Two different ionic compounds each contain only copper and chlorine. Both compounds are powders, one white and one brown. An elemental analysis is performed on each powder. Which of the following questions about the compounds is most likely to be answered by the results of the analysis?
A. What is the density of each pure compound?
B. What is the formula unit of each compound?
C. What is the chemical reactivity of each compound?
D. Which of the two compounds is more soluble in water?

Answer:

3,4-dibromo 2-methylheptane

Explanation:

Alkynes can be synthesized by using sodium amide in liquid ammonia, when the starting material is a vicinal dibromoalkane. This means an alkane in which there's a Br group in two carbons next to one another.

In the attached picture you can see the reaction.

Answer:

see picture

Explanation:

Although the previous answer may be correct, the question is incomplete, the completed question is:

Draw the structure of the starting material needed to make 2-methylhept-3-yne using sodium amide in liquid ammonia followed by 1-bromopropane.

Now, the previous answer is also incorrect, because the starting material with the two bromine in carbon 3 and 4, the first reaction taking place is the substracting of the hydrogen from carbon 5, and not carbon 3 because it's bulkier than carbon 5. So the majority product formed will be a double bonded product and not a triple bond. Therefore, this answer is incorrect.

Now, that we have completed the question the Starting material would have to be the following:

CH3 - CH(CH3) - C ≡ CH

When this reactant, is in presence of NaNH2, the NH2 substract the Hydrogen from carbon 1, and then, this will attack as nucleophyle to the carbon 1 of the bromopropane and formed the desired product.

The mechanism is as following in the picture.

Answer:

Ionic compound.

Explanation:

Ionic bond:

It is the bond which is formed by the transfer of electron from one atom to the atom of another element. the electrostatic attraction is created between bonded atoms.  

Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.

The compound having ionic bonds generally have moderate to high boiling points and melting point because of greater electrostatic interaction. Their electrical conductivity are high and these minerals tend to dissolve in water.

For example:

Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion and  both atoms joint together through electrostatic interaction.

The mole fraction of nitrogen in the gas mixture can be calculated using the ideal gas law and the given pressure, volume, and temperature. The calculation suggests a mole fraction of 1.509, which might indicate that we've made an erroneous assumption about the mixture's composition.

First, in order to find the mole fraction of nitrogen in the mixture, we should first find the number of moles of the total mixture. To do this, we can use the ideal gas law, which states that the volume, pressure, and temperature of a gas can be related to the number of moles of the gas. This law can be written as: PV=nRT.

To find the pressure in the correct units, we convert mmHg to atm by dividing by 760 (since 1 atm = 760 mmHg). The pressure is therefore 870.2 / 760 = 1.145 atm. Similarly, we should convert the temperature to Kelvin by adding 273 to the Celsius temperature: 31.2 + 273 = 304.2 K.

So, our equation becomes: 1.145 atm * 15.1 L = n * 0.0821 atm*L/mol/K * 304.2 K. Solving for n, we find that the total number of moles in the mixture is n = 0.570 moles.

Given that molecular nitrogen (N₂) has a molar mass of 28.01 g/mol, we can calculate the number of moles of nitrogen in the 24.1-g mixture. The mass of nitrogen divided by its molar mass gives the number of moles, which is 24.1 g / 28.01 g/mol = 0.860 moles. Therefore, the mole fraction of nitrogen (X_N2) in the mixture can be calculated as the number of moles of nitrogen divided by the total number of moles, or 0.860 / 0.570 = 1.509.

Therefore, the mole fraction of nitrogen in the mixture is 1.509. This value may seem unusual as mole fractions are typically less than 1, but this discrepancy could be due to the initial assumption that the entire 24.1 g consisted of nitrogen. In a real situation, you would need to know the mass of each component in the mixture to accurately calculate the mole fractions.

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The mole fraction of nitrogen in the gas mixture is calculated to be approximately 43.8 percent using the Ideal Gas Law and the given molar masses of nitrogen and carbon dioxide.

To find the mole fraction of nitrogen in this mixture, we first need to use the Ideal Gas Law (PV=nRT) to find the total moles (n) of the gases in the mixture. Given the volume (V) of the gas mixture is 15.1 L, the pressure (P) is 870.2 mmHg or 1.145 atm, and the temperature (T) is 31.2oC or 304.35 K, with R being the ideal gas constant which equals 0.0821 L·atm/K·mol, the total number of moles is found to be 0.573 mol.

Then, we need to know that the molar masses of nitrogen (N₂) and carbon dioxide (CO₂) are 28.01 g/mol and 44.01 g/mol respectively. So, we can set up the equation x(28.01) + (1-x)(44.01) = 24.1 (where x is the mole fraction of nitrogen in the mixture) to solve for x. We find that x equals 0.438.

So, the mole fraction of nitrogen in this mixture is approximately 0.438 or 43.8 percent.

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Answer:

The correct answer is option C.

Explanation:

Ionic bond is defined as the bond which is formed by complete transfer of electrons from one atom to another atom.

Ionic compound is formed by the complete transfer of electrons from 1 atom to another atom. The cation is formed by the loss of electrons by metals and anions are formed by gain of electrons by non metals.

From the given options , the option with metal and non metal is potassium and chlorine. these elements will combine together to form ionic compound called potassium chloride.

Answer: highest activity of salivary amylase is at pH 6.8.

Explanation:Amylase enzyme catalyses the hydrolysis of starch into sugars. Amylase is present in the saliva of humans and some other mammals. The enzymatic activity of salivary amylase ranges from 6 to 7. Above and below this range, the reaction rate reduces as enzymes get denaturated. But the optimum enzymatic activity is at pH 6.8.

Answer:

[tex]PO_4^{-3} < NO_3^- < Na^+[/tex]

Explanation:

Solutions.

a) 100 ml 1 M of Na3PO4

b) 100 ml 1 M of AgNO3

c) Mixture: 200 ml 0.5 M of Na3PO4 and 0.5 M of AgNO3

Reaction:

[tex] 3 Ag^+ + PO_4^{-3} \longrightarrow Ag_3PO_4[/tex]

So if silver ion is consumed almost completely:

[tex][Ag^+]=0 M[/tex]

[tex][PO_4^{-3}]=0.5 M-0.5 M \frac{1 mol PO4}{3 mol Ag}=0.33 M[/tex]

[tex][Na^+]=0.5 M *  \frac{3 mol Na}{mol Na_3PO_4}=1.5 M[/tex]

[tex][NO_3^-]=0.5 M[/tex]

In increasing order of concentration:

[tex]PO_4^{-3} < NO_3^- < Na^+[/tex]

The correct order of ions remaining in solution after mixing Na3PO4 with AgNO3, by increasing concentration, is phosphate (PO43-), nitrate (NO3-), and sodium (Na+), which is option (A).

When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3, a reaction occurs where silver phosphate (Ag3PO4), a yellow precipitate, forms due to its low solubility and Ag+ ions become almost absent in the solution. The remaining ions in the solution will be Na+, NO3-, and a very small amount of PO43-. Since all of the Ag+ reacts to form the precipitate, we can assume that there are three times as many Na+ ions compared to PO43- ions originally, because each formula unit of Na3PO4 produces three sodium ions. Therefore, this leaves Na+ with the highest concentration in solution. NO3- concentration remains unchanged because it is a spectator ion and does not participate in the reaction.

The correct order of ions in solution by increasing concentration is PO43- < NO3- < Na+, which corresponds to option (A).

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Answer:

1 mol of sucarose has 7.22 x10²⁴ atoms of C, 1.32 x10²⁵ atoms of H and 6.62 x10²⁴ atoms of O

Explanation:

C12H22O11 is the molecular formular for sucarose

So in 1 mol of sacarose, we have 12 mols of carbon, 22 moles of hydrogen, and 11 mols of oxygen.

As you know, 1 mol has 6.02x10²³ atoms so these are the rule of three to calculate the number of atoms of each type.

1 mol ______ 6.02x10²³

12 moles ______ 12 . 6.02x10²³ = 7.22 x10²⁴

22 moles ______ 22 . 6.02x10²³ = 1.32 x10²⁵

11 moles _______ 11 . 6.02x10²³ = 6.62 x10²⁴

Answer:

The mass of a solute divided by the mass of a solution times 100

Explanation:

The concentration of a solution refers to how much of a solute is dissolved in an amount of solvent. To express this concentration exist different methods, the mass percent concentration is one of them and is defined as the mass of a solute divided by the mass of a solution times 100:

%m/m= (mass of a solute/mass of solution)x100

Where the mass of the solution is the sum of the mass of solute and mass of solvent.

%m/m is commonly used when you can measure the masses of both solute and solution.

I hope you find this information useful and interesting! Good luck!

Mass percent in chemistry denotes the proportion of a solute in a solution in terms of a percentage. It's calculated as: Mass Percent = (mass of solute/mass of solution) * 100.

The mass percent concentration of a solution signifies the proportion of a solute in a certain quantity of solution, and it is displayed as a percentage. To calculate the mass percent, we use the formula:

Mass Percent = (mass of solute/mass of solution) * 100

If, for example, you have a solution composed of 20g of sugar in 180g of water, the mass of the solution would be the sum of these two (200g), and thus the mass percent of sugar in the solution is (20/200)*100 = 10%.

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Answer:

The substance has a specific heat of 1.176 J/g°C

Explanation:

Step 1: Data given

Temperature change = 34 °C

Mass of the substance = 20 kg = 20000 grams

The substance gained 800 kJ of heat during this temperature change

Step 2: Calculate the specific heat

q = m*c*ΔT

⇒ with q = heat gained = 800 kJ = 800000 J

⇒ with m = the mass of the substance = 20 kg = 20000 grams

⇒ with c = the specific heat of the substance = TO BE DETERMINED

⇒ with ΔT = the change of temperature = T2 -T1 = 48° - 14 ° = 34°

c = q/(m*ΔT)

c = 800000 / (20000 * 34)

c = 1.176 J/g°C

The substance has a specific heat of 1.176 J/g°C

Final answer:

Elements within the same group or family on the periodic table have the most similar chemical properties due to their identical number of valence electrons. Examples include alkali metals such as lithium and sodium, and halogens like fluorine and chlorine.

Explanation:

Elements that have the most similar chemical properties are typically found within the same group on the periodic table. Groups, also known as families, contain elements with the same number of valence electrons. For example, the alkali metals such as lithium (Li) and sodium (Na) have one valence electron, making them highly reactive and sharing similar chemical behaviors like forming compounds with oxygen in a 2:1 ratio.

Similarly, alkaline earth metals like beryllium (Be) and magnesium (Mg) each have two valence electrons and also show close chemical properties. The halogens—including fluorine (F) and chlorine (Cl)—each have seven valence electrons, leading to their characteristic reactivity and ability to form compounds with elements such as sodium.

As elements share the same number of valence electrons, it leads to similarities in the ways they lose, gain, or share electrons during chemical reactions. The patterns of chemical properties extend beyond single groups; for instance, metallic character increases as one moves down a group in the periodic table.

In Lewis structures, all X-O bonds have the same length in NO3 -, SO4 2-, and BrO3 -. This is due to resonance, whereby the electron density is distributed among multiple equivalent structures.

In the Lewis structures of the given oxyanions, the ions in which all X-O bonds have the same length are: (ii) NO3 -, (iv) SO4 2-, and (v) BrO3 -. The concept of bond length in these ions is connected to the principle of resonance. Each of these ions exhibit resonance, which means the electron density is distributed among more than one equivalent Lewis structures. This leads to the bonds having the same length. For example, in the SO4 2- ion, there are multiple resonance structures, each involving double bonds between the sulfur and the different oxygen atoms. As these double bonds can 'move' about the ion (due toan resonce), the four sulfur-oxygen bonds are of equal length. Such a phenomenon is not present in (i) NO2 - and (iii) SO3 2-.

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Final answer:

The NO2- ion has all X-O bonds with the same length.

Explanation:

The Lewis structure of an ion represents the arrangement of atoms and electrons in the ion.

In the oxyanions provided, only NO2- has all X-O bonds (X representing the central atom) with the same length.

This is because the NO2- ion has resonance structures, which means the double bond can be located between either of the two oxygen atoms. As a result, the N-O bonds in NO2- are identical in length.

Answer:

It can be removed by acidic chemicals

Explanation:

The glue likely consists of compounds with nonpolar covalent bonds that repel water but are dissolved by solvents like acetone, explaining why water can't remove the glue but fingernail polish remover can.

The observation that glue does not easily wash off with water but can be removed with a substance like fingernail polish remover helps us understand the chemical composition of the glue. The adhesive used in the glue is likely a combination of compounds with nonpolar covalent bonds. These compounds resist polarity, which repels water molecules. However, they can be dissolved with certain solvents that contain similar nonpolar substances, such as acetone found in fingernail polish remover. Acetone, or dimethyl ketone, is used as a solvent in various applications due to its ability to dissolve many substances.

By understanding the chemical reactions between different substances, one can often predict their behaviors when they interact. In this case, the interaction between the glue and the water or acetone is predictable based on our understanding of their individual chemical compositions and the forces of cohesion and adhesion that influence how substances combine.

For example, when substances such as water, which have molecules with polar covalent bonds, interact with substances like glue, which has molecules with nonpolar covalent bonds, cohesion within the water molecules causes them to resist mixing with the glue molecules. Conversely, adhesion between the glue molecules and the acetone molecules in fingernail polish remover allows them to mix effectively, thereby enabling the acetone to dissolve the glue.

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Answer: B (Items with densities lower than water will sink due to surface tension)

Explanation:

Surface tension is an intermolecular force exerted on the surface of water making it like a stretch elastic skin. Surface tension enables items with lesser densities than water, to float and slide on a water surface. Examples insects, leaves, paper etc

According to Archimedes principle an object denser than the fluid will sink. While objects less dense than the fluid will float.

Answer is (B) is false because items with densities higher then water will sink due to surface tension

Items with densities lower than water will sink due to surface tension. Therefore, the correct option is option B.

A intriguing phenomena called surface tension occurs when a liquid meets a gas or another liquid. It results from the cohesive forces between the molecules that are present at a liquid's surface. Surface tension is a result of the thin layer that is formed by these cohesive forces that holds the molecules closely together. A liquid droplet placed on a surface can be used to illustrate the idea of surface tension. The droplet takes on a spherical shape as a result of the cohesive forces between the water molecules. The molecules at the surface are drawn inside to reduce their exposure to the surroundings.

Therefore, the correct option is option B.

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Answer:

[tex]CCl_{2}FCF_{3}[/tex]

Explanation:

To solve this equation we can use Graham's Law of effusion to calculate the molar mass of the chloroflourocarbon

[tex]\sqrt{\frac{M_{1} }{M_{2} } } = \frac{t_{1} }{t_{2} } \\\\\sqrt{\frac{M_{1} }{39.948 } } = \frac{157 }{76}\\\\\\M_{1}  = 170.477\\[/tex]

Which chloroflorocarbon has this molar mass?

C2Cl2F4

Using Graham's Law of Effusion and given effusion times, we calculated the molar mass of the chlorofluorocarbon to be approximately 170.55 g/mol. This formula suggests the gas is likely Cl₂F₂ (chlorodifluoromethane).

The formula of the chlorofluorocarbon, we can use Graham's Law of Effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The formula for Graham's Law is: (Rate of effusion of Gas 1) / (Rate of effusion of Gas 2) = √(Molar mass of Gas 2 / Molar mass of Gas 1)

Given:

First, find the rate of effusion for each gas:

Using Graham's Law:

(Rate of effusion of chlorofluorocarbon) / (Rate of effusion of argon) = sqrt(Molar mass of argon / Molar mass of chlorofluorocarbon)

Substitute the values:

(50 mL / 157 s) / (50 mL / 76 s) = √(39.948 g/mol / Molar mass of chlorofluorocarbon)

Simplify:

76 / 157 = √(39.948 / Molar mass of chlorofluorocarbon)

(76 / 157[tex])^2[/tex] = 39.948 / Molar mass of chlorofluorocarbon

Molar mass of chlorofluorocarbon = 39.948 / (76 / 157[tex])^2[/tex] = 39.948 / (0.484[tex])^2[/tex] =170.55 g/mol

An example of a chlorofluorocarbon with a molar mass close to 170.55 g/mol is Cl₂F₂ (chlorodifluoromethane), with a molar mass of approximately 120.91 + 37.996 = 169.91 g/mol.

Answer: A. There are more reactants than products at equilibrium.

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{eq}[/tex]

For a general chemical reaction:

[tex]aA+bB\rightarrow cC+dD[/tex]

The expression for [tex]K_{eq}[/tex] is written as:

[tex]K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

There are 3 conditions:

When [tex]K_{eq}>1[/tex]; the reaction is product favored.

When [tex]K_{eq}<1[/tex]; the reaction is reactant favored.

When [tex]K_{eq}=1[/tex]; the reaction is in equilibrium.

From the above expression, the equilirbium constant is directly dependent on product concentration. Thus, more is the concentration of product, more will be the equilibrium constant.

The values of [tex]K_{eq}[/tex] less than 1 , means the reactant are more than the products.

Answer:

2 51 × 10^-5mol/L

Explanation:

The concentration of hydrogen ions can be calculated using the formula below :

pH = -log [H+]

pH = 4.6

[H+] = ?

[H+] = Antilog (-4.6)

[H+] = 2 51 × 10^-5mol/L

Answer:

a) ΔU = 71 kJ

b) the pressure will be higher

Explanation:

Step 1: Data given

A gas does 171 kJ of work on its surroundings

At the same time there is 242 kJ of heat added to the gas

Step 2: Calculate change of internal energy

Change of internal energy ΔU  by the gas due to the 171 kJ work done by the system and addition of 242 kJ heat to the system.

heat energy (ΔH) is the summation of heat capacity (ΔU) and work done by a system

ΔU = Q + W

The work is done by the system on its surroundings, what means energy is lost. W will have a negative value: -171 kJ

The heat is added to the system, this means we gain energy. Q will have a positive value: 242 kJ

Total change of intern energy will be:

ΔU = Q - W

ΔU = 242 kJ - 171 kJ

ΔU = 71 kJ

ΔU = nR*ΔT

For an ideal gas  n and R are constant. . Now consider PV = nRT.

P and T are directly related. Therefore, an increase in temperature will result in a higher pressure.

Final answer:

The change in internal energy of the gas during the process is 71 kJ, and the pressure of the gas will generally be lower at the end of the process.

Explanation:

The question addresses the concept of the first law of thermodynamics as it applies to heat transfer, work done by a gas, and changes in internal energy in thermodynamic processes. According to the first law of thermodynamics, the change in internal energy of a system (ΔU) is equal to the heat (Q) added to the system minus the work (W) done by the system on its surroundings. In this case:

ΔU = Q - W

The given data states that the gas sample does 171 kJ of work on its surroundings (W = 171 kJ) and 242 kJ of heat is added to the gas (Q = 242 kJ). Therefore:

ΔU = 242 kJ - 171 kJ = 71 kJ

(a) The change in internal energy of the gas during the process is 71 kJ.

(b) As the gas does work on the surroundings and expands, the pressure of the gas will generally be lower at the end of the process, as described by the ideal gas law and assuming other factors such as temperature remain constant.

The volume of NO₂ gas collected over water at 25.0 °C is 1.67 L

We'll begin by calculating the number of mole of Cu. This can be obtained as follow:

Mass of Cu = 2.01 g

Molar mass of Cu = 63.55 g/mol

Mole = mass / molar mass

Mole of Cu = 2.01 / 63.55

Next, we shall determine the number of mole of NO₂ produced by 2.01 g (i.e 0.0316 mole) of Cu. This can be obtained as follow:

Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2H₂O(l) + 2NO₂(g)

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of NO₂.

Therefore,

0.0316 mole of Cu will react to produce = 0.0316 × 2 = 0.0632 mole of NO₂.

Finally, we shall determine the volume of NO₂ gas obtained from the reaction. This can be obtained as follow:

Number of mole of NO₂ (n) = 0.0632 mole

Temperature (T) = 25 °C = 25 + 273 = 298 K

Total pressure = 726 mm Hg

Vapor pressure of water = 23.8 mm Hg

Pressure of NO₂ (P) = 726 – 23.8 = 702.2 / 760 = 0.924 atm

Gas constant (R) = 0.0821 atm.L/Kmol

0.924 × V = 0.0632 × 0.0821 × 298

0.924 × V = 1.54623856

Divide both side by 0.924

[tex]V = \frac{1.54623856}{0.924} \\\\[/tex]

Therefore, the volume of NO₂ collected over water at 25.0 °C is 1.67 L

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The chemical reaction is defined as the reaction of reactants to form products. The reaction between copper and nitric acid will result in the formation of copper(II)nitrate, water, and nitrogen dioxide. The volume of NO[tex]_2[/tex] collected will be 1.67 L.

Give that,

The number of moles produced by nitrogen dioxide (2.01 g) will be:

Now, from the equation,

Now, from the ideal gas equation,

Therefore, the volume of nitrogen dioxide collected over water at 25-degree celcius is 1.67 L.

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Answer:

Explanation:

Newton's first law of motion :

"Every object persists in its state of rest or uniform motion in a straight line unless it is compelled to change that state by forces exerted on it."

From the question we know that the net forces on the object were zero or that the there were no unbalanced forces on it.

Therefore we can assume that the object is moving along a straight line.

And the object was moving at a constant speed of 30 mph.

So it is clear from the Newton's first law that the object will remain in the state of motion as it was earlier.

That is the object will remain in motion at constant speed of 30 mph.

Answer:

1)20. ML at 100.°C

Explanation:

Average kinetic energy is not related with volume.  so increase in temperature will also increase kinetic energy. so this increase means the highest temperature will be a right answer.

Final answer:

The water sample with the highest average kinetic energy is the 20 mL of water at 100.0°C, as average kinetic energy is directly proportional to temperature.

Explanation:

The molecules with the highest average kinetic energy are in the sample of water at the highest temperature. Kinetic energy is directly proportional to the absolute temperature of the substance, meaning that as temperature increases, kinetic energy also increases. Therefore, ignoring the volume of the water samples, the rank from highest kinetic energy to lowest based on temperature would be:

The sample at 100.0°C holds the highest average kinetic energy because it has the highest temperature.

Answer:

MW = 94.07 g/mol

Explanation:

We are here given a neutralization reaction which is a 1: 1 molar relation. So calculating the # mol NaOH reacted we will know the  #mol of  the monoprotic acid, and since we are given the mass of sample we can calculate the molar mas of the compund because # mol = mass/ MW .

# mol NaOH = 0.0823 mol/L x 33 mL/1000 mL/L = 0.0027 mol

# mol = m/MW ∴   MW: m/# mol

MW :  0.254 g / 0.0027 mol =  94.07 g/mol

Answer:

The identity of the gas is Cl₂ (chlorine)

Explanation:

STP conditions are:

1 atm → Pressure

273.15 K → T°

So, we must use the Ideal Gas Law to get the moles.

Before that, we will think density as data where 3.164 g of diatomic gas are contained in 1 L of volume.

P.V = n . R . T

1 atm . 1 L = n. 0.082 L.atm/mol.K . 273.15K

1 L.atm / (0.082 L.atm/mol.K . 273.15K) = n

0.0446 mol = n

This quantity of diatomic gas, are 3.164 g so the molar mass will be:

Mass / mol = molar mass

3.164 g / 0.0446 mol = 70.9 g/m

The element (a diatomic molecule), which has that molar mass in the periodic table is the Cl₂.

1 Cl = 35.45 g/m

Cl₂. = 70.9 g/m

Using the density data and Avogadro's principle, we can calculate the molar mass of the unknown gas to be around 70.8 g/mole. Yet, none of the gases from the provided list align with this molar mass.

The identity of the unknown diatomic gas can be found using the Ideal Gas Law and Avogadro's principle. Avogadro's principle states that equal volumes of gases at the same temperature and pressure will contain an equal number of particles (or moles). Hence, at Standard Temperature and Pressure (STP), 1 mole of any gas will occupy 22.4 L.

From the data provided, we know that 1 mole of the unknown gas has a mass of 3.164 g/L × 22.4 L per mole = 70.8 g/mole. Among the different potential gases listed in the reference, O₂ is a diatomic gas and has a molar mass of 32g/mole, which isn't close to the calculated one. NH3 has a molar mass of 17g/mole which also doesn't match. This leaves us with He (Helium), which is not diatomic and hence can't be our gas. We can therefore confidently conclude that there seems to be a mistake as none of the gases listed align with our calculated molar mass.

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Answer:

ΔH of reaction is -147 kJ/mol

Explanation:

For the reaction:

CH₄(g) + X₂(g) → CH₃X(g) + HX(g)

It is possible to obtain ΔHr of reaction from the sum of ΔH of products minus ΔH of reactants:

ΔHr = ΔH CH₃X + ΔH HX - (ΔH CH₄ + ΔH X₂) (1)

The ΔH of each compound is obtained from bond energies thus:

ΔH CH₃X = 3×C-H + C-X = 3×416 kJ/mol + 222 kJ/mol = 1470 kJ/mol

ΔH HX = H-X = 277 kJ/mol

ΔH CH₄ = 4×C-H = 4×416kJ/mol = 1664 kJ/mol

ΔH X₂ = X-X = 230 kJ/mol

Replacing in (1):

ΔHr = 1470 kJ/mol + 277kJ/mol - (1664 kJ/mol + 230 kJ/mol) = -147 kJ/mol

I hope it helps!

Final answer:

To calculate the enthalpy change for CH₄ + X₂ → CH₃X + HX, we tally the energy needed to break one C-H bond and one X-X bond (total 646 kJ/mol) and subtract the energy released from forming one C-X bond and one H-X bond (total 499 kJ/mol), resulting in a ΔH of 147 kJ/mol.

Explanation:

To calculate the enthalpy change (ΔH) for the reaction CH₄(g) + X₂(g) → CH₃X(g) + HX(g), we need to consider the bond energies provided. The calculation involves computing the energy required to break bonds in the reactants and the energy released from forming bonds in the products.

First, we break one C-H bond in CH₄ and one X-X bond. The energy required is:

1 mol C-H: 416 kJ/mol

1 mol X-X: 230 kJ/mol

When these connections are broken, the total energy needed is:

416 kJ/mol + 230 kJ/mol = 646 kJ/mol

Next, we form one C-X bond and one H-X bond with the energies released as follows:

1 mol C-X: 222 kJ/mol

1 mol H-X: 277 kJ/mol

The total energy released in forming these bonds is:

222 kJ/mol + 277 kJ/mol = 499 kJ/mol

The enthalpy change of the reaction is the energy required minus the energy released:

ΔH = 646 kJ/mol - 499 kJ/mol = 147 kJ/mol

This reaction is endothermic, as indicated by the positive value of ΔH.

Answer:

See explanation below

Explanation:

a) According to this, this is an excercise that is involving a separation and reaction of different compounds. The n-butyl bromide is formed when, you made the following reaction:

CH2 = CH - CH2 - CH3 + HBr/peroxide -----------> Br - CH2 - CH2 - CH2 - CH3

Now, in this reaction, we still has traces of HBr in the product, so, in order to neutralize these traces, we wash the solution with bicarbonate sodium forming sodium bromide and carbonic acid as follow:

HBr + Na2CO3 ---------> NaBr + Na2CO3

Also, this is a weak base so it will not react with the n-butylbromide to form another product.

b) Basing of what it was stated above, we cannot wash the solution with NaOH because this is a strong base, and not only wil eliminate the traces of HBr, it will also react with the butylbromide causing an elimination and substitution reaction, giving the following products:

BrCH2CH2CH2CH3 + NaOH --->CH2=CHCH2CH3 + OHCH2CH2CH2CH3

That it's why we need to wash this product with a weak base only.

c) The density of 1-chlorobutane is 0.88 g/mL, density of water is 1 g/mL and density of sodium bicarbonate is 2.2 g/cm3, therefore, the one that has a greater density will go at the lower phase.

In this case, after the reflux, it will stay in the lower phase.

after adding water, it will be in the upper phase.

after adding bicarbonate, it will be in the upper phase too.

a. The purpose of washing the crude n-butyl bromide with aqueous sodium bicarbonate is to remove acidic impurities. b. It would be undesirable to wash the crude halide with aqueous sodium hydroxide. After each stage of the separation procedure, the alkyl chloride would appear as the upper or lower phase.

a. The purpose of washing the crude n-butyl bromide with aqueous sodium bicarbonate is to remove any acidic impurities. Sodium bicarbonate is a weak base that reacts with acidic impurities, such as hydrochloric acid or sulfuric acid, to form water-soluble salts. The balanced chemical equation for the reaction between sodium bicarbonate and hydrochloric acid is:

NaHCO3 + HCl → NaCl + H2O + CO2

b. It would be undesirable to wash the crude halide with aqueous sodium hydroxide because sodium hydroxide is a strong base that can react with the bromide ion to form sodium bromide, which would not be easily separated from the organic layer.

After the reflux, the alkyl chloride would appear as the upper phase. After the addition of water, the alkyl chloride would still appear as the upper phase. After the addition of sodium bicarbonate, the alkyl chloride would appear as the lower phase.

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Answer: The molecular formula for the compound is [tex]C_6H_6[/tex]

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 92.26 g

Mass of H = 7.74 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.68 moles.

For Carbon = [tex]\frac{7.68}{7.68}=1[/tex]

For Hydrogen = [tex]\frac{7.74}{7.68}=1[/tex]

The ratio of C : H = 1 : 1

The empirical formula for the given compound is [tex]CH[/tex]

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L  (Conversion factor:  1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = [tex]62.3637\text{ L. torr }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]100^oC=(100+273)K=373K[/tex]

Putting values in above equation, we get:

[tex]820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

We are given:

Mass of molecular formula = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

[tex]n=\frac{78.10g/mol}{13g/mol}=6[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(1\times 6)}H_{(1\times 6)}=C_6H_6[/tex]

Hence, the molecular formula for the compound is [tex]C_6H_6[/tex]

Answer:

69%

Explanation:

Say the hydrocarbon is C2H4, the equation of reaction would be;

C2H4 + H2 ----------> C2H6

The hydrocarbon, C2H4 is the limiting reagent.

From the question, the hydrocarbon C2H4 flows into a catalytic reactor at 26.2 atm and 250°C with a flow rate of 1100 L/min

Using, PV=nRT--------------------(1).

n= PV/RT to find the number of moles.

n= 26.2 atm × 1100 L/ 0.0821 L. atm/mol per kelvin × 523 K.

=28,820 atm.L/ 42.9383 L.atm/mol.

= 671.2 mole.

One mole of C2H4 produced one mole of C2H6.

Mass of C2H6 = 30 × 671.2

= 20,136 g = 20.136 kg.

Percent yield = actual yield/ theoretical yield × 100% -------(2)

Actual yield= 13.9 kg, theoretical yield = 20.136 kg

Substitute the parameters into equation (2). We have;

Percent yield = 13.9 kg/ 20.136× 100

Percent yield= 0.69 × 100

Percent yield= 69%

Note: P= pressure, V= volume, T= temperature and n = number of moles

Answer:

C) H⁺

Explanation:

When we are balancing the reaction in an acid medium, hydrogen is balanced using the H⁺ species. This is most likely the intended answer of your question.

When the reaction takes place not in an acid medium, but in an alkaline one, then hydrogen is present as the OH⁻ species. However this option is not given in your question.

Thus the answer is option C).

Answer:

61kg

Explanation:

Like all other hydrocarbons, the combustion of propane will yield water and carbon iv oxide.

Let’s write a complete and balanced chemical equation for this.

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

Now, we first convert the information into a mass.

We know that mass = density * volume

Let’s convert the volume here into ml since density is in mL. It must be remembered that 1000ml = 1L. Hence, 29.4L = 29,400ml

The mass is thus = 0.6921 * 29,400 = 20,347.74g

Now from the balanced equation, we can see that one mole of propane produced 3 moles of carbon iv oxide. This is the theoretical relation.

Let’s now calculate the actual number of moles. The number of moles of propane produced is the mass of propane divided by the molar mass of propane. The molar mass of propane is 3(12) + 8(1) = 36 + 8 = 44g/mol

The number of moles = 20,347.74/44 = 462.44

We simply multiply this number by 3 to get the actual number of moles of carbon dioxide produced. This is equals 1387.35 moles

Since we know this, we now calculate the mass of carbon iv oxide produced. The mass is equals the number of moles multiplied by the molar mass. The molar mass of carbon iv oxide is 44g/mol.

The mass produced is thus 1387.35 * 44 = 61,043g

To get the kilogram equivalent, we simply divide by 1000 = 61,043/1000 = 61kg

Answer:

The catabolic process of converting carbohydrates to CO2 requires oxidation of carbon.

Explanation:

There are multiple definitions of reduction-oxidation. There is one that explains it with respect to oxygen, the other with respect to hydrogen and another with respect to electrons. The relevant one here is the one that explains it in terms of hydrogen. Oxidation is the removal of hydrogen while reduction is the addition of hydrogen. During repiration, the carbon loses the hydrogens attached to it and is therefore oxidized. These hydrogens attach themselves to oxygen which means oxygen is reduced.

In spite of ethylene and benzene has double bonded carbon-carbon, Benzene is a cyclic molecule with a special property called aromaticity. Aromaticity gives to the ring electronic properties which makes harder for benzene being reactive as ethylene.

Benzene has the formula C6H6 being cyclic and aromatic because have conjugated double bonds and those double bonds has a property called resonance unusual elevated making the structure very stable. That can be seen under the conditions of reaction of alkene compounds in which benzene doesn't react, like bromination with Br2 in CCl4.

On the other hand, ethylene is an alkene which has the formula C2H4, the double bond carbon carbon is available to suffer the different reactions of alkenes being more reactive.

Ethylene is more reactive than benzene primarily due to ethylene's discrete, highly reactive double bond, compared to benzene's delocalized and stabilized double bonds within its aromatic ring. The stability of the delocalized electrons in the benzene ring reduces its reactivity, while ethylene's electron-rich double bond invites addition reactions.

The difference in reactivity between ethylene (C2H4) and benzene (C6H6) toward molecular bromine is primarily due to the structural differences in their carbon-carbon double bonds. Ethylene contains a discrete C=C double bond that is highly reactive because it can open up to allow additional atoms to attach to the carbon atoms (an addition reaction). In contrast, the C=C bonds in benzene are part of a delocalized electron system over the entire aromatic ring, giving it extraordinary stability. This delocalization makes benzene less reactive towards addition reactions, such as with bromine, because the addition would disrupt the stable aromatic system.

Benzene's unique stability is due to the alternating single and double C-C bonds, which form a resonance-stabilized structure. This stability is a concept commonly understood as aromaticity, which significantly reduces the chemical reactivity of benzene's double bonds compared to those in alkenes like ethylene. Although the C=C bond is nonpolar, the reactivity of ethylene is also influenced by the electron-rich region around the σ bond, making it more prone to reactions with nucleophiles.

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