). By the given data, Titanium (Ti) would be the most reactive given its high standard electrode potential for oxidation.

Biomedical Implant Materials

Titanium (Ti) and Gold (Au) are used for biomedical implants because they are biocompatible and highly resistant to corrosion in the body. Titanium forms a stable, protective oxide layer when exposed to air or bodily fluids, preventing further oxidation. While Aluminum (Al) has a high standard reduction potential, which may suggest it is resistive to corrosion, it is not used for implants due to potential toxicity and less stable oxide formation than Ti, which could lead to corrosion in the body.

Answer:

B) ) –1615.1 kJ mol^–1

Explanation:

since

SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1

the enhalpy of reaction will be

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants

therefore

∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr

4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]

4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol

∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol

therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol

Answer: 0.100 m [tex]K_2SO_4[/tex]

Explanation:

Elevation in boiling point is given by:

[tex]\Delta T_b=i\times K_b\times m[/tex]

[tex]\Delta T_b=T_b-T_b^0[/tex] = Elevation in boiling point

i= vant hoff factor

[tex]K_b[/tex] = boiling point constant

m= molality

1. For 0.100 m [tex]K_2SO_{4}[/tex]

[tex]K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}[/tex]  

, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be [tex]3\times 0.100=0.300[/tex]

2. For 0.100 m [tex]LiNO_3[/tex]

[tex]LiNO_3\rightarrow Li^{+}+NO_3^{-}[/tex]  

, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]2\times 0.100=0.200[/tex]

3.  For 0.200 m [tex]C_2H_8O_3[/tex]

, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be [tex]1\times 0.200=0.200[/tex]

4. For 0.060 m [tex]Na_3PO_4[/tex]

[tex]Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}[/tex]  

, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]4\times 0.060=0.24m[/tex]

Thus as concentration of solute is highest for [tex]K_2SO_4[/tex] , the elevation in boiling point is highest and thus has the highest boiling point.

The aqueous solution of K2SO4 has the highest boiling point among the given options.

The boiling point of a solution depends on the concentration and nature of the solute. In this case, we need to compare the boiling points of different aqueous solutions. Higher concentration and the presence of ionic solutes will generally result in a higher boiling point.

Comparing the given options, K2SO4 and Na3PO4 are both ionic compounds, and their solutions will have higher boiling points due to their strong ionic interactions. LiNO3 is also an ionic compound, but its concentration is lower than K2SO4 and Na3PO4. C3H8O3 is a molecular compound and does not dissociate into ions in solution, so its solution will have a lower boiling point than the ionic compounds. Therefore, the aqueous solution of K2SO4 (option A) will have the highest boiling point among the given options.

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Answer:

a) Nor Mg, neither I2 is the limiting reactant.

b) I2 is the limiting reactant

c) Mg is the limiting reactant

d) Mg is the limiting reactant

e) Nor Mg, neither I2 is the limiting reactant.

f) I2 is the limiting reactant

g) Nor Mg, neither I2 is the limiting reactant.

h) I2 is the limiting reactant

i) Mg is the limiting reactant

Explanation:

Step 1: The balanced equation:

Mg(s) + I2(s) → MgI2(s)

For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2

a. 100 atoms of Mg and 100 molecules of I2

We'll have the following equation:

100 Mg(s) + 100 I2(s) → 100MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

b. 150 atoms of Mg and 100 molecules of I2

We'll have the following equation:

150 Mg(s) + 100 I2(s) → 100 MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.

There will be produced 100 MgI2 molecules.

c. 200 atoms of Mg and 300 molecules of I2

We'll have the following equation:

200 Mg(s) + 300 I2(s) →200 MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.

There will be produced 200 MgI2 molecules.

d. 0.16 mol Mg and 0.25 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.

There will be produced 0.16 mol of MgI2.

e. 0.14 mol Mg and 0.14 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2

f. 0.12 mol Mg and 0.08 mol I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.

There will be produced 0.08 moles of MgI2.

g. 6.078 g Mg and 63.455 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles

Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles

This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.

There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2

h. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles

I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.

There will be produced 0.00788 moles of MgI2.

i. 1.00 g Mg and 2.00 g I2

We'll have the following equation:

Mg(s) + I2(s) → MgI2(s)

Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles

Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles

Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.

There will be produced 0.0411 moles of MgI2.

The limiting reagent in a chemical reaction is the reactant that is completely consumed first. To identify the limiting reagent, we need to compare the mole ratios of the reactants to the mole ratio of the products in the balanced chemical equation.

* Reaction mixture a: neither reactant is limiting.

* Reaction mixture b: Mg is the limiting reagent.

* Reaction mixture c: I2 is the limiting reagent.

* Reaction mixture d: Mg is the limiting reagent.

* Reaction mixture e: neither reactant is limiting.

* Reaction mixture f: I2 is the limiting reagent.

* Reaction mixture g: neither reactant is limiting.

* Reaction mixture h: I2 is the limiting reagent.

* Reaction mixture i: Mg is the limiting reagent.

To identify the limiting reagent in a reaction mixture, we need to compare the mole ratios of the reactants to the mole ratio of the products in the balanced chemical equation. The limiting reagent is the reactant that will be completely consumed before the other reactants are consumed.

**Balanced chemical equation:**

Mg(s) + I2 (s) → MgI2 (s)

**Mole ratios:**

Mg : I2 : MgI2 = 1 : 1 : 1

**a. 100 atoms of Mg and 100 molecules of I2**

Since we have the same number of atoms of Mg and molecules of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**b. 150 atoms of Mg and 100 molecules of I2**

Since we have more atoms of Mg than molecules of I2, Mg is the limiting reagent. I2 will be completely consumed before Mg is consumed.

**c. 200 atoms of Mg and 300 molecules of I2**

Since we have more molecules of I2 than atoms of Mg, I2 is the limiting reagent. Mg will be completely consumed before I2 is consumed.

**d. 0.16 mol Mg and 0.25 mol I2**

Since we have less moles of Mg than moles of I2, Mg is the limiting reagent. I2 will not be completely consumed.

**e. 0.14 mol Mg and 0.14 mol I2**

Since we have the same number of moles of Mg and moles of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**f. 0.12 mol Mg and 0.08 mol I2**

Since we have less moles of I2 than moles of Mg, I2 is the limiting reagent. Mg will not be completely consumed.

**g. 6.078 g Mg and 63.455 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 6.078 g / 24.31 g/mol = 0.250 mol

Moles of I2 = 63.455 g / 253.808 g/mol = 0.250 mol

Since we have the same number of moles of Mg and moles of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.

**h. 1.00 g Mg and 2.00 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 1.00 g / 24.31 g/mol = 0.0411 mol

Moles of I2 = 2.00 g / 253.808 g/mol = 0.00791 mol

Since we have less moles of I2 than moles of Mg, I2 is the limiting reagent. Mg will not be completely consumed.

**i. 1.00 g Mg and 20.00 g I2**

First, we need to convert the masses of Mg and I2 to moles:

Moles of Mg = 1.00 g / 24.31 g/mol = 0.0411 mol

Moles of I2 = 20.00 g / 253.808 g/mol = 0.0788 mol

Since we have more moles of I2 than moles of Mg, Mg is the limiting reagent. I2 will not be completely consumed.

Learn more about limiting reagent here:

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The final answer for the solution is Q= -8.8171*[tex]10^{-4}[/tex] kW

Explanation:

[tex]CO +H_2O  ⇄  CO_2+H_2[/tex]

basic:

100 moles of product gas

[tex]CO_2[/tex] = 40 moles

[tex]H_2[/tex] = 40 moles

[tex]H_20[/tex] = 20 moles

for 40 moles of[tex]CO_2[/tex] we need 40 moles of CO and 40 moles of [tex]H_2O[/tex]

therefore,

Inlet :

No of moles of CO = 40

No of moles of [tex]H_20[/tex] = 40

but [tex]H_20[/tex] is in the outlet stream = 20 modes

Excess = [tex]\frac{20}{40}\ \times 100[/tex]%

= 50%

50% excess stream is fed to the reactor

a) Rate of conversion:

ρ_water (density ) =[tex]\frac{1g}{cm^{3}}[/tex]

Given volume rate = 3.5 [tex]\frac{cm^{3} }{hr}[/tex]

3.5 * density = 3.5 * [tex]10^{-3}[/tex] [tex]\frac{kg}{hr}[/tex]

b) Rate (kW)

ΔH = ΔH_1 + ΔH_2 +ΔH_3

Rate of condensation Q° =n° ΔH

Q = n° (ΔH_1 + ΔH_2 +ΔH_3)

ΔH_1 is sensible heat

V =[tex]\int\limits^T_S {C_p(T) } \, dT[/tex]

C_p(T) = a + bT + CT^2 +dT^3

H_2O

a=32.218

b=o.192 * 10^-2

c= 1.055 * 10^-5

d= -3.593 * 10^-9

ΔH_2 = ∫(a+bT +a^2 +dT^3)dT (of limits 100 to 500 )

= -13497.6

ΔH_3 =∫(a +bT +T^2 +dT^3)dT (of limit 95 to 100)

= -2751.331

phase change

ΔH_2 =  - ΔH_v

ΔH_v = 7.3  @ 100° C

liquid → vapor ⇒ ΔH_v

vapor → liquid ⇒ -ΔH_v

Q= n°(-16324.231)

=350 *[tex]10^{-5}[/tex] (-16324.231)

=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) *[tex]\frac{kg}{hr}[/tex][tex]\frac{J}{mol}[/tex]

=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) * \frac{J}{mol}[/tex]

=[tex]\frac{350* 10^{-5} }{18}[/tex]  (-16324.231)  *kW

Q= -8.8171*[tex]10^{-4}[/tex] kW

The solution involves various calculations related to a chemical reactor, including determining the excess steam fed into the reactor, condensation rate, heat removal rate, and heat transfer rate.

This is a complex chemical engineering problem involving numerous calculations, including finding the excess steam fed into the reactor, condensation rate, heat removal rate, and heat transfer rate. Firstly, the percentage excess steam is found by comparing the stochiometric requirement of the reaction with actual steam flow. Next, we use the molar mass of water and the gas constant to find the condensation rate of water. For the heat removal rate, we need to consider the latent heat of condensation along with the energy balance of the condenser.

Lastly, for the heat transfer rate in the reactor, we need to take into account the enthalpy of reaction, heat capacities of reactants and products, and the reactor temperature profile to decide whether heat is added or removed from the reactor.

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Answer:

There are formed 98.05 g of MnCl₂

Explanation:

The reaction is this one:

MnO₂ + 4 HCl   →  MnCl₂ +  2 H₂O + Cl₂

First of all, determinate moles. Divide mass /molar mass

150 g / 36.45 g/m = 4.11 moles of HCl

Ratio between HCl and MnCl₂ is 4:1

4 moles of HCl produce 1 mol of Chloride

4.11 moles of HCl  'll produce (4.11 . 1)/ 4 =1.03 moles of chloride

Molar mass . Moles = Mass

Molar Mass MnCl₂ = 95.2 g/m

95.2 g/m  . 1.03 moles = 98.05 grams

Answer:

There are 129.4 grams of MnCl2 formed

Explanation:

Step 1: Data given

Mass of HCl = 150.0 grams

MnO2 = excess

Molar mass of HCl = 36.46 g/mol

Step 2: The balanced equation

MnO2+4HCl → MnCl2+2H2O+Cl2

Step 3: Calculate moles of HCl

Moles HCl = Mass HCl / molar mass HCl

Moles HCl = 150.0 grams / 36.46 g/mol

Moles HCl = 4.114 moles

Step 4: Calculate Moles of MnCl2

For 1 mol of MnO2 we need 4 moles of HCl to produce 1 mol of MnCl2, 2 moles of H2O and 1 mol Cl2

For 4.114 moles of HCl we'll have 4.114/4 = 1.0285 moles of MnCl2

Step 5: Calculate mass of MnCl2

Mass MnCl2 =moles MnCl2 * molar mass MnCl2

Mass MnCl2 = 1.0285 * 125.84 g/mol

Mass MnCl2 = 129.4 grams

There are 129.4 grams of MnCl2 formed

Answer:

15P) 3s² 3p³

19K) 4s¹

28Ni) 3d⁸ 4s²

35Br) 3d¹⁰ 4s² 4p⁵

58Ce) 4f¹ 5d¹ 6s²

Explanation:

1) 15P

Phosphorus has 15 electrons

The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . .

The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons

The electron configuration for 15P is: 1s2 2s2 2p6 3s2 3p3

3s² 3p³

2) 19K

Potassium has 19 electrons

The electron configuration for 19K is: 1s2 2s2 2p6 3s2 3p6 4s1

4s¹

3) 28Ni

Nickel has 28 electrons

The electron configuration for 28Ni is: 1 s 2  2 s 2  2 p 6  3 s 2  3 p 6  4 s 2  3 d 8

3d⁸ 4s²

4) 35Br

Bromine has 35 electrons

The electron configuration for 35Br is: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5

3d¹⁰ 4s² 4p⁵

5) 58Ce

Cerium has 58 electrons

The electron configuration for 58Ce is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f1 5d1

4f¹ 5d¹ 6s²

Answer:

The heat of combustion of liquid butane is: -2636 kJ/mol

Explanation:

This problem is solved by using Hess law of heats summation:

We have the heat of combustion for gaseous butane:

C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)     ΔHºcomb = -2658 kJ/mol

and we want the heat of combustion for liquid butane.

What we need to do is add the heat of vaporization for liquid butane to this equation:

C₄H₁₀ (l)   ⇒  C₄H₁₀ (g)   ΔHºvap =  22.44 kJ/mol

C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)  ΔHºcomb = -2658 kJ/mol

______________________________________________

C₄H₁₀ (l)   + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)  

So,

ΔH comb liq butane = -2658 kJ/mol + 22.44 kJ/mol = -2636 kJ/mol

Answer:

[tex]Br_2[/tex]

Explanation:

The compound [tex]X_2[/tex] is [tex]Br_2[/tex].

The reaction included are:

[tex]Br_2+NaCl--> No\ reaction[/tex]  (because  bromine is less reactive than chlorine)

But , because of hexane the solution get dilute and its color changes to orange.

Now, NaI is added and we know Br is more reactive than I.

Therefore it replace it.

Reaction: [tex]Br_2+NaI-->NaBr+I_2[/tex]

Purple color develop due to formation of Iodine.