We give 70 J as heat to a diatomic gas, which then expands at constant pressure. The gas molecules rotate but do not oscillate. By how much does the internal energy of the gas increase?

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

[tex](1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \,  (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}[/tex]

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

Final answer:

The molecular orbital diagram for noble gas compounds like KrF can be explained by the ground-state electron configuration and the expected bond strength of its cationic form.

Explanation:

Noble Gas Compounds Molecular Orbital Diagram:

The ground-state electron configuration of KrF is [Kr] 5s² 5p⁶. The cationic analog of KrF is likely to have a stronger bond due to the removal of an electron, leading to a decrease in repulsion between the nuclei and shared electrons.

Comparing Bond Strength in KrF vs. KrF⁺:

In KrF, the valence electrons fill the σ bonding and σ non-bonding orbitals, and the π bonding orbitals are empty.

When KrF loses an electron to become KrF⁺, it removes an electron from the σ non-bonding orbital (lone pair).

Losing this electron strengthens the bond because it removes electron density that opposes the bonding interaction in the σ bonding orbital. This is similar to losing a lone pair in other molecules.

Therefore, KrF⁺ is likely to have a stronger bond than KrF.

Answer: 48 Liters.

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 745 mmHg

[tex]P_2[/tex] = final pressure of gas =360 mmHg

[tex]V_1[/tex] = initial volume of gas = 26.5 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]23.3^oC=(273+23.3)K=296.3K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]-13.7^oC=273+(-13.7)K=259.3K[/tex]

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{745mmHg\times 26.5L}{296.3K}=\frac{360mmhg\times V_2}{259.3K}[/tex]

[tex]V_2=48L[/tex]

Therefore, the volume of the gas will be 48 Liters.

Answer : The initial rate of the reaction is, [tex]1.739\times 10^{-3}s^{-1}[/tex]

Explanation :

First we have to calculate the rate constant of the reaction.

Expression for rate law for first order kinetics is given by :

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,

k = rate constant  = ?

t = time taken for the process  = 44 s

[tex][A_o][/tex] = initial amount or concentration of the reactant  = 0.1108 M

[tex][A][/tex] = amount or concentration left time 44 s = 0.0554 M

Now put all the given values in above equation, we get:

[tex]k=\frac{2.303}{44}\log\frac{0.1108}{0.0554}[/tex]

[tex]k=0.0157[/tex]

Now we have to calculate the initial rate of the reaction.

Initial rate = K [A]

At t = 0, [tex][A]=[A_o][/tex]

Initial rate = 0.0157 × 0.1108 = [tex]1.739\times 10^{-3}s^{-1}[/tex]

Therefore, the initial rate of the reaction is, [tex]1.739\times 10^{-3}s^{-1}[/tex]

To find the molar mass of a gas collected over water, apply the ideal gas law to determine the moles of gas. Then, divide the mass of the gas by the number of moles.

To determine the molar mass of a gas from an experiment in a general chemistry laboratory, you first need to use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature in Kelvin.

The student collected a gas with a volume of 265 mL, which is equivalent to 0.265 L, at a pressure of 753 torr. Since 1 atm is equal to 760 torr, the pressure in atmospheres is 753 torr / 760 torr/atm = 0.9911 atm. The temperature must be converted from Celsius to Kelvin; thus, 27 °C is equal to 300.15 K (27 + 273.15 = 300.15 K).

To solve for n (the number of moles), you rearrange the ideal gas law to n = PV / RT. With the previously mentioned values and the gas constant R as 0.0821 L·atm/K·mol, n = (0.9911 atm × 0.265 L) / (0.0821 L·atm/K·mol × 300.15 K). After calculation, the number of moles of gas n is found.

Once n is calculated, the molar mass (M) can be found using the formula M = mass of gas (g) / number of moles (mol). Therefore, with the mass of the gas being 0.472 g, we calculate M = 0.472 g / n moles. By plugging in the value of n from our previous calculation, we can determine the molar mass.

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Final answer:

To calculate the molar mass of a gas, the ideal gas law is used with the given pressure, volume, and temperature. The molar mass is found to be approximately 44.22 g/mol after doing the conversions and calculations.

Explanation:

The student's question asks about determining the molar mass of a collected gas sample. To find out the molar mass of the gas, we use the ideal gas law formula (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature. Here's the step-by-step solution:

Convert the given pressure from torr to atm (753 torr / 760 torr/atm).

Convert the volume from mL to L (265 mL / 1000 mL/L).

Convert the temperature from Celsius to Kelvin (27 °C + 273.15).

Solve the ideal gas law equation for the number of moles (n).

Use the mass of the gas and the number of moles calculated to find the molar mass (molar mass = mass/n).

The detailed calculations show:

Solve for n: PV = nRT → n = PV / RT = (0.991 atm × 0.265 L) / (0.0821 L·atm/(mol·K) × 300.15 K) ≈ 0.01067 mol.

Finally, determine the molar mass: molar mass = mass/n = 0.472 g / 0.01067 mol ≈ 44.22 g/mol.

Hence, the molar mass of the gas is approximately 44.22 g/mol.

Answer:

Concentration of hydrogen ion, [tex][H^+]=5.0118*10^{-6} M[/tex]

Explanation:

pH is defined as the negative logarithm of hydrogen ion's concentration.

The lower the value of pH, the higher the acidic the solution is.

The formula for pH can be written as:

[tex]pH=-log[H^+][/tex]

Given,

pH of the saliva of Marco = 5.3

To calculate: Hydrogen ion concentration in the saliva

Thus, applying in the formula as:

[tex]pH=-log[H^+][/tex]

[tex]5.3=-log[H^+][/tex]

So,

[tex]log[H^+]=-5.3[/tex]

[tex][H^+]=10^{(-5.3)}[/tex]

[tex][H^+]=5.0118*10^{-6} M[/tex]

The hydrogen ion concentration of Marco's saliva is approximately 5.01 x 10^-6 M.

The pH scale ranges from 0 to 14, where anything below 7 is acidic and above 7 is alkaline. The hydrogen ion (H+) concentration of a solution can be determined using the pH value. A change of one unit on the pH scale represents a ten-fold change in the concentration of H+ ions. Given that Marco's saliva has a pH of 5.3, we can calculate the hydrogen ion concentration as follows:

Therefore, the hydrogen ion concentration of Marco's saliva is approximately 5.01 x 10⁻⁶M.

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Answer:

Explanation:

People who use marble to carve monuments and sculptures are always concerned because it can easily chemically weather.

Marble is a metamorphic rock derieved from limestone. The minerals that makes up limestone are calcite and dolomite. When a metamorphic transformation occurs, the rock is subjected to intense temperature and pressure. Marble is made of calcite and dolomite minerals.

Rain water dissolves carbon dioxide to form weak carbonic acid. This weak carbonic acid can easily dissolve marble thereby defacing the monument and the sculpture.

Acid rain can cause damage to marble monuments and sculptures because it reacts with the carbonate minerals in marble, leading to their dissolution and degradation over time.

People who care for monuments and sculptures made of marble are concerned about acid rain because marble is primarily composed of carbonate minerals such as calcite (CaCO3) and dolomite (Ca,Mg(CO3)2). When acid rain falls on marble, the weak acid reacts with the carbonate minerals, causing them to dissolve. This can lead to the degradation and erosion of the marble over time.

The reaction between acid rain and marble can be represented by the following equation: CaCO3(s) + H₂SO4 (aq) → CaSO4(s) + H₂O(1) + CO2(g). The reaction produces calcium sulfate, water, and carbon dioxide gas, which contribute to the dissolution of the marble.

Therefore, the concern for acid rain arises from the fact that it can cause significant damage to marble monuments and sculptures, affecting their aesthetic value and structural integrity.

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Hey there!:

* S,and Ne are non-metals.

* Si is a metalloid.Generally metalloids acts as semiconductor.

* Ag and Hg are metals because metals exhibits lustre property,  melleable and ductility  properties.

Hope this helps!

Explanation:

Metals are the substances which lose electrons to attain stability and hence they form cations.

Metals are good conductors of heat and electricity and generally they are solid at room temperature. Metals have a shiny surface and they are also malleable and ductile.

For example, Ag: Silver Shiny solid at room temperature that is also a good conductor of heat and electricity. Hence, silver is a metal.

Non-metals are the substances which gain electrons to gain stability and hence they form anions.

Non-metals are bad conductors of heat and electricity. They are brittle and non-shiny in nature.  

For example, S: Sulfur Powdery yellow solid is a non-metal.

Metalloids are the substances which show properties of both metals and non-metals.

Metalloids are moderately able to conduct heat and electricity. Generally metalloids are solid at room temperature. They are also able to react with other molecules.

For example, Si: Silicon Semiconductor is a metalloid.

Hence, the given substances are classified as follows.

Answer:

[tex]\boxed{\text{0.50 mol/L}}[/tex]

Explanation:

The balanced equation is

2COF₂ ⇌ CO₂+CF₄; Kc = 9.00

1. Set up an ICE table

[tex]\begin{array}{ccccc}\rm 2COF_{2} & \, \rightleftharpoons \, & \rm CO_{2} & +&\rm CF_{4}\\2.00& & 0& & 0\\-x& & +x & & +x\\2.00 - x& & x & &x \\\end{array}[/tex]

2. Solve for x

[tex]K_{c} = \dfrac{[\rm CO][ \rm CF_{4}]}{[\rm COF_{2}]^{2}} = 9.00\\\\\begin{array}{rcl}\dfrac{x^{2}}{(2.00 - x)^{2}} & = & 9.00\\\dfrac{x}{2.00 - x} & = & 3.00\\x & = &3.00(2.00 - x)\\x & = & 6.00 - 3.00x\\4.00x & = & 6.00\\x & = & \mathbf{1.50}\\\end{array}[/tex]

3. Calculate the equilibrium concentration of COF₂

c = (2.00 - x) mol·L⁻¹ = (2.00 - 1.50) mol·L⁻¹ = 0.50 mol

[tex]\text{The equilibrium concentration of COF$_{2}$ at equilibrium is $\boxed{\textbf{0.50 mol/L}}$}[/tex]

Check:

[tex]\begin{array}{rcl}\dfrac{1.50^{2}}{0.50^{2}} & = & 9.00\\\\\dfrac{2.25}{0.25}& = & 9.00\\\\9.00 & = & 9.00\\\end{array}[/tex]

OK.

The equilibrium concentration of COF2 can be found by solving a quadratic equation derived from the equilibrium constant expression and initial concentration, and then subtracting the amount that reacted from the initial concentration.

The student asked about the equilibrium concentration of carbonyl fluoride, COF2, in a specific chemical reaction where it is converted to carbon dioxide and carbon tetrafluoride. To determine the concentration of COF2 that remains at equilibrium starting with an initial concentration of 2.00 M, we must use the equilibrium constant (Kc) value provided, which is 9.00.

The equilibrium expression for the reaction 2COF2(g) ⇌ CO2(g) + CF4(g) is Kc = [CO2][CF4] / [COF2]2. Let the change in concentration of COF2 at equilibrium be 'x'. Thus, at equilibrium, we'll have [COF2] = 2.00 - 2x, [CO2] = x, and [CF4] = x. Plugging these into the equilibrium expression, we have 9.00 = x2 / (2.00 - 2x)2. Solving for x will give us the change in concentration of COF2, from which we can determine the equilibrium concentration of COF2 by subtracting 'x' from the initial concentration.

Therefore, to find the equilibrium concentration of COF2, one would need to solve the quadratic equation resulting from the substitution of equilibrium concentrations in terms of 'x' into the Kc expression, then calculate 2.00 M - 2x.

Answer: [Ni(en)3] > [Ni(en)(H2O)4] > [Fe(NH3)6] > [FeF6]

Explanation:

Generally chelating ligands stabilize the complex more than non chelated ligands and more the no of chelated ligands more the stability.

Here en (ethylenediamine) is a chelated ligand and stabilze the complex more by chelation.

And Strong field ligand (NH3) also stabilze the complex more than weak field ligand (F).

Hence F containing complex is least stable.

The rank of the coordination compounds in the increasing order of the stability is [tex]\rm [Ni(en)_3] > [Ni(en)(H_2O)_4] > [Fe(NH_3)_6] > [FeF_6][/tex].

The coordination compound are the complexes in which the central metal atom has been bounded by the nonmetal to the complexes through the chemical bond.

The stability of the coordination complexes is attributed to the number of chelating agents to the central atoms.

The increased number of atoms to the central atoms adds to the stability of the complex. Thus, the increasing order of the stability is [tex]\rm [Ni(en)_3] > [Ni(en)(H_2O)_4] > [Fe(NH_3)_6] > [FeF_6][/tex].

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Answer:

The correct option is d. Membranes, the brain and the nervous system.

Explanation:

The sphingolipids are a class of lipids which contain a sphingosine amino alcohol.

The main function of sphingolipid is the protection of the cells against harmful environment by forming a chemically resistant and mechanically stable membrane.

Sphingolipids are present in abundance in the brain and in the central nervous system where they are the important constituents of the plasma membranes. They are also important for the proper functioning, development and maintenance of the nervous system.

Therefore, Sphingolipids are particularly present in the membranes, the brain and the nervous system.

Calcium carbide (CaC₂) reacts with water to produce acetylene (C₂H₂): CaC₂ (s) + 2 H₂O (g) → Ca (OH)₂ (s) + C₂H₂ (g) Production of 6.5 g of C₂H₂ (MW = 26.036 g/mol) requires consumption of 2.3 gm. Hence, option B is the correct option.

The balanced chemical equation given is:

CaC₂ (s) + 2 H₂O (g) → Ca (OH)₂ (s) +C₂H₂ (g)

The molar ratio between CaC₂ and C₂H₂ is 1:1, and the molar ratio between H₂O and C₂H₂ is 2:1.

To determine the mass of H₂O required to produce 6.5 g of C₂H₂, one can use the following steps:

The number of moles of C₂H₂:

  Moles of C₂H₂ = Mass / Molar mass = 6.5 g / 26.036 g/mol

The molar ratio between  H₂O and C₂H₂ is 2:1, the number of moles of  H₂O required is half of the moles of C₂H₂:

  Moles of  H₂O = Moles of C₂H₂ / 2

The mass of  H₂O using its moles and molar mass:

  Mass of  H₂O = Moles of  H₂O × Molar mass of H₂O

1. Moles of C₂H₂ = 6.5 g / 26.036 g/mol ≈ 0.2494 mol

2. Moles of  H₂O = 0.2494 mol / 2 ≈ 0.1247 mol

3. Mass of  H₂O = 0.1247 mol × 18.016 g/mol ≈ 2.244 g

So, the consumption of H₂O required to produce 6.5 g of C₂H₂ is approximately 2.244 g.

The closest option to this value is 2.3 g, so the correct answer is:

B. 2.3

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Answer:

Explanation:

Writing the equation properly:

         (CH₃)₃N + H₂O ⇄ (CH₃)₃NH⁺ + OH⁻

         HNO₃ + H₂O  ⇄ H₂O + NO₃⁻

The bronsted-lowry theory defines an acid as a proton donor and a base as a proton acceptor.

In a bronsted-lowry acid-base reaction, the original acid gives up its proton and becomes a conjugate base. Also, the original base accepts a proton and becomes a conjugate acid. For every acid, there is a conjugate base and for every base there is a conjugate acid.

What differentiates an acid from its conjugate base is a proton. The difference between a base and its conjugate acid is a proton.

    (CH₃)₃N          +          H₂O          ⇄         (CH₃)₃NH⁺       +         OH⁻

Bronsted-lowry base     acid                conjugate acid        conjugate base

    HNO₃                       +      H₂O       ⇄           H₃O⁺        +                NO₃⁻

Bronsted-lowry acid           base             conjugate acid      conjugate base

In the reaction (a), (CH3)3N is the Brønsted-Lowry base, H2O is the Brønsted-Lowry acid, (CH3)3NH+ is the conjugate acid, and OH- is the conjugate base. In reaction (b), HNO3 is the Brønsted-Lowry acid, H2O is the Brønsted-Lowry base, H3O+ is the conjugate acid, and NO3- is the conjugate base.

In the reaction (a) (CH3)3N(aq) + H2O(l) <-> (CH3)3NH+(aq) + OH-(aq), (CH3)3N is the Brønsted-Lowry base, H2O is the Brønsted-Lowry acid, (CH3)3NH+ is the conjugate acid, and OH- is the conjugate base.

In the reaction (b) HNO3(aq) + H2O(l) <-> H3O+(aq) + NO3-(aq), HNO3 is the Brønsted-Lowry acid, H2O is the Brønsted-Lowry base, H3O+ is the conjugate acid, and NO3- is the conjugate base.

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When an electron jumps from higher energy level to a lower energy level it radiates or gives out energy in the form of radiation.

Electrons present in an atom revolve in different orbits which are stationary states and are also called as energy levels. The energy levels are numbered as integers which are also called  as principal quantum numbers.

Energy of the stationary state  is given as E= -R[tex]_h[/tex] 1/n² where R[tex]_h[/tex] is  the Rydberg's constant. When an electron is excited, and it moves from lower to higher energy levels there is absorption of energy, while when it moves from higher energy level to lower energy level it radiates or gives out energy in the form of radiation.

They can also be defined as the distances  between electron and nucleus of an atom . Electrons present in K energy level have least energy .Energy level diagrams are studied to understand nature of bonding , placement of electrons in orbits and  and elemental behavior under certain conditions.

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Answer:

100.71 cm H2O

Explanation:

In a fluid column barometer, the height of the fluid column is proportional to the pressure. The pressure is by definition:

[tex]P=\frac{F}{A}[/tex], where F is a force and A is the area.

In a column barometer the force is given by the weight of the fluid:

[tex]F=m*g[/tex], and the mass may be expressed as [tex]m=p*V[/tex], where [tex]p[/tex] is the density and V is the volume.

Replacing this in the pressure definition:

[tex]P=\frac{pVg}{A}[/tex]

In a constant cross section area column, the volume may be calculated as:

[tex]V=A*h[/tex] where A is the area and h the height. Replacing this in the previous equation:

[tex]P=\frac{pAhg}{A}=pgh[/tex]

Different columns may be over the same pressure, so:

[tex]P_{w}=P_{Hg}\\ p_{w}h_{w}g=p_{Hg}h_{Hg}g\\[/tex]

Dividing each part for gravity constant:

[tex]p_{w}h_{w}=p_{Hg}h_{Hg}[/tex]

And isolating hw:

[tex]h_{w}=\frac{p_{Hg}h_{Hg}}{p_{w} } \\h_{w}=\frac{13.5*746}{1.00}=10071[/tex] mm

It is equal to 1007,1 cm.

Answer:

The heat of reaction for the combustion of a mole of Ti in this calorimeter is [tex]7.769\times 10^4 kJ/mol[/tex].

Explanation:

[tex]Ti (s) + O_2 (g) \rightarrow TiO_2 (s)[/tex]

Moles of titanium =[tex]\frac{2.060 g}{47.867 g/mol}=0.04303 mol[/tex]

Heat absorbed by the bomb caloriometer on combustion of 0.04303 mol of titanium be Q

The heat capacity of the bomb caloriometer  =c = 9.84 kJ/K

Change in temperature of the bomb caloriometer :

=ΔT=91.60 °C-25.00 °C=66.6 °C = 339.75 K

Q = c × ΔT

[tex]Q= 9.84 kJ/K\times 339.75 K=3,343.14 kJ[/tex]

3,343.14 kJ of heat energy was released when 0.04303 moles of titanium undergone combustion.

So for 1 mol of titanium:

[tex]\frac{3,343.14 kJ}{0.04303 moles}=77,693.237 kJ/mol=7.769\times 10^4 kJ/mol[/tex]

The heat of reaction for the combustion of a mole of Ti in this calorimeter is [tex]7.769\times 10^4 kJ/mol[/tex].

Answer:

a. NaCl

b. Dissociation (the compound breaking down into ions)

Answer : The concentration of water vapor in a sample of air is, [tex]1.2\times 10^3ppm[/tex]

Explanation : Given,

Partial pressure of water = 0.91 torr

The total pressure of air = 735 torr

Parts per million (ppm) : It is defined as the mass of a solute present in one million [tex](10^6)[/tex] parts by the mass of the solution.

Now we have to calculate the concentration of water vapor in a sample of air.

[tex]\text{Concentration of water vapor}=\frac{\text{Partial pressure of water}}{\text{Total pressure of air}}\times 10^6[/tex]

Now put all the given values in this formula, we get  concentration of water vapor in a sample of air.

[tex]\text{Concentration of water vapor}=\frac{0.91torr}{735torr}\times 10^6=2.1\times 10^3ppm[/tex]

Therefore, the concentration of water vapor in a sample of air is, [tex]1.2\times 10^3ppm[/tex]

The concentration of water vapor in a sample of air that has a partial pressure of water of 0.91 torr and a total pressure of air of 735 torr can be calculated using Dalton's law of partial pressures. The concentration comes out to be approximately 1200 ppm.

To calculate the concentration of water vapor in a sample of air in parts per million (ppm), we first need to understand the relationship between partial pressures and total pressure in a mixture of gases. According to Dalton's law of partial pressures, the total pressure of a gas mixture is equal to the sum of the partial pressures of its components. Here, the partial pressure of water vapor is 0.91 torr and the total pressure of air is 735 torr.

To find the concentration of water vapor, divide the partial pressure of the water vapor by the total pressure of the air, and then multiply by 1,000,000 to convert the value to ppm. So, (0.91 torr / 735 torr) x 1,000,000 = 1238.7755 ppm. This value needs to be rounded off to the appropriate number of significant digits which in this case is two, so the concentration of water vapor in the air would be 1200 ppm in this scenario.

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Answer : The initial temperature of system 2 is, [tex]19.415^oC[/tex]

Explanation :

In this problem we assumed that the total energy of the combined systems remains constant.

[tex]-q_1=q_2[/tex]

[tex]m\times c_1\times (T_f-T_1)=-m\times c_2\times (T_f-T_2)[/tex]

The mass remains same.

where,

[tex]C_1[/tex] = heat capacity of system 1 = 19.9 J/mole.K

[tex]C_2[/tex] = heat capacity of system 2 = 28.2 J/mole.K

[tex]T_f[/tex] = final temperature of system = [tex]30^oC=273+30=303K[/tex]

[tex]T_1[/tex] = initial temperature of system 1 = [tex]45^oC=273+45=318K[/tex]

[tex]T_2[/tex] = initial temperature of system 2 = ?

Now put all the given values in the above formula, we get

[tex]-19.9J/mole.K\times (303-318)K=28.2J/mole.K\times (303-T_2)K[/tex]

[tex]T_2=292.415K[/tex]

[tex]T_2=292.415-273=19.415^oC[/tex]

Therefore, the initial temperature of system 2 is, [tex]19.415^oC[/tex]

Answer : The heat of combustion is, 15642 J

Solution :

Formula used :

[tex]q=m\times c\times \Delta T[/tex]

where,

[tex]q[/tex] = heat of combustion = ?

[tex]m[/tex] = mass of fructose = 1.0 g

[tex]c[/tex] = heat capacity of the calorimteter  = [tex]9.90KJ/g^oC=9900J/g^oC[/tex]

conversion used : 1 KJ = 1000 J

[tex]\Delta T[/tex] = change in temperature = [tex]1.58^oC[/tex]

Now put all the given values in the above formula, we get

[tex]q=1.0g\times 9900J/g^oC\times 1.58^oC[/tex]

[tex]q=15642J[/tex]

Therefore, the heat of combustion is, 15642 J

Answer:

q combustion = -15.6 kJ (exothermic)

Explanation:

Bomb calorimeter questions are interesting because we are usually given the heat capacity of the bomb calorimeter not the specific heat capacity (which includes grams in the unit). Therefore we don't actually need to include the mass of the fructose as long as we know how much the temperature of the calorimeter changed (ΔT) and the heat capacity of the calorimeter.

We know that:

-q (combustion) = q (calorimeter) and we have enough information to calculate q (calorimeter):

   q (calorimeter) = (Heat Capacity)(Change in Temp.)

⇒ q(cal) = (9.90 kJ/°C)(1.58 °C) = 15.642 kJ

                                                  = 15.6 kJ (3 sig figs)

Since -q (combustion) = q (calorimeter), then:  

q (combustion) = -15.6 kJ (negative sign simply means heat released)

Answer:

D)the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A

Explanation:

As mentioned the possible chemical formula for a compound can be predicted from electron configurations of the elements.

The first element 'A' has electronic configuration as "[core]ns²

It means it has two valence electrons and it can give those valence electrons to attain noble gas or full filled stability.

So the element is most likely to form a dipositive ion.

Now for element B the configuration is :[Core]ns²np⁵

thus it has seven valence electrons and it will most likely to accept an electron to gain full filled stability. So B will form a mono negative ion.

The possible formula will be AB₂ [Like MgCl₂]

So the false statements are:

D)the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A : the first ionization energy of A will be lower than B as B cannot give an electron easily.

The other statements are correct:

A) element A serves as the reducing agent in the reaction to form AxBy : As A will lose electrons so it will acts as reducing agent and itself gets oxidized

B) yes it is whole number as calculated above.

C) x + y = 1+2 = 3

E) Already calculated that B has oxidation state of -1.

The false statement is that; "the first ionization energy (IE1) of B is most probably lower than the first ionization energy (IE1) of A"

The first ionization energy is the energy required to remove the first electron from an atom. This is a periodic trend that increases across the period but decreases down the group.

Looking at the electron configurations of A and B it is clear that A is an alkaline earth metal while B is a halogen. As such, A is a reducing agent and B will have an oxidation state of -1 as expected of halogens.

The compound will have the formula AX2 hence x + y = 3. However, since ionization energy increases across the period, the first ionization energy of B must certainly be greater than that of A.

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Answer: Option (a) is the correct answer.

Explanation:

A chemical bonding in a substance occurs when there is exchange of electrons between the atoms. These electrons are present in the outermost orbital are actually away from the nucleus of the atom.

Due to which there is no force of attraction between nucleus or valence electrons. Hence, it is easy for the outermost orbital to lose electron.

Whereas nucleus of an atom contains neutrons and protons. And, changes in the nucleus of an atom leads to nuclear reactions.

Therefore, we can conclude that the configuration of the outermost orbital determines chemical bonding characteristics.

Explanation:

Gram is unit which is used to expressed mass of a substance. Many other units can be used to express the mass of a substance. All the units are inter changeable.

Conversion factors used to convert the units are:

(a) Gram to mega gram

[tex]1g=1\times 10^{-6}Mg[/tex]

(b) Gram to kilo gram

[tex]1g=1\times 10^{-3}kg[/tex]

(c) Gram to deci gram

[tex]1g=1\times 10^{1}1dg[/tex]

(d) Gram to centi gram

[tex]1g=1\times 10^{2}cg[/tex]

(e) Gram to mili gram

[tex]1g=1\times 10^{3}mg[/tex]

(f) Gram to micro gram

[tex]1g=1\times 10^{6}\mu g[/tex]

(g) Gram to nano gram

[tex]1g=1\times 10^{9}ng[/tex]

(h) Gram to pico gram

[tex]1g=1\times 10^{12}pg[/tex]

Hence, the conversion factors for gram equivalents are given above.

To convert 1 gram to different metric prefixes in scientific notation, we use the powers of ten specific to each prefix, ensuring the numeric value is between 1 and 1000. The result is megagrams (1e-6), kilograms (1e-3), decigrams (1e1), centigrams (1e2), milligrams (1e3), micrograms (1e6), nanograms (1e9), and picograms (1e12).

The task involves converting the mass of 1 gram into various metric prefixes and expressing them in scientific notation, where the numeric value is greater than one but less than 1000. The prefixes include mega, kilo, deci, centi, milli, micro, nano, and pico. Below are the conversions using the appropriate powers of ten to express 1 gram in the specified units:

Answer: The spontaneous cell reaction having smallest [tex]E^o[/tex] is [tex]I_2+Cu\rightarrow Cu^{2+}+2I^-[/tex]

Explanation:

We are given:

[tex]E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V[/tex]

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

[tex]E^o_{cell}=E^o_{oxidation}+E^o_{reduction}[/tex]

The combination of the cell reactions follows:

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

[tex]I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)[/tex]

Oxidation half reaction:  [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-[/tex]   [tex]E^o_{oxidation}=0.45V[/tex]

Reduction half reaction:  [tex]I_2(s)+2e^-\rightarrow 2I_-(aq.)[/tex]   [tex]E^o_{reduction}=0.54V[/tex]

[tex]E^o_{cell}=0.45+0.54=0.99V[/tex]

Thus, this cell will not give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

[tex]I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)[/tex]

Oxidation half reaction:  [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-[/tex]   [tex]E^o_{oxidation}=-0.34V[/tex]

Reduction half reaction: [tex]I_2(s)+2e^-\rightarrow 2I_-(aq.)[/tex]   [tex]E^o_{reduction}=0.54V[/tex]

[tex]E^o_{cell}=-0.34+0.54=0.20V[/tex]

Thus, this cell will give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

[tex]Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)[/tex]

Oxidation half reaction:  [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-[/tex]   [tex]E^o_{oxidation}=0.45V[/tex]

Reduction half reaction:  [tex]Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)[/tex]   [tex]E^o_{reduction}=0.34V[/tex]

[tex]E^o_{cell}=0.45+0.34=0.79V[/tex]

Thus, this cell will not give the spontaneous cell reaction with smallest [tex]E^o_{cell}[/tex]

Hence, the spontaneous cell reaction having smallest [tex]E^o[/tex] is [tex]I_2+Cu\rightarrow Cu^{2+}+2I^-[/tex]

Answer:

The expected ratio of half-lives for a reaction will be 5:1.

Explanation:

Integrated rate law  for zero order kinetics is given as:

[tex]k=\frac{1}{t}([A_o]-[A])[/tex]

[tex][A_o][/tex] = initial concentration

[A]=concentration at time t

k = rate constant

if, [tex][A]=\frac{1}{2}[A_o][/tex]

[tex]t=t_{\frac{1}{2}}[/tex], the equation (1) becomes:

[tex]t_{\frac{1}{2}}=\frac{[A_o]}{2k}[/tex]

Half life when concentration was 0.05 M=[tex]t_{\frac{1}{2}}[/tex]

Half life when concentration was 0.01 M=[tex]t_{\frac{1}{2}}'[/tex]

Ratio of half-lives will be:

[tex]\frac{t_{\frac{1}{2}}}{t_{\frac{1}{2}}'}=\frac{\frac{[0.05 M]}{2k}}{\frac{[0.01 M]}{2k}}=\frac{5}{1}[/tex]

The expected ratio of half-lives for a reaction will be 5:1.

Answer: The mass of sodium carbonate reacted is 0.205 g.

Explanation:

To calculate the moles of a solute, we use the equation:

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

We are given:

Volume of nitric acid = 25 mL = 0.025 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.155 moles/ L

Putting values in above equation, we get:

[tex]0.155mol/L=\frac{\text{Moles of nitric acid}}{0.025L}\\\\\text{Moles of nitric acid}=0.003875mol[/tex]

For the given chemical reaction:

[tex]Na_2CO_3+2HNO_3\rightarrow H_2CO_3+2NaNO_3[/tex]

By Stoichiometry of the reaction:

If 2 moles of nitric acid reacts with 1 mole of sodium carbonate.

So, 0.003875 moles of nitric acid will react with [tex]\frac{1}{2}\times 0.003875=0.0019375moles[/tex] of sodium carbonate.

To calculate the mass of sodium carbonate, we use the equation:

Molar mass of sodium carbonate = 105.98 g/mol

Moles of sodium carbonate = 0.0019375 moles

Putting values in above equation, we get:

[tex]0.0019375mol=\frac{\text{Mass of sodium carbonate}}{105.98g/mol}\\\\\text{Mass of sodium carbonate}=0.205g[/tex]

Hence, the mass of sodium carbonate reacted is 0.205 g.

To find out how many grams of sodium carbonate are required to completely react with 25.0 mL of 0.155 M nitric acid, we used stoichiometry. First, we calculated the number of moles of nitric acid, then used the balanced equation to find out the moles of sodium carbonate, which is equal to the moles of nitric acid. Finally, we multiplied this by the molar mass of sodium carbonate to get the weight in grams, which is 0.411 grams.

The subject of this question is related to a chemical reaction between sodium carbonate and nitric acid. In order to determine the amount of sodium carbonate required to completely react with the given amount of nitric acid, we need to use stoichiometry.

Here are the steps:

So, we need 0.411 grams of sodium carbonate to completely react with 25.0 mL of 0.155 M nitric acid.

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Answer:

1. Multiply the equivalence point volume by the NaOH concentration.

2. Divide that result by the volume taken from the formic acid solution for titration.

Explanation:

The reaction between NaOH and formic acid is one to one, it means that a NaOH mol reacts for each mole of formic acid in the solution. So the number of moles who react in the NaOH titrator is the same number of moles that were in the volume taken of the formic acid solution for titration. So, the number of formic acid moles can be calculated as:

[tex]n_{CHOOH}=V_{NaOH}*C_{NaOH}[/tex]

Then, divide that number of moles by the volume taken from the formic acid solution:

[tex]C_{CHOOH}=\frac{V_{NaOH}*C_{NaOH}}{V_{CHOOH}}[/tex]

The concentrations must be in Molar units (Mol/Liter) and the volume in liters, so:

[tex]C_{CHOOH}=\frac{0.02666*C_{NaOH}}{V_{CHOOH}}[/tex]

The concentration of the formic acid solution can be determined using the titration curve provided by the student.

To determine the concentration of the formic acid solution using the student's titration curve, follow these steps:

1. Identify the equivalence point on the titration curve. This is the point where the pH changes rapidly and corresponds to the addition of 26.66 mL of NaOH.

2. At the equivalence point, the moles of NaOH added are equal to the moles of formic acid in the initial solution.

3. Calculate the moles of NaOH added using the concentration of the NaOH solution[tex](C_NaOH)[/tex] and the volume of NaOH added [tex](V_NaOH)[/tex] at the equivalence point:

[tex]\[ \text{moles of NaOH} = C_{\text{NaOH}} \times V_{\text{NaOH}} \][/tex]

4. Since the stoichiometry of the reaction between formic acid (HCOOH) and NaOH is 1:1, the moles of NaOH added will be equal to the moles of formic acid (HCOOH) in the initial solution.

5. Calculate the concentration of the formic acid solution using the moles of formic acid and the initial volume of the formic acid solution [tex](V_HCOOH):[/tex]

[tex]\[ \text{Concentration of formic acid} = \frac{\text{moles of HCOOH}}{V_{\text{HCOOH}}} \][/tex]

6. The concentration of formic acid can be expressed as:

[tex]\[ C_{\text{HCOOH}} = \frac{C_{\text{NaOH}} \times V_{\text{NaOH}}}{V_{\text{HCOOH}}} \][/tex]

7. Plug in the known values to find the concentration of formic acid:

[tex]\[ C_{\text{HCOOH}} = \frac{C_{\text{NaOH}} \times 26.66 \text{ mL}}{25.00 \text{ mL}} \][/tex]

8. If the concentration of NaOH is not given, it can be determined from the pH of the solution before the equivalence point where the pH starts to increase sharply. At this point, the concentration of [tex]OH^-[/tex] ions can be calculated, and since[tex][OH^-] = \(\frac{K_w}{[\text{H}_3\text{O}^+]}\),[/tex] the concentration of NaOH can be found.

9. Once the concentration of NaOH is known, use it in the equation from step 7 to find the concentration of formic acid.

Answer:

Kgoal = 8.15 X 10⁻²⁸

Explanation:

The goal reaction is

4PCl₅(g)⇌P₄(s)+10Cl₂(g)   Kgoal = ?

The given reactions are

P₄(s)+6Cl₂(g)⇌4PCl₃(g), K₁ = 2.00×10¹⁹

PCl₅(g)⇌PCl₃(g)+Cl₂(g), K₂ = 1.13×10⁻²

We can obtain the goal equation by

i) multiplying the second equation with four

ii) subtracting the equation one from above equation

We know that

i) If we multiply an equation with a number the equilbirium constant increases that times (we have to raise the power of equilibrium constant by that number)

ii) if we subtract two equations the equilibrium constants are divided

Kgoal = (K₂)⁴ / K₁

Kgoal = 8.15 X 10⁻²⁸

To determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)⇌P4(s)+10Cl2(g), we can use the given equilibrium constants for the two reactions involving the same species. By multiplying the equations for the given reactions, we can obtain the equation for the desired reaction and calculate the value of Kgoal. The value of Kgoal is found to be 2.26×1017.

To determine the equilibrium constant, Kgoal, for the reaction 4PCl5(g)↔P4(s)+10Cl2(g), we can use the given equilibrium constants for the two reactions involving the same species. We start by writing the equation for the desired reaction as the sum of the two given reactions:

By multiplying these two equations together, we can obtain the equation for the desired reaction and calculate the value of Kgoal. Multiplying the equations gives:
4PCl5(g)+24Cl2(g)↔16PCl3(g)+4P(s)+(4)(10)Cl2(g)

Since Kgoal is the product of the equilibrium constants for the forward and reverse reactions, we can calculate it as:

Kgoal = K1 × K2 = (2.00×1019)(1.13×10-2) = 2.26×1017

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Answer:

1.08 atm is the total pressure in the flask at equilibrium.

The value of the [tex]K_p=1.44[/tex].

Explanation:

Partial pressure of A at equilibrium = [tex]p_a=0.37 atm[/tex]

Partial pressure of B at equilibrium = [tex]p_b=2x[/tex]

                           A(g)      ⇄    2B(g)

at t=0                 0.73 atm           0

At equilibrium  (0.73- x)         2x

[tex]p_a=0.37 atm=(0.73- x)[/tex]

x = 0.36 atm

[tex]p_b=2x=2\times 036 atm=0.72 atm[/tex]

Total pressure in the flask at equilibrium = P

[tex]P=p_a+p_b=0.36 atm+0.72 atm = 1.08 atm[/tex]

The expression of equilibrium constant will be given as:

[tex]K_p=\frac{p_{b}^2}{p_a}=\frac{(0.72 atm)^2}{0.36 atm}=1.44[/tex]

The partial pressure is the pressure exerted by individual gases in the volume it is present. 1.08 atm is the total pressure and 1.44 is the equilibrium constant.

The total static and velocity pressure of a system is called total pressure. While the equilibrium constant is the proportion of the partial pressure of products and reactants.

Given,

From the reaction,

[tex]\begin{aligned}p_{a} &= 0.37 \rm\; atm \\\\\\&= (0.73 - x)\\\\\rm x &= 0.36 \;\rm atm\end{aligned}[/tex]

Solving the partial pressure of B:

[tex]\begin{aligned}p_{b} &= 2\rm x\\\\&= 2 \times 0.36\\\\&= 0.72\;\rm atm\end{aligned}[/tex]

Total pressure (P) in the flask will be:

[tex]\begin{aligned} \rm P &= p_{a}+ p_{b}\\\\&=0.36 + 0.72\\\\&= 1.08 \;\rm atm\end{aligned}[/tex]

Also, the equilibrium constant  will be calculated as:

[tex]\begin{aligned} K_{p} &= \dfrac{p_{b}^{2}}{p_{a}}\\\\&=\dfrac{(0.72)^{2}}{0.3.6}\\\\&=1.44 \end{aligned}[/tex]

Therefore, A. 1.08 atm is the total pressure and B. 1.44 is the equilibrium constant.

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Answer:

Explanation:

It's false, but not because it is not a decomposition equation. It does resemble one.

As near as I can tell, it should be

2H2O >>> 2H2 + O2

Hey there!:

the chemical reaction between 2-iodoctane and sodium nitrite is as follows:

Answer:

On the attached picture.

Explanation:

Hello,

In the case, the reaction between 2-iodooctane and sodium nitrite, leads to the formation of an alkyl nitrite and a nitro alkane as shown on the attached picture. Once the reaction began, the salt breaks and the sodium bonds with the iodine from the 2-iodooctane to form sodium iodide, in such a way, a free radical in the second carbon is formed so the NO₂ could bond both as a nitrite and as a nitro radical; therefore, the formed species are octyl 2-nitrite and 2-nitrooctane.

Best regards.