A 4.00 gram sample of a solution of sodium chloride in water was heated until all the water had evaporated. The sodium chloride that remained weighed 1.22 grams. Calculate the percentage of water in the original 4.00 grams of solution.

The answer & explanation for this question is given in the attachment below.

Ozone, according to the VSEPR theory, has a bent or 'V' shaped geometry due to the repulsion of electron pairs. This is because it has one lone pair and two bonding domains.

      O

    /    \

   O    O

The structure of ozone, or O₃, can be drawn according to the VSEPR theory. The central atom is one oxygen atom while the other two oxygen atoms are attached to the central one. Then, there is one lone pair on the central atom, creating a 'bent' or 'V' shape in its geometry.

The VSEPR (Valence Shell Electron Pair Repulsion) theory suggests that electron pairs will repel each other as much as possible, resulting in specific molecular geometries. Ozone is a molecular geometry example of a molecule with 3 total sites of electrons, 2 bonding domains, and one non-bonding domain. This leads to a 'bent' or 'V' shape because the non-bonding pair pushes the two bonding domains closer together.

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Answer:

The least energetic spectral line in the infrared series of the H atom is 656.1 nm

Explanation:

Photon wavelength is inversely proportional to energy. To obtain the least energetic spectral line of the hydrogen atom (H), we determine the longest wavelength possible.

[tex]\frac{1}{\lambda} = R_H[\frac{1}{n_f^2} -\frac{1}{n^2}][/tex]

Where;

nf = 2

n = 3

RH is Rydberg constant = 1.09737 × 10⁷m⁻¹

λ is the wavelength of the least energetic spectral line

Substituting the above values into the equation, we will have

[tex]\frac{1}{\lambda} = 1.09737 X 10^7[\frac{1}{2^2} -\frac{1}{3^2}][/tex]

[tex]\frac{1}{\lambda} = 1.09737 X 10^7[\frac{1}{4} -\frac{1}{9}][/tex]

[tex]\frac{1}{\lambda} = 1.09737 X 10^7[0.25 -0.1111][/tex]

[tex]\frac{1}{\lambda} = 1.09737 X 10^7[0.1389][/tex]

[tex]\frac{1}{\lambda} = 1524246.93[/tex]

[tex]\lambda} = \frac{1}{1524246.93}[/tex]

[tex]\lambda} = 6.561 X10^{-7} m[/tex]

λ = 656.1 X10⁻⁹ m

In (nm): λ = 656.1 nm

Therefore, the least energetic spectral line in the infrared series of the H atom is 656.1 nm

The wavelength of the least energetic spectral line in the infrared series of the hydrogen atom is approximately 18,400 nanometers (nm).

To find the wavelength of the least energetic spectral line in the infrared series of the hydrogen atom, we can use the Rydberg formula for the hydrogen atom:

1 / λ = R_H * (1/n₁² - 1/n₂²)

Where:

λ is the wavelength of the spectral line.

R_H is the Rydberg constant for hydrogen, approximately 1.097 x 10^7 m⁻¹.

n₁ is the principal quantum number of the initial energy level.

n₂ is the principal quantum number of the final energy level.

For the least energetic line in the infrared series, we need to consider the transition where the electron moves from a higher energy level (n₂) to a lower energy level (n₁). In this case, n₂ > n₁.

Since we are interested in the infrared series, we'll consider transitions ending in the n₁ = 3 energy level. We want the least energetic line, so we'll choose the smallest value for n₂.

Let's take n₂ = 4 and calculate:

1 / λ = 1.097 x 10^7 m⁻¹ * (1/3² - 1/4²)

1 / λ = 1.097 x 10^7 m⁻¹ * (1/9 - 1/16)

1 / λ = 1.097 x 10^7 m⁻¹ * (7/144)

Now, solve for λ:

λ = 144 / (1.097 x 10^7 m⁻¹ * 7)

λ ≈ 0.0184 meters or 18.4 millimeters

To express the wavelength in nanometers (nm), we can convert millimeters to nanometers:

λ ≈ 18.4 mm * 1,000,000 nm/mm = 18,400 nm

So, the wavelength of the least energetic spectral line in the infrared series of the hydrogen atom is approximately 18,400 nm.

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Answer:

6.6 is the [tex]pK_a[/tex] of the weak acid.

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant =?

The ratio of conjugate base to acid is = [tex]\frac{[salt]}{acid}=4[/tex]

pH = 7.2

Putting values in above equation, we get:

[tex]7.2=pK_a+\log(4)[/tex]

[tex]pK_a=7.2-\log(4)=6.598\approx 6.6[/tex]

6.6 is the [tex]pK_a[/tex] of the weak acid.

Answer:

[H₃O⁺] = 2.63×10⁻¹⁰ M

As pH = 9.57, the solution is basic

Explanation:

We must know this knowledge:

[OH⁻] . [H₃O⁺] = 1×10⁻¹⁴

3.8×10⁻⁵ . [H₃O⁺] = 1×10⁻¹⁴

[H₃O⁺] = 1×10⁻¹⁴ / 3.8×10⁻⁵ → 2.63×10⁻¹⁰ M

Let's determine the pH to state if the solution is acidic or basic

pH < 7 → acidic ; pH > 7 → basi

pH = - log [H₃O⁺]

pH = - log 2.63×10⁻¹⁰ → 9.57

Answer:

The concentration is [-1 + sqrt(1+0.11t)]/0.1542 M

Explanation:

Let the concentration of CH3CHO after selected reaction times be y

Rate = Ky^2 = change in concentration of CH3CHO/time

K = 0.0771 M^-1 s^-1

Change in concentration of CH3CHO = 0.358 - y

0.0771y^2 = 0.358-y/t

0.0771ty^2 = 0.358 - y

0.0771ty^2 + y - 0.358 = 0

The value of y must be positive and is obtained in terms of t using the quadratic formula

y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M

Final answer:

The question involves calculating the instantaneous rate of a second order decomposition reaction of acetaldehyde using the given rate constant and concentration values.

Explanation:

The question deals with a second order reaction describing the decomposition of acetaldehyde (CH3CHO) into methane (CH4) and carbon monoxide (CO). A second order reaction rate is dependent on the square of the concentration of one reactant or the product of two reactants concentrations. The rate constant provided (0.0771 M-1 s-1 or 4.71 × 10-8 L mol-1 s-1) is used alongside the concentration of acetaldehyde to determine the instantaneous rate of reaction or, in some scenarios, to deduce the remaining concentration of acetaldehyde at a given time.

Let the concentration of CH3CHO after selected reaction times be y

Rate = Ky^2 = change in concentration of CH3CHO/time

K = 0.0771 M^-1 s^-1

Change in concentration of CH3CHO = 0.358 - y

0.0771y^2 = 0.358-y/t

0.0771ty^2 = 0.358 - y

0.0771ty^2 + y - 0.358 = 0

The value of y must be positive and is obtained in terms of t using the quadratic formula

y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M

Answer:

the range of d metal radii is not very great.

Explanation:

The difference in metallic radii are not great hence the metallic ions are almost similar in size across the series. As a result of this, they can easily take up positions in the lattice of other transition metals leading to the formation of transition metal alloys. This explains the wide range of transition metal alloys used for various purposes in industry.

Answer:

Answer in explanation

Explanation:

a. Boron , element 5

Helium has 2 electrons, add to the other 3 to give 5.

Other group members are : Aluminum Al, Gallium Ga, Indium In , Thallium Tl and Nihonium Nh

b. Sulphur, element 16

Neon is 10 , add other 6 electrons to make 16

Other group members are: Oxygen O, selenium Se , Tellurium Te and Polonium Po

c. Lanthanum, element 57

Xenon is 54, add the other 3 electrons to give 57.

Other elements in group : Scandium Sc , Yttrium Y , Actinium Ac, Lutetium Lu and/or Lawrencium Lr

Final answer:

The molality of the sodium hydroxide (NaOH) solution is 0.54 mol/kg.

Explanation:

The molality of a solution is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is sodium hydroxide (NaOH) and the solvent is water.

To calculate the molality, we need to first convert the given mass of NaOH to moles using its molar mass, which is 40.0 g/mol. Then, we need to convert the mass of the solution to kilograms using the density of the solution, which is 1.15 g/mL.

Using the given information:

Mass of NaOH = 24.8 g/L

Density of solution = 1.15 g/mL

Molar mass of NaOH = 40.0 g/mol

The molality can be calculated as follows:

Answer: Reversible competitive inhibition

Explanation:

In the case of reversible competitive inhibition, an inhibitor molecule competes with the substrate for binding to the active site of the enzyme. The inhibitor blocks the active site of the enzyme. Thus the enzyme substrate complex do not form. The structure of the inhibitor is similar to the substrate thus also have the binding affinity with the enzyme. The process is reversible because the inhibitor will leave the enzyme it exerts no permanent effect on the enzyme.

The given situation is the example of reversible competitive inhibition as substrate remain unchanged and the enzyme was not able to act on the substrate chemically may be due to inhibition of the function of the enzyme.

Answer: The bonds are intermediate between double and single bonds

Explanation:

A closer look at the diagram below shows that the bonds in sulphur IV oxide are intermediate between double and single bonds. Hence they do not have the exact bond angle of single bonds. This is why the bond angle is not exactly 120°. There are two resonance structures in the diagram that clearly show this point.

Explanation:

a) Bohr model is perfect for atoms that have single electron and fortunately both Be3+ ion and H atom have one electron so, Bohr model can easily and accurately applied to predict the spectrum of Be3+ and H atom.

b) The energy of an atom in  Bohr model is given by

[tex]E= \frac{-13.6z^2}{n^2}[/tex]

the values of z for H atom and Be3+ ion are 1 and 4 respectively. Hence, energy of atoms would be different for both atoms. Hence, line spectra to be identical is not possible.

Answer:

Light as a wave

1. Young's Double Slit Experiment

2. Davisson-Germer Experiment

Light as a particle

1. Einsteins Photoelectric Effect Phenomenon

2.  Diffraction Phenomenon of Particles

Evidence for light as a wave includes diffraction and interference patterns, while evidence for the particle model includes the photoelectric effect and emission spectra.

The mid-17th century debate on whether light is a wave or a particle extended well into the 20th century until revolutionary concepts by Planck and Einstein. There is evidence for both the wave model and the particle model of light. Two pieces of evidence for the wave nature of light are diffraction and interference patterns, as seen in Thomas Young's double-slit experiment.

Diffraction occurs when light encounters an obstacle, spreading out as a result, and interference patterns occur when waves overlap and combine in constructive or destructive ways. Conversely, evidence for the particle nature includes phenomena like the photoelectric effect, as explained by Einstein, where light knocks electrons from a material, and the behavior of emission spectra, where individual energy quanta or "photons" are emitted from atoms.

Answer:

MG is less polar  

Explanation:

The structures of crystal violet (CV) and malachite green (MG) are shown in Figures 1 and 2, respectively.  

The obvious difference is that CV has an extra dimethylamino group (polar).

Alumina is a polar adsorbent, so it retains the more polar substances more strongly and they are eluted last.

MG is less polar than CV, so it is retained less strongly and is eluted first.

Answer:

Ethical Behavior

Explanation:

It's the ethical behavior how companies work or do business that have the positive impact on the community. They not only think about making money but also about the welfare of the society. They are concerned about the products they made and it's impact on the environment. Ethical behaviour is based on the human perception of right and wrong. That kind of behaviour whichis  not required by the law, but is done for the betterment of the society is ethical behaviour.

Silver Mining's voluntary decision to install environmentally friendly devices, despite no legal requirement, exemplifies business ethical behavior.

In this instance, Silver Mining's decision to install scrubbers on the smokestacks of their new facility, even though not legally required, is a clear example of business ethical behavior. This action demonstrates the company prioritizing environmental protection over potential costs. While the act also aligns with legal behavior, it is not driven by legal necessity. It's a voluntary measure taken for the greater good, thus making it an ethical business decision.

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Explanation:

A. CeCl

Cerium chloride.

Metal: Ce+

B. SrBr2

Strontium chloride.

Metal: Sr2+

C. K2O

Potassium oxide.

Metal: K+

D. LiF

Lithuim fluoride.

Metal: Li+

Chlorine (Cl) creates an anion with a -1 charge, but the metal cerium (Ce) only forms one type of ion with a +3 charge. As a result, the substance is known as cerium(III) chloride.

a) CeCl: Cerium(III) chloride

Whereas bromine (Br) generates an anion with a -1 charge, the metal strontium (Sr) only forms one sort of ion with a +2 charge in this combination. As a result, the substance is known as strontium bromide.

b) SrBr2: Strontium bromide

In this molecule, oxygen (O) generates an anion with a -2 charge whereas the metal potassium (K) only produces one sort of ion with a +1 charge. The substance is referred to as potassium oxide as a result.

c) K2O: Potassium oxide

Lithium is the only element present in this combination (Li). As lithium only ever produces an ion with a positive charge, it is known simply as lithium.

d) Li: Lithium

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Answer:

HCl solution - H30+ and Cl ions. pH 3

NaCl - Na+ and Cl-. pH 7

Explanation:

a) HCl solution - the hydrogen ion combines with water molecule to form the hydronium molecule which is responsible for acidity. The chloride ion is also found in solution.

pH = -log [H+] = -log(10^-3) = 3

b) NaCl 10-3 g. The solid dissocates in water forming the Na+ and Cl- ions. None of these ions affect pH

Answer:

The new volume is 5.37 L

Explanation:

Step 1: Data given

Initial volume = 10.0 L

Initial temperature = 22.0 °C

Initial pressure = 1.00 atm

Final temperature = 202 °C

Final pressure = 3.00 atm

Step 2: Calculate final volume

(P1*V1)/T1  = (P2*V2)/T2

⇒ with P1 = The initial pressure = 1.00 atm

⇒ with V1 = The initial volume = 10.0 L

⇒ with T1 = The initial temperature = 22 °C = 295 Kelvin

⇒ with  P2 = The final pressure = 3.00 atm

⇒ with V2 = The final volume = TO BE DETERMINED

⇒ with T2 = The final temperature = 202 °C = 475 Kelvin

(1.00 * 10.0) / 295 = (3.00 * V2) / 475

10 / 295 = 3V2/ 475

3V2 = 4750/295

V2 = 5.37 L

The new volume is 5.37 L

Answer:

The trend in metallic character as a function of position in the periodic table is that the metallic character increases as you go down a group. Since the ionization energy decreases going down a group (or increases going up a group), the increased ability for metals lower in a group to lose electrons makes them more reactive.

This is not the same for the atomic size, as you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting a larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus.

Similarly, it is also different for the ionization energy trend, as you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus.

Final answer:

The metallic character in the periodic table decreases across a period and increases down a group. It trends similarly to atomic size but oppositely to ionization energy.

Explanation:

Metallic character: The metallic trend follows the trend of the atomic radius. It increases within a group of the periodic table from the top to the bottom and decreases within a period from left to right. Metallic character relates to the ease of losing an electron in a chemical reaction and is opposite to the trend of ionization energy.

Atomic size shows a trend that parallels the metallic character. Atomic size increases down a group because of the increase in electron shells, which makes the valence electrons less tightly held. Conversely, atomic size decreases from left to right within a period due to an increasing effective nuclear charge, which draws electrons closer to the nucleus, reducing the size of the atom.

In summary, the metallic character and atomic size increase from right to left in a period and from top to bottom in a group, while ionization energy generally shows the opposite trend. Hence, the trend in metallic character is similar to the trend in atomic size but opposite to the trend in ionization energy.

The question is incomplete , complete question is:

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

(a) n = 2 to n = 4

(b) n = 2 to n = 1

(c) n = 2 to n = 5

(d) n = 4 to n = 3

Answer:

Hence the order of the transition will be : d < a < c < b

Explanation:

[tex]E_n=-13.6\times \frac{Z^2}{n^2}ev[/tex]

where,

[tex]E_n[/tex] = energy of [tex]n^{th}[/tex] orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen atom:

[tex]E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV[/tex]

Energy of n = 2 in an hydrogen atom:

[tex]E_2=-13.6\times \frac{1^2}{2^2}eV=-3.40eV[/tex]

Energy of n = 3 in an hydrogen atom:

[tex]E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV[/tex]

Energy of n = 4 in an hydrogen atom:

[tex]E_4=-13.6\times \frac{1^2}{4^2}eV=-0.85 eV[/tex]

Energy of n = 5 in an hydrogen atom:

[tex]E_5=-13.6\times \frac{1^2}{5^2}eV=-0.544 eV[/tex]

a) n = 2 to n = 4 (absorption)

[tex]\Delta E_1= E_4-E_2=-0.85eV-(-3.40eV)=2.55 eV[/tex]

b) n = 2 to n = 1 (emission)

[tex]\Delta E_2= E_1-E_2=-13.6 eV-(-3.40eV)=-10.2 eV[/tex]

Negative sign indicates that emission will take place.

c) n = 2 to n = 5 (absorption)

[tex]\Delta E_3= E_5-E_2=-0.544 eV-(-3.40eV)=2.856 eV[/tex]

d) n = 4 to n = 3 (emission)

[tex]\Delta E_4= E_3-E_4=-1.51 eV-(-0.85 eV)=-0.66 eV[/tex]

Negative sign indicates that emission will take place.

According to Planck's equation, higher the frequency of the wave higher will be the energy:

[tex]E=h\nu [/tex]

h = Planck's constant

[tex]\nu [/tex] frequency of the wave

So, the increasing order of magnitude of the energy difference :

[tex]E_4<E_1<E_3<E_2[/tex]

And so will be the increasing order of the frequency of the of the photon absorbed or emitted. Hence the order of the transition will be :

: d < a < c < b

Final answer:

The order of increasing frequency of photon absorbed or emitted for the H atom electron transitions is (a) n = 2 to n = 4, (c) n = 2 to n = 5, (b) n = 2 to n = 1, and (d) n = 2 to n = 1.

Explanation:

The frequency of a photon absorbed or emitted by a hydrogen atom is related to the energy difference between the initial and final energy levels. The energy difference decreases as the value of n increases, so (a) n = 2 to n = 4 has the lowest frequency, followed by (c) n = 2 to n = 5, (b) n = 2 to n = 1, and finally, (d) n = 2 to n = 1 has the highest frequency.

Answer : The concentration of a solution with an absorbance of 0.420 is, 0.162 M

Explanation :

Using Beer-Lambert's law :

[tex]A=\epsilon \times C\times l[/tex]

As per question, at constant path-length there is a direct relation between absorbance and concentration.

[tex]\frac{A_1}{A_2}=\frac{C_1}{C_2}[/tex]

where,

A = absorbance of solution

C = concentration of solution

l = path length

[tex]A_1[/tex] = initial absorbance = 0.350

[tex]A_2[/tex] = final absorbance = 0.420

[tex]C_1[/tex] = initial concentration = 0.135 M

[tex]C_2[/tex] = final concentration = ?

Now put all the given value in the above relation, we get:

[tex]\frac{0.350}{0.420}=\frac{0.135}{C_2}[/tex]

[tex]C_2=0.162M[/tex]

Thus, the concentration of a solution with an absorbance of 0.420 is, 0.162 M

Answer:

emits (radiates) , energy difference

Explanation:

According to the Bohr theory, when an electron jumps from higher orbital to the lower orbital, it radiates energy which is equal to the energy difference between the orbitals.

Mathematically, it can be shown as:-

The expression for Bohr energy is shown below as:-

[tex]E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules[/tex]

For transitions:

[tex]Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]

[tex]\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]

Also, [tex]\Delta E=\frac {h\times c}{\lambda}[/tex]

Where,  

h is Plank's constant having value [tex]6.626\times 10^{-34}\ Js[/tex]

c is the speed of light having value [tex]3\times 10^8\ m/s[/tex]

According to the Bohr model of the atom, when an electron transitions from a higher-energy orbit to a lower-energy orbit, it emits electromagnetic energy equal to the energy difference between the two orbits.

The Bohr model of the atom states that when an electron transitions from a higher-energy orbit to a lower-energy orbit, it emits electromagnetic energy with an energy equal to the difference between the two orbits. This energy is released in the form of a photon.

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We can calculate the vapor pressure of liquid cobalt at a given temperature by using the Clausius-Clapeyron equation and the given vapor pressure at another temperature. This involves substituting known values into the equation and solving for the desired vapor pressure.

The question is asking for the calculated vapor pressure of liquid cobalt at a certain temperature based on its known vapor pressure at another temperature. This involves using the Clausius-Clapeyron equation, which describes the relationship between the vapor pressure of a substance and its temperature. Let's denote the initial conditions (i.e., 400 mm Hg at 3.03x10^3 K) as P1 and T1, and the conditions we want to find (i.e., vapor pressure at 3.07x10^3K) as P2 and T2.

First, convert the molar heat of vaporization from kJ/mol to J/mol by multiplying by 1000, which gives 450000 J/mol. Next, the Clausius-Clapeyron equation can be rearranged to solve for P2:

P2 = P1 * exp [ -ΔHvap (1/T2 - 1/T1) / R ]

where ΔHvap is the molar heat of vaporization, R is the ideal gas constant (8.314 J/mol·K). Substituting all known values into this equation will give the vapor pressure of liquid Co at the desired temperature.

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The vapor pressure of liquid cobalt at 3.07x10^3 K is approximately 3748.64 mm Hg.

The vapor pressure of liquid cobalt at a temperature of 3.07x10^3 K can be determined using the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature. The equation is given by:

[tex]\[ \ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \][/tex]

First, we need to convert [tex]\( \Delta H_{\text{vap}} \)[/tex] from kJ/mol to J/mol to match the units of [tex]\( R \)[/tex]:

[tex]\[ \Delta H_{\text{vap}} = 450 \text{ kJ/mol} \times 1000 \text{ J/kJ} = 450,000 \text{ J/mol} \][/tex]

Now we can plug the values into the Clausius-Clapeyron equation:

[tex]\[ \ln\left(\frac{P_2}{400 \text{ mm Hg}}\right) = -\frac{450,000 \text{ J/mol}}{8.314 \text{ J/(mol·K)}}\left(\frac{1}{3.07x10^3 \text{ K}} - \frac{1}{3.03x10^3 \text{ K}}\right) \][/tex]

Solving for [tex]\( P_2 \):[/tex]

[tex]\[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{1}{3.07x10^3} - \frac{1}{3.03x10^3}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{3.03x10^3 - 3.07x10^3}{(3.07x10^3)(3.03x10^3)}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{-40}{3.07x10^3x3.03x10^3}\right) \] \[ \ln\left(\frac{P_2}{400}\right) = -\frac{450,000}{8.314}\left(\frac{-40}{9.3051x10^6}\right) \][/tex]

[tex]\[ P_2 \approx 3748.64 \text{ mm Hg} \][/tex]

Therefore, the vapor pressure of liquid cobalt at 3.07x10^3 K is approximately 3748.64 mm Hg.

Answer:

9.57

Explanation:

Given that:

[tex]pK_{b}=-\log\ K_{b}=-\log(8.0\times 10^{-5})=4.1[/tex]

Considering the Henderson- Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:

[tex]pOH=pK_b+log\frac{[conjugate\ acid]}{[base]}[/tex]

So,  

[tex]pOH=4.1+\log\frac{0.540}{0.250}=4.43[/tex]

pH + pOH = 14  

So, pH = 14 - 4.43 = 9.57

Answer:

The answer would be A. the number of electrons in the outermost electron shell.

Explanation:

These are called valence electrons which are transferred, shared, and rearranged by creating covalent bonds producing new substances.

Hope this helped! :)

The type of chemical reaction an atom chooses is determined by the number of the outermost electrons in the outermost shell of an atom.

About the importance of electrons the below points should be noted;

Therefore the answer is the number of electrons in the outermost electron shell.

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Explanation:

Ksp of NiCO3 = 1.4 x 10^-7

Ksp of CuCO3 = 2.5 x 10^-10

Ionic equations:

NiCO3 --> Ni2+ + CO3^2-

CuCO3 --> Cu2+ + CO3^2-

[Cu2+][CO3^2-]/[Ni2+][CO3^2-]

= (2.5* 10^-10)/(1.4* 10^-7)

= 0.00179.

[Cu2+]/[Ni2+]

= 0.00179

= 0.00179*[Ni2+]

If all of Cu2+ is precipitated before Na2CO3 is added.

= 0.00179 * (0.25)

The amount of Cu2+ not precipitated = 0.000448 M

The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100

= 0.000448/0.25 * 100

= 0.18%

Therefore, percentage precipitated = 100 - 0.18

= 99.8%

The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.

The metal ions can be separated by slowly adding Na2CO3 based on the relative solubilities of their carbonates. Nickel carbonate (NiCO3) is less soluble than copper carbonate (CuCO3), allowing the selective precipitation of nickel ions before copper ions.

To determine if the metal ions can be separated by slowly adding Na2CO3, we need to consider the solubility of the metal carbonates. Nickel carbonate (NiCO3) and copper carbonate (CuCO3) both have low solubilities, but it is crucial to examine their relative solubilities. If one carbonate is significantly less soluble than the other, it can be selectively precipitated first.

In this case, NiCO3 is less soluble than CuCO3. Therefore, by slowly adding Na2CO3 to the solution, we can precipitate the majority of the Ni2+ ions as NiCO3 before CuCO3 begins to precipitate. This satisfies the condition that 99% of the metal ion must be precipitated before the other metal ion begins to precipitate.

Therefore, it is possible to separate the nickel and copper ions in the solution by slowly adding Na2CO3.

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Answer:

The percent yield of a reaction is 48.05%.

Explanation:

[tex]WO_3+3H_2\rightarrow W+3H_2O[/tex]

Volume of water obtained from the reaction , V= 5.76 mL

Mass of water = m = Experimental yield of water

Density of water = d = 1.00 g/mL

[tex]M=d\times V = 1.00 g/mL\times 5.76 mL=5.76 g[/tex]

Theoretical yield of water : T

Moles of tungsten(VI) oxide = [tex]\frac{51.5 g}{232 g/mol}=0.2220 mol[/tex]

According to recation 1 mole of tungsten(VI) oxide gives 3 moles of water, then 0.2220 moles of tungsten(VI) oxide will give:

[tex]\frac{3}{1}\times 0.2220 mol=0.6660 mol[/tex]

Mass of 0.6660 moles of water:

0.666 mol × 18 g/mol = 11.988 g

Theoretical yield of water : T = 11.988 g

To calculate the percentage yield of reaction , we use the equation:

[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

[tex]=\frac{m}{T}\times 100=\frac{5.76 g}{11.988 g}\times 100=48.05\%[/tex]

The percent yield of a reaction is 48.05%.

Answer:The percent yield of a reaction is 48.05%.

Explanation:

Answer :

(a) Number of electrons in an atoms is, 2

(b) Number of electrons in an atoms is, 10

(c) Number of electrons in an atoms is, 10

Explanation :

There are 4 quantum numbers :

Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as . The value of this quantum number ranges from . When l = 2, the value of

Spin Quantum number : It describes the direction of electron spin. This is represented as . The value of this is  for upward spin and  for downward spin.

Number of electrons in a sublevel = 2(2l+1)

(a) 2s

n = 2

Value of 'l' for 's' orbital : l = 0

Number of electrons in an atoms = 2(2l+1) = 2(2×0+1) = 2

(b) n = 3, l = 2

Number of electrons in an atoms = 2(2l+1) = 2(2×2+1) = 10

(c) 6d

n = 2

Value of 'l' for 'd' orbital : l = 2

Number of electrons in an atoms = 2(2l+1) = 2(2×2+1) = 10

In quantum chemistry, the 2s sublevel can hold 2 electrons, the n = 3, l = 2 sublevel (3d) can hold 10 electrons, and the 6d sublevel can also hold 10 electrons.

In quantum chemistry, specific sublevels or orbitals within an atom can hold certain numbers of electrons.

(a) The 2s sublevel can hold a maximum of 2 electrons. This is because 's' orbitals can contain up to 2 electrons.

(b) For n = 3, l = 2, this refers to the 3d orbital. 'd' orbitals can hold a maximum of 10 electrons, so 10 electrons can exist in this energy level.

(c) For the 6d orbital, it can also hold up to 10 electrons, as it is also a 'd' orbital.

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Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%

Mass of the solution , m= 5.125 g

Mass of the hydrogen peroxide = x

[tex]w/w\% = \frac{x}{m}\times 100[/tex]

[tex]3\%=\frac{x}{5.125 g}\times 100[/tex]

[tex]x=\frac{3\times 5.125 g}{100}=0.15375 g[/tex]

Mass of hydregn pervade in the solution = 0.15375 g

Moles of hydregn pervade in the solution :

[tex]=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol[/tex]

0.004522 moles of hydrogen peroxide molecules are present.

To determine the number of moles of hydrogen peroxide molecules present in the solution, multiply the mass of the solution by the mass percent of hydrogen peroxide. Then, convert the mass of hydrogen peroxide to moles using its molar mass. The number of moles of hydrogen peroxide molecules is approximately 0.00452.

To determine the number of moles of hydrogen peroxide molecules present, we first need to calculate the mass of hydrogen peroxide in the solution. We can do this by multiplying the mass of the solution (5.125 g) by the mass percent of hydrogen peroxide (3.00% or 0.03).

Mass of hydrogen peroxide = 5.125 g × 0.03 = 0.15375 g

Next, we need to convert the mass of hydrogen peroxide to moles using its molar mass. The molar mass of hydrogen peroxide (H2O2) is 34.0146 g/mol.

Moles of hydrogen peroxide = 0.15375 g ÷ 34.0146 g/mol ≈ 0.00452 mol

Therefore, there are approximately 0.00452 moles of hydrogen peroxide molecules present.

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Answer:

m = 3 moles/kg

Explanation:

This is a problem of freezing point depression, and the formula or expression to use is the following:

ΔT = i*Kf¨*m (1)

Where:

ΔT: Change of temperature of the solution

i: Van't Hoff factor

m: molality of solution

Kf: molal freezing point depression of water (Kf = 1.86 °C kg/mol)

Now, the value of i is the number of moles of particles obtained when 1 mol of a solute dissolves. In this case, we do not know what kind of solution is, so, we can assume this is a non electrolyte solute, and the value of i = 1.

Let's calculate the value m, which is the molality solving for (1):

m =  ΔT/Kf (2)

Finally, let's calculate ΔT:

ΔT = T2 - T1

ΔT = 0 - (-5.58)

ΔT = 5.58 °C

Now, let's replace in (2):

m = 5.58/1.86

m = 3 moles/kg

This is the molality of solution.

The other data of mass, can be used to calculate the molecular mass of this unknown solid, but it's not asked in the question.