A 35.0-ml sample of 1.00 m kbr and a 60.0-ml sample of 0.600 m kbr are mixed. the solution is then heated to evaporate water until the total volume is 50.0 ml. how many grams of silver nitrate are required to precipitate out silver bromide in the final solution?

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.

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.

Answer:

F⃗ mag  =

−cosθj^ + sinθi^

Explanation:

use cross product

Final answer:

The magnetic force acting on a positive charge moving in the xy plane with a velocity vector can be determined using the right-hand rule and the cross-product formula.

Explanation:

The direction of the magnetic force acting on a positive charge moving in the xy plane with velocity v⃗ =vcos(θ)i^+vsin(θ)j^, when the local magnetic field is in the +z direction, can be determined using the right-hand rule. First, join the tails of the velocity vector and the magnetic field vector. Then, curl your right fingers from the velocity vector to the magnetic field vector. The direction in which your right thumb points is the direction of the force. In this case, the magnetic force would be directed into the page.

If the velocity and magnetic field are parallel to each other, there is no orientation of the hand that will result in a force direction. Therefore, the force on the charge is zero.

If the velocity vector is given as v = (2.0î – 3.0ĵ + 1.0k) × 10^2 m/s, the force can be calculated using the cross-product formula.

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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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.

Learn more:

1. Rank the elements according to first ionization energy: https://brainly.com/question/1550767

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

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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.

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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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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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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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.

The balanced molecular equation is [tex]\boxed{{\text{Al}}\left(s\right)+3{\text{AgN}}{{\text{O}}_3}\left({aq}\right)\to {\text{Al}}{{\left({{\text{N}}{{\text{O}}_{\text{3}}}}\right)}_3}\left({aq}\right)+3{\text{Ag}}\left(s\right)}[/tex]

The balanced net ionic equation is [tex]\boxed{{\text{Al}}\left( s\right)+3{\text{A}}{{\text{g}}^+}\left({aq}\right)\to{\text{A}}{{\text{l}}^{3+}}\left({aq}\right)+3{\text{Ag}}\left(s\right)}[/tex]

The reduction half-cell reaction is [tex]\boxed{{\text{Ag}}+{e^-}\to{\text{Ag}}}[/tex]

The oxidation half-cell reaction is [tex]\boxed{{\text{Al}}\to{\text{A}}{{\text{l}}^{3+}}+3{e^-}}[/tex]

Further Explanation:

The three types of equations that are used to represent the chemical reaction are as follows:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The reactants and products remain in undissociated form in the molecular equation. In the case of total ionic equation, all the ions that are dissociated and present in the reaction mixture are represented while in the case of overall or net ionic equation only the useful ions that participate in the reaction are represented.

The steps to write the molecular equation and net ionic reaction are as follows:

Step 1: Write the molecular equation for the reaction with the phases in the bracket.

In the reaction,1 mole of Al reacts with 3 moles of [tex]{\text{AgN}}{{\text{O}}_3}[/tex] to form 1 mole of [tex]{\text{Al}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_3}[/tex] and 3 moles of Ag. The balanced molecular equation of the reaction is as follows:

 [tex]{\text{Al}}\left(s\right)+3{\text{AgN}}{{\text{O}}_3}\left( {aq}\right)\to{\text{Al}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}}\right)_3}\left({aq}\right)+3{\text{Ag}}\left(s\right)[/tex]

Step2: Dissociate all the compounds with the aqueous phase to write the total ionic equation. The compounds with solid and liquid phases remain same. The total ionic equation is as follows:

 [tex]{\text{Al}}\left(s\right)+3{\text{A}}{{\text{g}}^+}\left({aq}\right)+{\text{NO}}_3^-\left( {aq}\right)\to{\text{A}}{{\text{l}}^{3+}}\left({aq}\right)+{\text{NO}}_3^-\left({aq}\right)+3{\text{Ag}}\left(s\right)[/tex]

Step3. The common ions on both sides of the reaction get cancelled out to get the net ionic equation.

[tex]{\text{Al}}\left(s\right)+3{\text{A}}{{\text{g}}^+}\left({aq}\right)+\boxed{{\text{NO}}_3^-\left({aq}\right)}\to{\text{A}}{{\text{l}}^{3+}}\left({aq}\right)+\boxed{{\text{NO}}_3^ - \left({aq}\right)}+3{\text{Ag}}\left(s\right)[/tex]

Therefore, the net ionic equation is as follows:

[tex]{\text{Al}}\left(s\right)+3{\text{A}}{{\text{g}}^+}\left({aq}\right)\to{\text{A}}{{\text{l}}^{3 + }}\left({aq}\right)+3{\text{Ag}}\left(s\right)[/tex]

Redox reaction:

It is a type of chemical reaction in which the oxidation states of atoms are changed. In this reaction, both reduction and oxidation are carried out at the same time. Such reactions are characterized by the transfer of electrons between the species involved in the reaction.

The process of gain of electrons or the decrease in the oxidation state of the atom is called reduction while that of loss of electrons or the increase in the oxidation number is known as oxidation. In redox reactions, one species lose electrons and the other species gain electrons. The species that lose electrons and itself gets oxidized is called as a reductant or reducing agent. The species that gains electrons and gets reduced is known as an oxidant or oxidizing agent. The presence of a redox pair or redox couple is a must for the redox reaction.

The general representation of a redox reaction is,

[tex]{\text{X}}+{\text{Y}}\to{{\text{X}}^+}+{{\text{Y}}^-}[/tex]

The oxidation half-reaction can be written as:

[tex]{\text{X}}\to{{\text{X}}^+}+{e^-}[/tex]

The reduction half-reaction can be written as:

[tex]{\text{Y}}+{e^-}\to{{\text{Y}}^-}[/tex]

Here, X is getting oxidized and its oxidation state changes from  to +1 whereas B is getting reduced and its oxidation state changes from 0 to -1. Hence, X acts as the reducing agent whereas Y is an oxidizing agent.

Ag in silver nitrate forms solid silver during the reaction so it is getting reduced. The reduction half-cell reaction is as follows:

[tex]{\text{Ag}}+{e^-}\to{\text{Ag}}[/tex]

Aluminium gets converted to [tex]{\text{A}}{{\text{l}}^{3+}}[/tex] by oxidizing itself. The oxidation half-cell reaction is as follows:

[tex]{\text{Al}}\to{\text{A}}{{\text{l}}^{3+}}+3{e^-}[/tex]

Learn more:

1. Balanced chemical equation: https://brainly.com/question/1405182

2. Oxidation and reduction reaction: https://brainly.com/question/2973661

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: net ionic equation, Ag, Al, NO3-, Al3+, e-, Ag+, redox, oxidizing, reducing, oxidation half-cell reaction, reduction half-cell reaction, molecular equation, AgNO3, Al(NO3)3.

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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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.

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.

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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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]

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The face-centered cubic unit cell of nickel has an edge length of 352.4 pm. The density of nickel is 8.9 gcm⁻³.

The mass of a material per unit of volume is its density. Density is most frequently represented by the symbol, however the Latin letter D may also be used. Density is calculated mathematically by dividing mass by volume.

Because it enables us to predict which compounds will float and which will sink in a liquid, density is a crucial notion. As long as an object's density is lower than the liquid's density, it will often float.

Four atoms altogether make up a face-centered cubic unit cell. As a result, a face-centered cubic unit cell has a density of 4 x M / A3 x Na. There are two types of density, absolute and relative density.

Thus, The density of nickel is 8.9 gcm⁻³.

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Final answer:

Nickel does not crystallize in a simple cubic structure because the calculated density for a simple cubic arrangement is much lower than the actual density of nickel. The actual density of nickel (8.90 g/cm³) indicates that nickel has a denser crystal structure, specifically a face-centered cubic lattice, which contains more atoms per unit cell.

Explanation:

No, nickel does not crystallize in a simple cubic structure. If nickel were to crystallize in a simple cubic structure, we would calculate its density based on the volume occupied by one atom in the unit cell. To find the volume of the unit cell, we would cube the edge length of the unit cell, which is given as 0.3524 nm (convert to cm: 3.524 x 10−8 cm). The volume of the unit cell would then be (3.524 x 10−9 cm)3 = 4.376 x 10−23 cm3.

Knowing that there is one atom per unit cell in a simple cubic structure and using the molar mass of nickel (58.6934 g/mol) along with Avogadro's number (6.022 x 1023 atoms/mol), we can determine the mass of one atom of nickel. The mass of one mole of nickel atoms divided by Avogadro's number gives the mass of a single atom to be 9.746 x 10−23 g. Therefore, the density of nickel in a simple cubic structure would be the mass of one atom divided by the volume of the unit cell, equating to 2.23 g/cm3.

However, the actual density of Ni is 8.90 g/cm3, which is significantly higher than what we would expect for a simple cubic structure. This implies that nickel crystallizes in a denser structure, such as a face-centered cubic (fcc) lattice, where more atoms are present per unit cell compared to a simple cubic structure. In an fcc lattice, there are effectively 4 atoms per unit cell.

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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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.

Learn more:

1. Which statement is true for Boyle’s law: https://brainly.com/question/1158880

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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.

The car would have traveled an estimated distance of approximately 185 miles.

The formula for distance which is:

Distance = Speed  x Time

In this case, the estimated distance would be:

Distance = 57.9 miles/hour x 3.2 hours

Now, do the multiplication:

Distance ≈ 57.9 x 3.2

Distance ≈ 185.28 miles

Therefore, the umber of miles driven is 185 miles.