A chemist prepares a solution of iron chloride by measuring out of into a volumetric flask and filling to the mark with distilled water. Calculate the molarity of anions in the chemist's solution.

Answer:

The answer to your question is below

Explanation:

a) Acid dissociation reaction for HCl

                  HCl(g)  +   H₂O(l)    ⇒    H⁺¹  (aq)  +  Cl⁻¹ (aq)

b) pH = -log [H⁺¹]

Concentration = 5 x 10⁻⁴

                   pH = -log [5 x 10⁻⁴]

                    pH = 3.3

c) Acid dissociation reaction

                       NaOH(s)  + H₂O   ⇒   Na⁺¹(aq)   +   OH⁻¹(aq)

d) pH

                      pOH = -log [7 x 10⁻⁵]

                      pOH = 4.2

                      pH = 14 - 4.2

                     pH = 9.8

The acid dissociation and pH for HCl and NaOH were detailed. For HCl, the reaction and solution pH were HCl -> H+ + Cl- and 3.3 respectively. For NaOH, the dissociation and solution pH were NaOH -> Na+ + OH- and 9.8 respectively.

a) The acid dissociation reaction for hydrochloric acid (HCl) is as follows:

HCl → H+ + Cl-

b) The pH of a solution of hydrochloric acid can be calculated using the formula:

pH = -log[H+].

As HCl is a strong acid, it will completely ionize in water, meaning the concentration of H+ ions equals the concentration HCl. Therefore, the pH for a 5.0 x 10^-4 M HCl solution is -log(5.0 x 10^-4) = 3.3.

c) Sodium hydroxide is a base, not an acid, so its ionization does not involve releasing protons into the solution, but rather attracting them. It has the following dissociation reaction:

NaOH → Na+ + OH-.

d) As for the pH of a solution of 7.0 x 10^-5 M NaOH, we first need to find the pOH using the formula pOH = -log[OH-], since NaOH being a strong base completely ionizes. This equals -log(7.0 x 10^-5) = 4.2. Given that pH and pOH are related by the expression pH + pOH = 14 at 25°C, we can solve for the pH of the NaOH solution = 14 - pOH = 14 - 4.2 = 9.8.

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Answer:

1. 2.04 W/m²

2. 1.63°C

Explanation:

The radiative force that the Earth receives comes from the Sun. When the Sun rays come to the surface, some of them are absorbed and then it is reflected in the space. The greenhouse gases (like CO2) blocks some of these rays, and then the surface stays warm. The excessive amount of these gases makes the surface warmer, which unbalance the climate on Earth.

1. The variation of the radiative forcing can be calculated based on the concentration of the CO2 by the equation:

ΔF = 5.35*ln(C/C0)

Where C is the final concentration, and C0 is the initial concentration.

ΔF = 5.35*ln(410/280)

ΔF = 2.04 W/m²

2. The temperature change in the Earth's surface caused by the variation of the radiative forcing can be calculated by:

ΔT = 0.8*ΔF

ΔT = 0.8*2.04

ΔT = 1.63 K = 1.63°C

The increase in CO2 concentration from 280 ppm to 410 ppm today results in an extra radiative forcing of 2.2 W/m² on Earth's surface. The equivalent temperature change is approximately 1.7°C.

The increase in CO2 concentration from 280 ppm to 410 ppm today has resulted in an extra radiative forcing on Earth's surface. The radiative forcing is calculated using the formula ΔF (Wm^(-2)) = α ln(C/C0), where α = 5.35. By plugging in the values, the radiative forcing is approximately 2.2 W/m².

The equivalent temperature change can be calculated using the formula ΔT(K) = λ*ΔF, where λ = 0.8 per (Wm^(-2)). By multiplying the radiative forcing by the climate sensitivity, we get a temperature increase of about 1.7°C.

It is important to note that the temperature increase may catch up to a new equilibrium once oceans warm and ice melts, resulting in a higher temperature increase in the future.

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To calculate the amount of heat needed to boil ethanol, use the formula Q = m * c * ΔT. Convert the temperature to Kelvin, calculate the change in temperature, and then apply the formula to find the amount of heat. The amount of heat needed to boil 183.g of ethanol is 20 kJ.

To calculate the amount of heat needed to boil ethanol, we can use the equation:

Q = m * c * ΔT

Where Q is the amount of heat, m is the mass of ethanol, c is the specific heat capacity of ethanol, and ΔT is the change in temperature.

First, we need to calculate the change in temperature:

Now we can calculate the amount of heat:

Therefore, the amount of heat needed to boil 183.g of ethanol is 20 kJ.

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Answer:

Radius of platinum atom is [tex]1,38\times10^{-8} \ cm.[/tex]

Explanation:

We know, Density in solid state is also given by , [tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]

Here, Z= total no of atom or molecule per unit cell. ( Z = 4 for FCC structure )

M = atomic weight = 195.08 gm/mol.

N_A= Avogadro's number = [tex]6.023\times 10^{23} .[/tex]

a= size of side of cube.

Also for, FCC crystal [tex]a=\sqrt2\times 2r[/tex]    ....1

Since,[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}\\[/tex]

Therefore, [tex]a^3=\dfrac{Z\times M}{N_A\times \rho}[/tex]

Putting all values , [tex]a=3.9\times 10^{-8} \ cm.[/tex]

Putting value of a in equation 1 we get,

[tex]r=\dfrac{a}{\sqrt2\times2}=1.38\times 10^{-8} \ cm.[/tex]

Hence, this is the required solution.

The radius of a platinum atom can be found using the density, atomic weight, and the nature of the crystal structure. The calculation involves finding the volume of an atom from the atomic weight, density and Avogadro's number and converting this volume to a radius by considering the atom as a sphere.

The radius of a platinum atom can be calculated using the density, atomic weight, and the information about the crystal structure (in this case, face-centered cubic or FCC). Platinum (Pt) has a density (d) of 21.4 g/cm3 and an atomic weight (AW) of 195.08 g/mol.

First, let's calculate the number of atoms (n) in the unit cell of an FCC structure. An FCC unit cell has 4 atoms. Avogadro's number (Na) is 6.022 x 1023 mol-1.

Keep in mind that the values have to be converted to cm in the end.

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Answer: The density of phosphorus is [tex]3.81g/cm^3[/tex]

Explanation:

To calculate the density of phosphorus, we use the equation:

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]

where,

[tex]\rho[/tex] = density

Z = number of atom in unit cell = 1  (CCP)

M = atomic mass of phosphorus = 31 g/mol

[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]

a = edge length of unit cell = [tex]238pm=238\times 10^{-10}cm[/tex]    (Conversion factor:  [tex]1cm=10^{10}pm[/tex]  )

Putting values in above equation, we get:

[tex]\rho=\frac{1\times 31}{6.022\times 10^{23}\times (238\times 10^{-10})^3}\\\\\rho=3.81g/cm^3[/tex]

Hence, the density of phosphorus is [tex]3.81g/cm^3[/tex]

The question is incomplete, here is the complete question:

A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Answer: The concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]

Explanation:

To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the stock magnesium fluoride solution

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of chemist's magnesium fluoride solution

We are given:

[tex]M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL[/tex]

Putting values in above equation, we get:

[tex]0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M[/tex]

Hence, the concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]

Answer:

a) K = 5.3175

b) ΔG = 3.2694

Explanation:

a) ΔG° = - RT Ln K

∴ T = 25°C ≅ 298 K

∴ R = 8.314 E-3 KJ/K.mol

∴ ΔG° = - 4.140 KJ/mol

⇒ Ln K = - ( ΔG° ) / RT

⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))

⇒ Ln K = 1.671

⇒ K = 5.3175

b) A → B

∴ T = 37°C = 310 K

∴ [A] = 1.6 M

∴ [B] = 0.45 M

∴ K = [B] / [A]

⇒ K = (0.45 M)/(1.6 M)

⇒ K = 0.28125

⇒ Ln K = - 1.2685

∴ ΔG = - RT Ln K

⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )

⇒ ΔG = 3.2694

Explanation:

The important quantum-mechanical concepts associated with the Bohr model of atom are :

1. Electrons are nothing but particles that revolve around the nucleus in discrete orbitals.

2. Energy is associated with each orbital is quantised. Meaning electron in each shell will have energy in multiple of a fixed quanta.

Final answer:

The Bohr model introduced two key concepts of quantum mechanics: the quantization of energy levels and the quantization of angular momentum, both crucial for understanding atomic structure and the emission spectra of atoms.

Explanation:

The Bohr model of the atom represents an early understanding of quantum mechanics, specifically applicable to the hydrogen atom. It introduced two fundamental quantum-mechanical concepts that underpin the modern understanding of atomic structure and behavior, despite its ultimate replacement by more comprehensive theories.

Quantized Energy Levels

The first important concept is the quantization of energy levels of electrons within an atom. According to the Bohr model, electrons can only occupy certain allowed energy levels, and transitions between these levels result in the absorption or emission of photons with specific energies. This concept matched the observed discrete spectra of hydrogen, although the model assumed circular orbits for electron paths, which was later found to be incorrect.

Quantized Angular Momentum

The second concept is the quantization of angular momentum. Bohr proposed that the angular momentum of an electron in orbit is quantized and can only take on certain values, each associated with a different orbit. Although Bohr's assumptions about specific, circular orbits were eventually supplanted by the probabilistic nature of electron positioning in quantum mechanics, the idea that angular momentum is quantized remains a fundamental aspect of quantum theory.

Answer: 319 ml

Explanation:

Given : 86 g of [tex]H_3PO_4[/tex] is dissolved in 100 g of solution.

Density of solution = 1.71 g/ml

Volume of solution=[tex]\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.71g/ml}=58.5ml[/tex]

moles of [tex]H_3PO_4=\frac{\text {given mass}}{\text {Molar mass}}=\frac{86g}{98g/mol}=0.88mol[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex]     .....(1)

Molarity of standard [tex]H_3PO_4[/tex] solution =[tex] \frac{0.88\times 1000}{58.5ml}=15.04M[/tex]  

To calculate the volume of acid, we use the equation given by neutralization reaction:

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of standard acid which is [tex]H_3PO_4[/tex]

[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted acid which is [tex]H_3PO_4[/tex]

We are given:

[tex]M_1=15.04M\\V_1=?mL\\\\M_2=4.00M\\V_2=1200mL[/tex]

Putting values in above equation, we get:

[tex]15.04\times V_1=4.00\times 1200\\\\V_1=319mL[/tex]

Thus 319 ml of the standard [tex]H_3PO_4[/tex] solution are required.

To prepare a 1200 mL of 4.00 M H3PO4 solution from an 86% w/w H3PO4 solution with a specific density of 1.71 g/mL, 320 mL of the standard solution is required.

To determine how many milliliters of a standard 86% H3PO4 solution (with a specific density of 1.71 g/mL) are required to prepare a 1200 mL 4.00 M H3PO4 solution, we can use the concept of molarity and the equation M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the diluted solution, respectively.

First, we need to calculate the molarity of the stock solution (86% H3PO4). We know that 86% (w/w) means there are 86 grams of H3PO4 in 100 grams of solution. As the density of the solution is 1.71 g/mL, we can find the volume that 100 grams of solution will occupy:

Volume = mass / density = 100 g / 1.71 g/mL = 58.48 mL

Now, we calculate the mols of H3PO4 in 86 grams:

Moles H3PO4 = mass (g) / molar mass (g/mol) = 86 g / 97.99 g/mol ≈ 0.877 mol

The molarity (M1) of the stock solution is:

Molarity = moles of solute / volume of solution (L) ≈ 0.877 mol / 0.05848 L ≈ 15.00 M

Next, using the dilution equation M1V1 = M2V2, where M2 = 4.00 M and V2 = 1200 mL (1.2 L), we can solve for V1:

V1 = (M2V2) / M1 = (4.00 M * 1.2 L) / 15.00 M = 0.32 L

Since we need the volume in milliliters:

V1 = 0.32 L * 1000 mL/L = 320 mL

Therefore, 320 mL of the standard 86% H3PO4 solution is required to prepare a 1200 mL of 4.00 M H3PO4 solution.

Answer:

4.7 × 10¹⁴ g/cm³

Explanation:

The mass of the sun (and of the neutron star) is 1.98 × 10³⁰ kg = 1.98 × 10³³ g.

The diameter of the neutron star is 20 km and its radius is 1/2 × 20 km = 10 km = 10⁴ m = 10⁶ cm

If we consider the neutron star to be approximately spherical, we can calculate its volume using the following expression.

V = 4/3 × π × r³

V = 4/3 × π × (10⁶ cm)³

V = 4.19 × 10¹⁸ cm³

The density (ρ) is equal to the quotient between the mass and the volume.

ρ = 1.98 × 10³³ g / 4.19 × 10¹⁸ cm³ = 4.73 × 10¹⁴ g/cm³ ≈ 4.7 × 10¹⁴ g/cm³

The rate constants for the forward and reverse steps given that the forward step is first-order in A, and the reverse step is first-order in both B and C is 6.67 x 10⁻⁹ sec⁻¹.

Rate constant is defined as the proportionality constant in the equation expressing the association between a chemical reaction's pace and its constituents' concentrations. The dependence of the molar concentration of the reactants on the rate of reaction can be determined by knowing the rate constant.

Given rt = 3μs = 3 x 10⁻⁶ s

K eq = 2 x 10⁻¹⁶

B = 2 x 10⁻⁴

C = 2 x 10⁻⁴

rt = 1 / Kf + Kr

3 x 10⁻⁶ = 1 / Kf + Kr ------- ( 1 )

K eq = Kf / Kr

2 x 10⁻¹⁶ =  Kf / Kr ----------- ( 2 )

From equation 1 and 2

3 x 10⁻⁶ =  1 / Kf + Kf / 2 x 10⁻¹⁶

3 x 10⁻⁶ =  1 / 2 x 10⁻¹⁶ x Kf + Kf  / 2 x 10⁻¹⁶

Kf =  2 x 10⁻¹⁶ / 3 x 10⁻⁶ ( 2 x 10⁻¹⁶ + 1)

Kf = 6.67 x 10⁻⁹ sec⁻¹

Thus, the rate constants for the forward and reverse steps given that the forward step is first-order in A, and the reverse step is first-order in both B and C is 6.67 x 10⁻⁹ sec⁻¹.  

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Answer:

volume =0.9645cm³

Explanation:

density refers to the ratio of mass to volume of a substance. It gives us a clue about how heavy an object is.

density = mass/volume

volume = mass/density

mass=21.8g

density=22.6g

volume =21.8/22.6

volume =0.9645cm³

The volume occupied by a 21.8 g sample of osmium with a density of 22.6 g/cm³ is approximately 0.9646 cm³, calculated by dividing the mass by the density.

To find the volume (in cm³) occupied by a 21.8 g sample of osmium, you can use the formula:

Volume = Mass / Density

Given:

- Mass (m) = 21.8 g

- Density (D) = 22.6 g/cm³

Now, plug these values into the formula:

Volume = 21.8 g / 22.6 g/cm³

Volume ≈ 0.9646 cm³

So, the volume occupied by the 21.8 g sample of osmium is approximately 0.9646 cm³.

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Answer: Thus chopping a log into sawdust is not an example of a chemical reaction.

Explanation:

Physical reaction is a reaction in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.

Example: chopping a log into sawdust as the shape and size of wood changes.

Chemical reaction is a reaction in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.

Example: burning a plastic water bottle : as the plastic would combine with oxygen to give oxidised products.

the production of hydrogen gas from water: the chemical composition changes

the tarnishing of a copper penny: oxidation of copper takes place to give copper oxide.

charging a cellular phone: the electrical energy is used for a chemical reaction

Chopping a log into sawdust is an example of a physical change because it changes the size not the composition of the substance.

Chemical reactions involve such kind change that leads to the formation of a new product (compound).

For example- the production of hydrogen gas from water, the burning of plastic produce gases.

These reactions refer to the rearrangements of the atoms in a substance but do not to the formation of a new product such as temperature, phase, size, etc.

For example- chopping a log into sawdust

Therefore, chopping a log into sawdust is an example of a physical change because it changes the size not the composition of the substance.

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Answer: density =3.377g/cm³

Explanation:

Density =( molecular weight × effective number of atoms per unit cell) / (volume of unit cell × avogadro constant)

D= (M ×n) /(V×A)

M= 137g/mol

n= 2 (For BCC)

V=a³ , where a= 4r/√3

a= (4×222)/√3

a=512.69pm

a= 512.69×10^-10cm

V= ( 512.69×10^-10)^3

V= 1.3476×10^-22cm³

D= (137×2)/(1.3476×10^-22 × 6.02^23)

D= 3.377g/cm³

Therefore the density of barium is 3.377g/cm³

Given:

Radius of barium,

We know,

→ [tex]a = \frac{4r}{\sqrt{3} }[/tex]

     [tex]= \frac{4\times 222}{\sqrt{3} }[/tex]

     [tex]= 512.69 \ pm[/tex]

     [tex]= 512.69\times 10^{-10} \ cm[/tex]

Now,

→ [tex]V = a^3[/tex]

      [tex]= (512.69\times 10^{-10})^3[/tex]

      [tex]= 1.3476\times 10^{-22} \ cm^3[/tex]

hence,

The density will be:

→ [tex]D = \frac{M\times n}{V\times A}[/tex]

      [tex]= \frac{137\times 2}{1.3476\times 10^{-22}\times 6.02\times 10^{23}}[/tex]

      [tex]= 3.377 \ g/cm^3[/tex]

Thus the response above is right.

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Answer:

Kinetic Energy of the particle is half the mass of the particle multiplied by the square of the velocity of the particle

Explanation:

Kinetic Energy is given by 1/2mv^2

Where, m is the mass of the object and v is the velocity of the object

Answer: kinetic energy is equal to 1/2mv^2

Answer: 0.39mol of the steam are in the cooker.

Explanation:Please see attachment for explanation

Given:

                                   = 115+273

                                   = 388 K

We know the relation,

→ [tex]PV = nRT[/tex]

or,

→     [tex]n = \frac{PV}{RT}[/tex]

By substituting the values, we get

          [tex]= \frac{4\times 3.1}{0.082\times 388}[/tex]

          [tex]= 0.39 \ mol[/tex]

          [tex]= 0.39 \ mol[/tex]

Thus the above answer is correct.

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Answer:

51.2g of CO2

Explanation:

The first step is to balance the reaction equation as shown in the solution attached. Without balancing the reaction equation, one can never obtain the correct answer! Then obtain the masses of octane reacted and carbon dioxide produced from the stoichiometric equation. After that, we now compare it with what is given as shown in the image attached.

To calculate the mass of carbon dioxide produced from the complete combustion of octane, you can use the balanced chemical equation and the molar masses of octane and CO2. First, calculate the number of moles of octane, then use the ratio of CO2 to octane to find the moles of CO2. Finally, multiply the moles of CO2 by the molar mass of CO2 to find the mass of CO2 produced.

To calculate the mass of carbon dioxide produced from the complete combustion of octane (C8H18), you need to use the balanced chemical equation: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O. From this equation, you can see that for every 2 molecules of octane burned, 16 molecules of CO2 are generated. To find the mass of CO2 produced, you can use the molar mass of CO2 (44.01 g/mol) and the molar mass of octane (114.22 g/mol).

First, calculate the number of moles of octane: moles of octane = mass of octane / molar mass of octane = 16.6 g / 114.22 g/mol = 0.145 mol

Since 2 molecules of octane produce 16 molecules of CO2, you can multiply the number of moles of octane by the ratio of CO2 to octane: moles of CO2 = moles of octane x (16 mol CO2 / 2 mol octane) = 0.145 mol x (16 mol CO2 / 2 mol octane) = 1.16 mol CO2

Finally, calculate the mass of CO2 produced: mass of CO2 = moles of CO2 x molar mass of CO2 = 1.16 mol x 44.01 g/mol = 51.08 g

Therefore, the mass of carbon dioxide produced from the complete combustion of 16.6 g of octane is approximately 51.08 g.

While metallic bonds have strong attractions and ionic bonds involve the transfer and acceptance of electrons from the valence shell, covalent bonds involve sharing electrons.

Chemical bonds are defined as a link made by two surfaces or items that have been brought together, often by heat, pressure, or an adhesive agent. Chemical bonds can have many different types, but covalent and ionic bonds are the most well-known. When one atom has less energy, the other has enough thanks to these bonds. Atoms are held together by the force of attraction, which enables the electrons to unite in a bond.

In contrast to covalent bonding, which involves atoms sharing their additional electrons locally, metallic bonding involves all atoms giving off their extra electrons and forming a sea of them. The ability of an atom to stick together and fill the orbits of its outermost electrons in order to arrange itself in a form that is the most stable.

Thus, while metallic bonds have strong attractions and ionic bonds involve the transfer and acceptance of electrons from the valence shell, covalent bonds involve sharing electrons.

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Answer:

ρ = 0.439 g/mL

Explanation:

Given data

0.000285 L × (1000 mL/ 1 L) = 0.285 mL

The density (ρ) of a liquid is equal to its mass (m) divided by its volume (V).

ρ = m/V

ρ = 0.125 g/0.285 mL

ρ = 0.439 g/mL

The density of the liquid is 0.439 g/mL.

The density of a liquid, defined as mass per unit volume, with given mass of 0.125 grams and volume of 0.000285 liters, can be calculated using the formula: Density = mass/volume. After converting liters to milliliters, the density is approximately 0.439 g/mL.

The question is asking for the calculation of the density of a liquid given its mass and volume. Density is a property of matter defined as mass per unit volume. It can be calculated using the formula: Density = mass/volume. In this case, the mass of the liquid is 0.125 grams and the volume is 0.000285 liters. But please note that the density should be presented in g/mL. Therefore, we need to convert the volume from liters to milliliters. 1 liter equals 1000 milliliters, so 0.000285 liters equals 0.285 milliliters. Using the formula, Density = 0.125 g / 0.285 mL which gives us a density of approximately 0.439 g/mL.

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Answer:

energy required is 0.247kJ

Explanation:

The formula to use is Energy = nRdT;

Where n is number of mole

R is the molar gas constant

dT is the change in temperature

n = reacting mass of mercury / molar mass of mercury = 27.4/200.59 = 0.137

dT = final temperature - initial temperature = 376.20 - 158.30 = 217.90K

R = 8.314Jper mol per Kelvin

Energy = 0.137 x 8.314 x 217.90 = 247.12J

Energy in kJ= 247.12/1000= 0.247kJ

Explanation:

Atomic number of carbon is 6 and its electronic distribution is 2, 4. As there are 4 valence electrons present in a carbon atom. Hence, it belongs to group 14 of the periodic table.

So, being a non-metal carbon atom tends to form covalent bonds. Also, the electronegativity value of carbon atom is mid-way between those of metals and non-metals.

In order to complete its octet, carbon atom tends to gain or share its 4 valence electrons. Therefore, it forms four bonds in all its compounds.

The Si-Si bond will be less stronger than C-C bond due to the larger size of silicon atom. As a result, there will be less overlapping between the silicon atoms due to which the bond formed will be weak in nature.

Whereas C-C bond is more stronger as a carbon atom is smaller in size as compared to silicon atom. As a result, more will be the overlapping between the carbon atoms. Hence, more stronger will be C-C bond than Si-Si bond.

As there will be covalent bonds present in a carbon chain and these bonds are weak as compared to ionic bonds. Hence, relatively little heat is released when a C chain reacts and one bond replaces the other.

Thus, we can conclude that out of the given options statements of I, II, III and V are true.

Answer:

P(total) = 46.08 atm

Explanation:

Given data:

Mass of NH₂COONH₄ = 6.8 g

Kc = 0.008

Temperature = 202°C = 202+273 = 475 K

Total pressure at equilibrium = ?

Solution:

Equilibrium equation:

NH₂COONH₄   ⇄   2NH₃ + CO₂

Formula:

Kp = Kc(RT)³

Kp = 0.008 (0.0821 atm.L/mol.K )³(475K)³

Kp = 471.56

                                                                  NH₃       CO₂

            Initial concentration                       0          0

            Change in concentration             2x           x

            equilibrium concentration            2x           x

AS

471.56 = 2x (x)

471.56 = 2x²

x² = 471.56 /2

x² = 235.78

x = 15.36

Pressure of ammonia = 2x = 2(15.36) = 30.72 atm

Pressure of carbon dioxide = x = 15.36 atm

Total pressure:

P(total) = P(NH₃) + P(CO₂)

P(total) = 30.72 atm +  15.36 atm

P(total) = 46.08 atm

The question is incomplete, here is the complete question:

At 20°C the vapor pressure of benzene [tex](C_6H_6)[/tex] is 75 torr, and that of toluene [tex](C_7H_8)[/tex] is 22 torr. Assume that benzene and toluene form an ideal solution.

What is the mole fraction of benzene in the solution that has a vapor pressure of 38 torr at 20°C? Express your answer using two significant figures.

Answer: The mole fraction of benzene is 0.302

Explanation:

Let the mole fraction of benzene be 'x' and that of toluene is '1-x'

To calculate the total pressure of the mixture of the gases, we use the equation given by Raoult's law, which is:

[tex]p_T=\sum_{i=1}^n(\chi_{i}\times p_i)[/tex]

We are given:

Vapor pressure of benzene = 75 torr

Vapor pressure of toluene = 22 torr

Vapor pressure of solution = 38 torr

Putting values in above equation, we get:

[tex]38=[(75\times x)+(22\times (1-x))]\\\\x=0.30[/tex]

Hence, the mole fraction of benzene is 0.30

Final answer:

The mole fraction of benzene in a solution with a vapor pressure of 38 torr at 20 C is calculated using Raoult's law with the vapor pressure of pure benzene. The calculation yields a value of approximately 0.5086, which is rounded to 0.51 when expressed in two significant figures.

Explanation:

Calculation of Mole Fraction of Benzene in a Solution

To calculate the mole fraction of benzene in a solution based on vapor pressure data, we use Raoult's law. According to Raoult's law, the vapor pressure of a solution component is equal to the product of the mole fraction of that component in the solution and the vapor pressure of the pure component. Since the question does not specify the total vapor pressure or the vapor pressure of pure benzene at 20°C, we will refer to the additional information provided that indicates at 20°C, the vapor pressures of pure benzene and toluene are 74.7 mmHg and 22.3 mmHg, respectively, to guide the calculation. Thus, the mole fraction (X) of benzene can be found by rearranging the Raoult's law equation: P = XP0, where P is the vapor pressure of the solution, and P0 is the vapor pressure of pure benzene. This rearrangement gives us X = P / P0.

Given that the solution has a vapor pressure of 38 torr (mmHg), and using the provided vapor pressure of pure benzene ( 74.7 mmHg), the calculation for the mole fraction of benzene (XC6H6) is: XC6H6 = 38 mmHg / 74.7 mmHg

From this calculation, the mole fraction of benzene is approximately 0.5086, which can be rounded to two significant figures to give a final answer of 0.51.

Answer:

See explanation below for answer

Explanation:

We'll do this question in two parts:

a) Weight percent of the alloy:

This is pretty easy, all we have to do is to sum the mass of each element in the alloy and then, divide each mass between the total to get the percentage in weight. In other words:

%W = m/mtotal * 100

The total mass:

mtotal = 98 + 65 = 163 g

%Sn = 98/163 * 100 = 60.12%

%Pb = 65/163 * 100 = 39.88%

b) atom percent of the alloy:

In this case, we do something similar that in part a) with the difference that instead of using mass, we will use moles of each element. The molar mass of Tin and Lead reported are 118.71 g/mol and 207.2 g/mol. The expression to use will be:

C = n/ntotal * 100

so for each element, the moles are:

moles Sn = 98/118.71 = 0.8255 moles

moles Pb = 65/207.2 = 0.3137 moles

the total moles:

ntotal = 0.8255 + 0.3137 = 1.1392 moles

So the composition in atoms for each element is:

C Sn = 0.8255/1.1392 * 100 = 72.46%

C Pb = 0.3137/1.1392 * 100 = 27.54%

The weight percent of tin in an alloy of 98 g tin and 65 g lead is 60.12%, and the weight percent of lead is 39.88%. The atom percent of tin is 71.26% and the atom percent of lead is 28.74%.

The question is asking for the composition in weight percent of an alloy that contains 98 g tin and 65 g lead, and also its composition in atom percent.

To find the weight percent of a component in an alloy, the formula is: (mass of component/total mass) x 100%. Therefore, the weight percent of tin in this alloy is: (98/(98+65)) x 100% = 60.12% and the weight percent of lead is 100% - 60.12% = 39.88%.

The composition in atom percent refers to the number of atoms of a component over the total number of atoms. Lead and tin both have different atomic masses, so we can't use the same weights. The atomic mass of tin is 118.71 g/mol and lead is 207.2 g/mol. Therefore, the atom percent of tin is: (98/118.71)/(98/118.71 + 65/207.2) x 100% = 71.26%. The atom percent of lead is 100% - 71.26% = 28.74%.

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Answer:

CaAsHO₄

Explanation:

The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.

Now, to calculate the empirical formula, we can do this in three steps.

Step 1. Calculate the amount in moles of each element.

In these case, we just divide the percent composition with the molar mass of each one of them:

Ca: 22.3 / 40.078 = 0.5564

As: 41.6 / 74.9216 = 0.5552

O: 35.6 / 15.9994 = 2.2251

H: 0.6 / 1.00794 = 0.5953

Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:

Ca: 0.5564/0.5552 = 1.0022

As: 0.5552/0.5552 = 1

O: 2.2251/0.5552 = 4.0077

H: 0.5953/0.5552 = 1.0722

Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula

Step 3. Write the empirical formula with the rounded numbers obtained

In this case we will have:

Ca: 1

As: 1

O: 4

H: 1

The empirical formula would have to be:

CaAsHO₄

Answer: From the attached document

1. Complete structure

2. Skeletal structure

The complete structural formula for isooctane (2,2,4-trimethylpentane) is (CH3)3CCH2CH(CH3)2, while the skeletal formula is CH3-C-(CH3)2-CH2-CH(CH3)2. These represent its molecular structure in detail and a simplified version, respectively.

The complete structural formula and skeletal formula for 2,2,4-trimethylpentane, commonly known as isooctane, are as follows:

Complete Structural Formula:

In the complete structural formula, we represent each atom in the molecule explicitly. Isooctane has the following structure:

```

     H

      |

H - C - (CH3)2

|     |

H - C - CH3

     |

     CH3

```

In this representation, each line represents a chemical bond, and we've labeled carbon (C) and hydrogen (H) atoms to show their positions in the molecule. Isooctane is highly branched, containing multiple methyl (CH3) groups attached to a central carbon atom.

Skeletal Formula:

The skeletal formula is a simplified way of representing the molecular structure. It focuses on the carbon framework, and hydrogen atoms bonded to carbon are usually omitted. Isooctane's skeletal formula is as follows:

```

  CH3

  |

CH3-C-(CH3)2

  |

  CH3

```

In this skeletal formula, each vertex represents a carbon atom, and the lines between them represent carbon-carbon bonds. The methyl (CH3) groups are indicated as branches off the carbon atoms.

Isooctane is an important component in the determination of gasoline's octane rating because it has a high resistance to engine knocking, making it a reference standard with an octane rating of 100. Gasoline blends with a higher percentage of isooctane are less prone to premature ignition and have higher octane ratings, indicating their suitability for high-performance engines.

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Answer:

0.189

Explanation:

Aqueous solution only has ammonia and water in it. Assume total mass of 100 g

Meaning mass of ammonia is 18% of 100 g = 18 g

Mass of water is 82% of 100 g = 82g

Number of moles of ammonia  = mass / molar mass ammonia =18/17 = 1.059

Number of moles of water = mass / molar mass water = 82/18 =4.556

Mole fraction ammonia = 1.059 / (1.059+4.556) = 0.189

Final answer:

The mole fraction of ammonia in an 18.0% by mass aqueous solution of ammonia is calculated based on the ratio of moles of ammonia to total moles in the solution, resulting in a mole fraction of 0.188.

Explanation:

Mole fraction is the ratio of the moles of solute to the total moles in the solution. Given the percentage by mass, we can calculate the mole fraction of ammonia, assuming the mass of the solution is 100 grams for simplicity.

An 18.0% by mass solution means there are 18.0 grams of NH₃ and 82.0 grams of water (H₂O). To find the moles of each component, we divide their mass by their respective molar masses: NH₃ (17.031 g/mol) and H₂O (18.015 g/mol).

Moles of NH₃ = 18.0 g / 17.031 g/mol = 1.057 moles
Moles of H₂O = 82.0 g / 18.015 g/mol = 4.551 moles
The total moles in solution = 1.057 moles (NH₃) + 4.551 moles (H₂O) = 5.608 moles

The mole fraction of NH₃ = Moles of NH₃ / Total moles in solution = 1.057 / 5.608 = 0.188

Therefore, the mole fraction of ammonia in the solution is 0.188.

Answer:

a. 5.0 x 10⁷ brown grains = 50 million

b. 5.0 x 10³ brown grains = 5000

Explanation:

The concentration of 2 % brown sand means we have for every 100 grains of sand 2 are brown.

We need to calculate the number of brown sand in the bucket as follows:

= 2.5 x 10⁹  billion grains of sand x  (2 brown grains/ 100 grains of sand)

= 5.0 x 10⁷ brown grains

Likewise if the concentration of brown sand is 2.0 ppm, it mean that we have 2 brown grain per every million grains of sand.

= 2.5 x 10⁹ billion grains of sand x ( 2.0 brown grains/10⁶ grains of sand )

= 5.0 x 10³ brown grains

The answers make sense since a concentration of 1 part per million is ten thousandths of  a 1 percent

If the concentration of brown sand is 2.0%, there are 5 x 10^7 brown sand grains in the bucket.

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Answer:

                     Triplet at 1.15 ppm

                     Singlet at 2.02 ppm

                     Quartet at 4.10 ppm

Explanation:

                     In ethyl acetate there are three different sets of protons having different surrounding as compared to each other. The predicted ¹H-NMR of ethyl acetate is attached below and following are the signals shown by this compound,

(i)  A singlet around 2.02 ppm is a characteristic peak shown by protons at carbon a. This is because there is no proton available on the adjacent carbon so there is no coupling.

(ii) A triplet around 1.15 ppm is a peak shown by protons at carbon c. This is because there are two proton present on the adjacent carbon so, according to (n+1) rule it will give a triplet peak.

(iii) A quartet around 4.10 ppm is a peak shown by protons at carbon b. This is because there are three proton present on the adjacent carbon so, according to (n+1) rule it will give a quartet peak. Also, as it is directly attached to oxygen atom hence, the chemical shift is downfielded or deshielded.

Answer:

a) No, it couldn't

b) 78.81 g

Explanation:

When a nonvolatile solute is added to a pure solvent, the melting point of the solvent must decrease, which is called cryoscopy. The temperature variation (melting point of pure solvent - the melting point of solution) can be calculated by:

ΔT = Kc.W.i

Where Kc is the cryoscopy constant (for water it's 1.86 °C/mol.kg), W is the molality, and i is the van't Hoff factor of the solute.

W =m1/m2*M1

Where m1 is the mass of the solute (in g), m2 is the mass of the solvent (in kg), and M1 is the molar mass of the solute (CaCl2 = 111.0 g/mol). The van't Hoff factor indicates how much of the solute ionizes in the solution.

a) The melting point of water is 0°C, so let's calculate the new melting point with the given information:

m1 = 70.1 g

m2 = 100 g = 0.1 kg

W = 70.1/(0.1*111) = 6.31 mol/kg

ΔT = 1.86*6.31*2.5

ΔT = 29.34°C

0 - T = 29.34

T = -29.34°C

Thus, at -33°C, the ice will not melt yet.

b) Let's then found out the value of m1 in this case:

ΔT = Kc.W.i

0 - (-33) = 1.86*W*2.5

4.65W = 33

W = 7.1 mol/kg

W = m1/M*m2

7.1 = m1/(111*0.1)

m1 = 78.81 g

Calcium chloride cannot melt the ice at -33°C. Cryoscopy is the method of determining a decrease in melting point due to dissolved substances.

The decrease in the melting point of a pure solvent when a non-volatile solute is added to the solution.

[tex]\Delta T = K_c. W.i[/tex]

Where

[tex]K_c[/tex]  - cryoscopy constant= 1.86 °C/mol.kg for water

[tex]W[/tex]  - molality = 6.31 mol/kg

[tex]i[/tex]  - Van't Hoff factor of the solute = 2.5

Put the values in the formula,

[tex]\Delta T = 1.86\times 6.31\times 2.5\\\\\Delta T = \rm \ 29.34^oC[/tex]

Therefore, Calcium chloride cannot melt the ice at -33°C.

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