Write the ground state electron configuration of w using the noble-gas shorthand notation.

Final answer:

Violet light has a shorter wavelength and higher photon energy compared to red light, with violet having the shortest wavelengths and red the longest within the visible spectrum.

Explanation:

Considering light at the two ends of the visible light spectrum, violet light has a shorter wavelength and a higher photon energy than red light. In the visible light spectrum, violet light has the shortest wavelengths (approximately 400 nm) and thus carries the most energy. Conversely, red light has the longest wavelengths (approximately 700 nm) and carries the least amount of energy.

Sunlight, for example, which is blackbody radiation, peaks in the visible spectrum and has more intensity in the red than in the violet, giving the sun a yellowish appearance. The high energy of violet photons is why dyes that absorb violet light fade more quickly, and when you observe faded posters, the blues and violets are the last to fade.

The molar concentration of an unknown copper (ii) sulfate solution can be determined by reacting it with excess zinc, calculating the moles of copper obtained and hence the moles of copper sulfate, and subsequently the molar concentration.

An alternate method for determining the molar concentration of an unknown sample of copper (ii) sulfate solution involves a series of calculative steps. Firstly, we must know the stoichiometric factor between the copper (ii) sulfate and a known substance. In this case, we can use the reaction of copper sulfate with excess zinc metal as a reference in a standard data.

Here's how to calculate: Upon reaction of a known mass of copper sulfate with excess zinc metal, a certain mass of copper metal is obtained. Using this equation:CuSO4 (aq) + Zn (s).

Step 1: Calculate the number of moles of copper obtained from the mass using the molar mass of copper. Step 2: This number of moles is the same as the moles of copper sulfate in your sample because of the 1:1 stoichiometry in the reaction. Step 3: Determine the molar concentration (M) of the solution by using the formula M = moles of solute / volume of solution (in liters). If the volume of the solution is unknown, you can use other identifying tests, such as a titration.

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To determine the molar concentration of copper (II) sulfate solution, an alternate method can be used. This method involves finding the mass of CuSO4, converting it to moles using Avogadro's number, and then dividing the moles by the volume of the solution to calculate the molar concentration.

An alternate method for determining the molar concentration of the unknown copper (II) sulfate solution can be done using the standard data. One way to do this is by finding the mass of CuSO4 and using Avogadro's number to convert it to moles. Then, divide the moles of CuSO4 by the volume of the solution in liters to calculate the molar concentration.

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The system volume of 1 gram-mole of methyl chloride vapor at 100°C and 10 atm is approximately 3.06 liters.

To estimate the system volume of 1 gram-mole of methyl chloride vapor under given conditions, we use the Ideal Gas Law equation:

PV = nRT

Here, P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the universal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

Given:

Using the equation, we solve for V:

V = (nRT)/P

Substituting the values, we get:

[tex]V = \frac{(1 \, \text{mol} \times 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \times 373.15 \, \text{K})}{10 \, \text{atm}}[/tex]

[tex]V = \frac{(1 \times 0.0821 \times 373.15) \, \text{L} \cdot \text{atm} / \text{K}}{10 \, \text{atm}}[/tex]

[tex]V = \frac{30.634115 \, \text{L} \cdot \text{atm}}{10 \, \text{atm}}[/tex]

[tex]V = 3.0634115 \, \text{L}[/tex]

V ≈ 3.06 L

So, the system volume is approximately 3.06 liters.

Final answer:

On Mars, the acceleration due to gravity is about one-third of that on Earth, which means an object weighs significantly less on Mars compared to its weight on Earth.

Explanation:

The question pertains to the acceleration due to gravity on the surface of Mars compared to Earth. On Mars, the acceleration due to gravity is about one-third of that on Earth. Specifically, the gravitational acceleration on Mars is approximately 3.71 m/s², while on Earth, it is about 9.81 m/s². Thus, an object on Mars weighs significantly less than it does on Earth. For example, if a space probe weighs 100 pounds on Earth, on Mars, it would weigh roughly 38 pounds because the acceleration due to gravity on Mars is 0.38 that of Earth's gravity. This difference significantly impacts how objects move and respond to forces on Mars compared to Earth.

By replacing one 60W incandescent lightbulb with an 18W compact fluorescent lightbulb, you would save 420 watt-hours of energy over 10 hours.

To calculate the amount of energy saved over 10 hours when one 60W incandescent lightbulb is replaced with an 18W compact fluorescent lightbulb, we need to find the energy consumed by each type of bulb and then calculate the difference.

Energy consumed by a bulb can be calculated using the formula:

Energy (in watt-hours) = Power (in watts) × Time (in hours)

Let's calculate the energy consumed by each bulb:

For the 60W incandescent lightbulb:

Energy consumed = 60W × 10 hours = 600 watt-hours

For the 18W compact fluorescent lightbulb:

Energy consumed = 18W × 10 hours = 180 watt-hours

Now, let's calculate the energy saved:

Energy saved = Energy consumed by incandescent bulb - Energy consumed by compact fluorescent bulb

Energy saved = 600 watt-hours - 180 watt-hours

Energy saved = 420 watt-hours

So, by replacing one 60W incandescent lightbulb with an 18W compact fluorescent lightbulb, you would save 420 watt-hours of energy over 10 hours.

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Chlorine has higher ionization energy

Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated neutral gaseous atom, molecule or ion. It is quantitatively expressed in symbols as

X + energy → X+ + e−

Where X is any atom, molecule or ion capable of being ionized, X+ is that atom or molecule with an electron removed, and e− is the removed electron. This is generally an endothermic process.

The ionization energy of Sodium (alkali metal) is 496KJ/mol whereas Chlorine's first ionization energy is 1251.1 KJ/mol.  

Alkali metals (IA group) have small ionization energies, especially when compared to halogens.  Because as we move across the period from left to right, in general, the ionization energy increases. The atoms become smaller which causes the nucleus to have greater attraction for the valence electrons. Therefore, the electrons are more difficult to remove.

[tex]\boxed{{\text{Chlorine}}}[/tex] has higher ionization energy than sodium.

Further Explanation:

Ionization energy:

It is the amount of energy that is required to remove the most loosely bound valence electrons from the isolated neutral gaseous atom. It is denoted by IE. The value of IE is related to the ease of removing the outermost valence electrons. If these electrons are removed so easily, small ionization energy is required and vice-versa. It is inversely proportional to the size of the atom.

Ionization energy trends in the periodic table:

1. Along the period, IE increases due to the decrease in the atomic size of the succeeding members. This results in the strong attraction of electrons and hence are difficult to remove.

2. Down the group, IE decreases due to the increase in the atomic size of the succeeding members. This results in the lesser attraction of electrons and hence are easy to remove.

Sodium and chlorine are present in period 3 of the periodic table. Sodium lies to the left region of the period while chlorine lies to the right.

The atomic number of sodium atom [tex]\left({{\text{Na}}}\right)[/tex] that lies in left region of the period 3 is 11 and its electronic configuration is [tex]{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{p}}^{\mathbf{6}}}{\mathbf{3}}{{\mathbf{s}}^{\mathbf{1}}}[/tex] . The atomic number of chlorine is 17 and its electronic configuration is [tex]{\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{2}}{{\mathbf{p}}^{\mathbf{6}}}{\mathbf{3}}{{\mathbf{s}}^{\mathbf{2}}}{\mathbf{3}}{{\mathbf{p}}^{\mathbf{5}}}[/tex] . Sodium has only one electron in its outermost valence shell that can be removed easily in order to achieve the nearest stable noble gas configuration of He, resulting in its low ionization energy. Chlorine is one electron short of noble gas so it can gain an electron easily, but its removal requires a large amount of energy. So the ionization energy of chlorine is higher than that of sodium.

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2. Write the chemical equation for the first ionization energy of lithium: https://brainly.com/question/5880605

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Periodic classification of elements

Keywords: ionization energy, sodium, atomic number, electron, neutral, isolated, gaseous atom, IE, chlorine, group, period, higher.

The number of moles of P2O5 produced from the given amounts of phosphorus and oxygen is equal to the number of moles of phosphorus or oxygen used.

To determine the number of moles of P2O5 produced from the given amounts of phosphorus and oxygen, you need to compare the amounts of each reactant used in Part A and Part B. Based on the given information, it is stated that in Part A, 1.80 mol of P2O5 is formed from a given amount of phosphorus and excess oxygen. In Part B, 1.40 mol of P2O5 is formed from a given amount of oxygen and excess phosphorus. Since the stoichiometry of the reaction is a 1:1 ratio between P2O5 and phosphorus, we can conclude that 1.80 mol of phosphorus is required to produce 1.80 mol of P2O5. Similarly, 1.40 mol of oxygen is required to produce 1.40 mol of P2O5. Therefore, the number of moles of P2O5 produced from the given amounts of phosphorus and oxygen is equal to the number of moles of phosphorus or oxygen used, which is 1.80 mol and 1.40 mol respectively.

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The [tex]{{\text{H}}^+}[/tex] concentration of vinegar solution is [tex]\boxed{{\text{0}}{\text{.0000708 M}}}[/tex]

Further Explanation:

An acid is a substance that has the ability to donate [tex]{{\mathbf{H}}^{\mathbf{+}}}[/tex]ions or can accept electrons from the electron-rich species. The general dissociation reaction of acid is as follows:

[tex]{\text{HA}}\to{{\text{H}}^+}+{{\text{A}}^-}[/tex]

Here, HA is an acid.

The acidic strength of an acid can be determined by pH value. The negative logarithm of hydronium ion concentration is defined as pH of the solution. Lower the pH value of an acid, the stronger will be the acid. Acidic solutions are likely to have pH less than 7. Basic or alkaline solutions have pH more than 7. Neutral solutions have pH equal to 7.

Vinegar contains acetic acid [tex]\left({{\text{C}}{{\text{H}}_3}{\text{COOH}}}\right)[/tex], water and some traces of other chemicals and flavors.

The formula to calculate pH is as follows:

[tex]{\text{pH}}=-{\text{log}}\left[{{{\text{H}}^+}}\right][/tex]                                   …… (1)

Here,

[tex]\left[{{{\text{H}}^+}}\right][/tex] is hydrogen ion concentration.

On rearranging equation (1), we get:

[tex]\left[{{{\text{H}}^+}}\right]={10^{-{\text{pH}}}}[/tex]                                           …… (2)

The pH of vinegar is 4.15.

Substitute 4.15 for pH in equation (2)

[tex]\begin{gathered}\left[{{{\text{H}}^+}}\right]={10^{-4.15}}\\=0.0000707946\\\approx0.0000708\;{\text{M}}\\\end{gathered}[/tex]

So the concentration of [tex]{{\mathbf{H}}^{\mathbf{+}}}[/tex] ion in vinegar is 0.0000708 M.

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2. Reason for the acidic and basic nature of amino acid. https://brainly.com/question/5050077

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Acid, base and salts.

Keywords: pH, neutral, acidic, basic, alkaline, 4.15, vinegar, acetic acid, water, chemicals, negative logarithm, H+, 0.0000708 M, pH more than 7, pH less than 7, pH equal to 7.

5.862 moles of O₂

Given:

Combustion of 54.7 g of C₂H₄ to form CO₂ and H₂O.

Question:

How many moles of O₂ are required for the complete reaction of combustion of C₂H₄?

The Process:

Let us convert mass to mole for C₂H₄.

[tex]\boxed{ \ n = \frac{mass}{Mr} \ } \rightarrow \boxed{ \ n = \frac{54.7}{28} = 1.954 \ moles \ }[/tex]

The combustion reaction of  C₂H₄ (ethylene, also named ethene) can be expressed as follows:

[tex]\boxed{ \ C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O \ }[/tex] (the reaction is balanced)

According to chemical equation above, proportion between C₂H₄ and O₂ is 1 to 3. Therefore, we can count the number of moles of O₂.

[tex]\boxed{ \ \frac{n(O_2)}{n(C_2H_4)} = \frac{3}{1} \ }[/tex]

[tex]\boxed{ \ n(O_2) = \frac{3}{1} \times n(C_2H_4) \ }[/tex]

[tex]\boxed{ \ n(O_2) = \frac{3}{1} \times 1.954 \ moles \ }[/tex]

Thus, the number of moles of O are required for the complete reaction of the combustion of C₂H₄ is 5.862 moles.

_ _ _ _ _ _ _ _ _

Notes:

If we want to calculate the mass of O₂, then we use the number of moles of O₂ that have been obtained.

Actually, a methylene group is simply any compound which contains a C=C double bond group and the rest are single bonded carbon groups. Some example of the isomers of C6H12 which contains methylene group is:

1-hexene

2,3-dimethyl-2-butene

2,3-dimethyl-1-butene

2-methyl-2-pentene

trans-2-hexene

4-methyl-1-pentene

cis-2-hexene

trans-3-hexene

2-ethyl-1-butene

2-methyl-1-pentene

3,3-dimethyl-1-butene

4-methyl-cis-2-pentene

cis-3-methyl-2-pentene

trans-3-methyl-2-pentene

Answer:

[tex]7.6LNOBr[/tex]

Explanation:

Hello,

In this case, since no temperature and pressure are known, one could develop the stoichiometric relationship for 1 mole of [tex]Br_2[/tex] per 2 moles of [tex]NOBr[/tex] in terms of volume as shown below because of the Avogadro's law (change in mole proportional to the change in volume at constant both pressure and temperature):

[tex]3.8LBr_2*\frac{2LNOBr}{1LBr_2} =7.6LNOBr[/tex]

Best regards.

CO mass = 4,435. 18 = 79,839 grams

Stokiometry in Chemistry learns about chemical reactions mainly emphasizing quantitative, such as calculation of volume, mass, number, which is related to the number of ions, molecules, elements etc.

In chemical calculations, the reaction can be determined, the number of substances that can be expressed in units of mass, volume, mole, or determine a chemical formula, for example the substance level or molecular formula of hydrate.

In stockiometry therein includes

relative atomic mass (Ar) and relative molecular mass (Mr)

Mr. AxBy = (x.Ar A + y. Ar B)

Reactions that occur:

Bi₂O₃ (s) + 3C (s) → 2Bi (s) + 3CO (g)

Mr Bi₂O₃ = 2. ar bi + 3. Ar O

Mr Bi₂O₃ = 2. 209 + 3. 16

Mr. Bi₂O₃= 466

mole Bi₂O₃ = gram / Mr

mole = 689/466

mole 1.4785

Comparison of Bi reaction coefficients:  Bi₂O₃  = 1: 2,

Mr. CO = 12 + 16 = 18

mass CO = mole. Mr

CO mass = 4,435. 18 = 79,839 grams

how many moles of water you can produce

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Excess reactant

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The percentage yield

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Limiting reactant

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No, I believe this is not an example of a chemical reaction. What we actually see here is a physical change of the solution. Since we are adding more water to an aqueous solution which is also made up mostly of water, what we are simply basically doing is dilution. Since the solution is being diluted, so definitely the color turned lighter.

According to the forces of attraction, the carboxylic acid with  lowest boiling point is methanoic acid.

Forces of attraction is a force by which atoms in a molecule  combine. it is basically an attractive force in nature.  It can act between an ion  and an atom as well.It varies for different  states  of matter that is solids, liquids and gases.

The forces of attraction are maximum in solids as  the molecules present in solid are tightly held while it is minimum in gases  as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.

The physical properties such as melting point, boiling point, density  are all dependent on forces of attraction which exists in the substances.

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Concentration of caffeine is 430 mg/L = 0.43g/L 
The molar mass of caffeine is 194.19 g/mol 

Therefore the molarity is:

Molarity = (0.43/194.19) mol/L 
Molarity = 0.002214 mol/L 
Molarity = 0.002214 M 

Given pKb = 10.4:

Kb = 10^-pKb = 10^ -10.4 = 3.981 x 10^ -11 

Kb is equivalent to:
Kb = [caffeine H+][OH-] / [caffeine] 
3.981 x 10^ -11 = [caffeine H+][OH-] / (0.002214) 
[caffeineH+][OH-] = 8.815 x 10^ -14 

But since:
[caffeineH+] = [OH-] 

Hence,

[OH-]^2 = 8.815 x 10^ -14 
[OH-] = 2.969 x 10^ -7 

The formula for pH is:

pH = 14 + log [OH-]

The pH of a solution containing a caffeine concentration of 430 mg/L is 4.75.

To calculate the pH of a caffeine solution, we can first use the provided pKb (10.4) to find the Kb, using the equation Kb = 10^(-pKb). We can then use the Kb to find the concentration of OH-, represented by the equilibrium C8H10N4O₂ (aq) + H₂O(1) ⇒ C8H10N4O₂H+ (aq) + OH¯ (aq). By inserting the equilibrium concentrations into the Kb expression and solving, we can find the OH- concentration.

The pH of a solution containing a caffeine concentration of 430 mg/L can be calculated using the equilibrium constant expression for caffeine. The equilibrium equation is: C8H10N4O2(aq) + H2O(l) ⇌ C8H10N4O2H+(aq) + OH-(aq). By substituting the given concentrations into the expression, the pH can be determined. The equation gives a pH of 4.75 for the solution.

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the final temperature of the water is approximately [tex]\(4.009 C\)[/tex].

To find the final temperature of the water after adding the heat produced from burning 8.800 g of [tex]\(C_6H_6\)[/tex] (benzene), we'll use the concept of heat transfer and the specific heat capacity of water.

The heat released from the combustion of [tex]\(C_6H_6\)[/tex] will be transferred to the water, causing its temperature to increase. We'll use the equation:

Q = mcΔT

Where:

- Q is the heat transferred (in Joules)

- m is the mass of the water (in grams)

- c is the specific heat capacity of water (4.18 J/g°C)

- ΔT  is the change in temperature of the water (in °C)

First, we need to calculate the heat released from burning [tex]\(C_6H_6\)[/tex].

Given:

- Mass of [tex]\(C_6H_6\)[/tex] burned, [tex]\(m_{C_6H_6} = 8.800 \, g\)[/tex]

- Heat of combustion of [tex]\(C_6H_6\)[/tex], [tex]\(ΔH_{comb} = -3263 \, kJ/mol\)[/tex]

Using the molar mass of [tex]\(C_6H_6\) (\(M_{C_6H_6} = 78.11 \, g/mol\))[/tex], we can find the number of moles of [tex]\(C_6H_6\)[/tex] burned and then calculate the heat released.

Next, we'll use the equation for heat transfer to find the change in temperature of the water, and then add this change to the initial temperature of the water to get the final temperature.

Let's calculate step by step.

Step 1: Calculate the heat released from burning [tex]\(C_6H_6\)[/tex].

1. Find the number of moles of [tex]\(C_6H_6\)[/tex]:

[tex]\[n_{C_6H_6} = \frac{m_{C_6H_6}}{M_{C_6H_6}} = \frac{8.800 \, g}{78.11 \, g/mol} \approx 0.1128 \, mol\][/tex]

2. Calculate the heat released from burning [tex]\(C_6H_6\)[/tex] using its molar enthalpy of combustion:

[tex]\[Q_{comb} = n_{C_6H_6} \times ΔH_{comb} = 0.1128 \, mol \times (-3263 \, kJ/mol)\][/tex]

[tex]\[Q_{comb} = -368.112 \, kJ\][/tex]

Step 2: Calculate the change in temperature of the water.

1. Use the equation for heat transfer:

[tex]\[Q_{water} = mcΔT\][/tex]

Where [tex]\(Q_{water}\)[/tex] is the heat absorbed by water, \(m\) is the mass of water, c is the specific heat capacity of water, and \(ΔT\) is the change in temperature of water.

2. Rearrange the equation to solve for [tex]\(ΔT\)[/tex]:

[tex]\[ΔT = \frac{Q_{comb}}{mc}\][/tex]

Given:

- [tex]\(m_{water} = 5691 \, g\)[/tex]

- [tex]\(c_{water} = 4.18 \, J/g°C\)[/tex]

3. Substitute the values and calculate \(ΔT\):

[tex]\[ΔT = \frac{-368.112 \times 10^3 \, J}{(5691 \, g) \times (4.18 \, J/g°C)}\][/tex]

[tex]\[ΔT \approx -16.991°C\][/tex]

Step 3: Find the final temperature of the water.

Given:

- Initial temperature of water, [tex]\(T_{initial} = 21°C\)[/tex]

The final temperature [tex](\(T_{final}\))[/tex] of the water can be found by adding the change in temperature [tex](\(ΔT\))[/tex] to the initial temperature [tex](\(T_{initial}\))[/tex]:

[tex]\[T_{final} = T_{initial} + ΔT\][/tex]

[tex]\[T_{final} = 21°C - 16.991°C\][/tex]

[tex]\[T_{final} \approx 4.009°C\][/tex]

Therefore, the final temperature of the water is approximately [tex]\(4.009 C\)[/tex].

Answer: The correct answer is Tyndall effect.

Explanation:

Colloids are defined as the mixtures where the size of the particle is within the range of 2nm to 1000 nm. In these mixtures, physical boundary is seen between the dispersed phase and dispersed medium.

Tyndall effect is defined as the effect in which scattering of light takes place by the particles present in a colloid or in very fine suspension.

For Example: Scattering of sunlight by clouds

Thus, the correct answer is Tyndall effect.

The scattering of light by a colloidal suspension is known as the Tyndall effect.

When a beam of light passes through a colloidal solution or a suspension, the suspended particles disperse and scatter the light.

This scattering is more pronounced when the suspended particles are larger in size compared to the wavelength of the light. The scattered light becomes visible, creating a cone or beam of light that is observable in the direction of the incident light. The Tyndall effect is often used to study and characterize colloidal systems, as it provides valuable information about particle size, concentration, and overall dispersion.

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Ne (g) effuses at a rate that is [tex]\boxed{{\text{2}}{\text{.6}}}[/tex] times that of Xe (g) under the same conditions.

Further Explanation:

Graham’s law of effusion:

Effusion is the process by which molecules of gas travel through a small hole from high pressure to the low pressure. According to Graham’s law, the effusion rate of a gas is inversely proportional to the square root of the molar mass of gas.

The expression for Graham’s law is as follows:

[tex]\boxed{{\text{R}}\propto\dfrac{1}{{\sqrt {{\mu }} }}}[/tex]

Here,

R is the rate of effusion of gas.

[tex]{{\mu }}[/tex] is the molar mass of gas.

Higher the molar mass of the gas, smaller will be the rate of effusion and vice-versa.

The rate of effusion of Ne is expressed as follows:

[tex]{{\text{R}}_{{\text{Ne}}}} \propto \dfrac{1}{{\sqrt {{{{\mu }}_{{\text{Ne}}}}} }}[/tex]

                                                 ......(1)

Here,

[tex]{{\text{R}}_{{\text{Ne}}}}[/tex] is the rate of effusion of Ne.

[tex]{{{\mu }}_{{\text{Ne}}}}[/tex] is the molar mass of Ne.

The rate of effusion of Xe is expressed as follows:

[tex]{{\text{R}}_{{\text{Xe}}}}\propto\dfrac{1}{{\sqrt{{{{\mu }}_{{\text{Xe}}}}}}}[/tex]

                                     ......(2)

Here,

[tex]{{\text{R}}_{{\text{Xe}}}}[/tex] is the rate of effusion of Xe.

[tex]{{{\mu }}_{{\text{Xe}}}}[/tex] is the molar mass of Xe.

On dividing equation (1) by equation (2),

[tex]\dfrac{{{{\text{R}}_{{\text{Ne}}}}}}{{{{\text{R}}_{{\text{Xe}}}}}}=\sqrt {\dfrac{{{{{\mu }}_{{\text{Xe}}}}}}{{{{{\mu }}_{{\text{Ne}}}}}}}[/tex]     ......(3)

Rearrange equation (3) to calculate  [tex]{{\text{R}}_{{\text{Ne}}}}[/tex].

[tex]{{\text{R}}_{{\text{Ne}}}}=\left( {\sqrt {\dfrac{{{{{\mu }}_{{\text{Xe}}}}}}{{{{{\mu }}_{{\text{Ne}}}}}}} } \right){{\text{R}}_{{\text{Xe}}}}[/tex]     ......(4)

The molar mass of Ne is 20.17 g/mol.

The molar mass of Xe is 131.29 g/mol.

Substitute these values in equation (4).

[tex]\begin{aligned}{{\text{R}}_{{\text{Ne}}}}&= \left({\sqrt {\frac{{{\text{131}}{\text{.29}}}}{{{\text{20}}{\text{.17}}}}} } \right){{\text{R}}_{{\text{Xe}}}}\\&= \left( {\sqrt {6.50917} } \right){{\text{R}}_{{\text{Xe}}}}\\&= 2.5513{{\text{R}}_{{\text{Xe}}}}\\&\approx 2.6{{\text{R}}_{{\text{Xe}}}}\\\end{aligned}[/tex]

Therefore the rate of effusion of Ne is 2.6 times the rate of effusion of Xe.

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: Effusion, rate of effusion, molar mass, Ne, Xe, 2.6 times, Graham’s law, inversely proportional, square root.

Neon effuses faster than xenon due to its lighter molar mass, and the effusion rate for neon will be larger than that for xenon, resulting in a smaller effusion time for neon.

The student's question pertains to the comparison of the effusion rates of neon (Ne) and xenon (Xe) gases under the same conditions. The effusion rate of a gas is inversely proportional to the square root of its molar mass, according to Graham's law of effusion. Given that neon is lighter than xenon, it will effuse at a faster rate. Using the provided effusion time calculations, if it takes 243 seconds for xenon to effuse, then by solving for the time it would take for the same amount of neon to effuse using the ratio of the square roots of their molar masses, we determine the time for neon to be approximately 95.3 seconds.

This result is expected because the lighter a gas is, the faster it should effuse, making the effusion rate for neon larger than that for xenon, and consequently, the time for effusion is smaller for neon than xenon as presented in the example calculation.

Answer: "2" should be placed infront of [tex]Na_3PO_4[/tex] to have the same number of phosphate ions on each side.

Explanation: For a given reaction in the question,

The phosphate ions on product side are 2, so there should be 2 phosphate atoms on the reactant side as well.

And the number of Magnesium atoms on product side is 3, there should be 3 magnesium atoms on the reactant side as well.

To balance out Phosphate and magnesium atoms, 6 sodium and chlorine atoms each are formed on reactant side, so to balance these atoms, 6 atoms of each should be present on product side.

Now, the balanced Chemical equation becomes:

[tex]2Na_3PO_4+3MgCl_2\rightarrow Mg_3(PO_4)_2+6NaCl[/tex]

By placing the coefficient '2' in front of Na3PO4 in the given chemical equation, we can ensure that the same number of phosphate ions are present on both sides of the equation.

In the given chemical equation, we see PO4 appearing twice: once in Na3PO4 and again in Mg3(PO4)2 on the other side of the equation. It's crucial to balance this equation so that the same number of phosphate ions are present on both sides. Each Mg3(PO4)2 molecule contains two phosphate ions (PO4 units), which means if we have one Mg3(PO4)2 on the right side of the equation, we need two PO4 units on the left side as well.

To achieve this, place the coefficient '2' in front of Na3PO4, which dictates that we have two Na3PO4 molecules (each containing one PO4 unit) allowing us to have two phosphate ions on the left. Hence, the balanced chemical equation will look like: 2Na3PO4 + MgCl2 → Mg3(PO4)2 + NaCl.

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