Predict the products of the decomposition of lithium nitride, Li3N.

Answer:

335 Joules kJ of heat will be released

Explanation:

Given the balanced equation:

8 Al (s) + 3 Fe3O4 (s) ----------> 4 Al2O3 (s) + 9 Fe(s),

ΔH = -3350*KJ/mol rxn

This is the heat  released when 8 moles of Al react with 3 mol Fe3O4.

We then need to calculate the moles of reactants, verify if there is a limiting reagent and proceed to answer the question based on the soichiometry of the reaction.

Atomic weight Al = 26.98 g/mol  Molecular Weight Fe3O4 = 231.53 g/mol

mol Al = 47.6 g/26.98 g/mol = 1.76 mol

mol Fe3O4 = 69.12 g/ 231.53 g/mol = 0.30 mol

Limiting reagent calculation:

8 mol Al / 3 mol Fe3O4  x 0.30 mol Fe3O4 =  0.80  mol Al are required and we have 1.76 mol, therefore  Fe3O4 is the limiting reagent

Amount of Heat

-3350 kJ/ 3 mol Fe3O4  x 0.30 mol Fe3O4 = -335.00 kJ

Answer:

Check it below

Explanation:

1) The dots in Lewis Notation represent the Electronic Valence, in other words, the amount of Valence Electrons. The Lewis structures has the advantage of pictorially displaying the valence electrons around the symbol of the Atom.

2) We can easily find in a Periodic Table the number of bonds. Notice that we must rearrange the dots to observe the rule of the Octets. Each atom tends to have more stability and behave as noble gas, with eight electrons in its outer shell. So, in some examples, we'll have to rearrange the dots in order to have follow the Octet rule.

Check below each Lewis Structure

2.1[tex]NH_{3}[/tex] Ammonia

[tex]Valence:\\ N=5,H=1[/tex] Given by each group number.

2.2[tex]SO_{2}[/tex]  Sulfur Dioxide

[tex]Valence:\\ S=6,O=2[/tex]

2.3) [tex]CH_{3}OH[/tex] Methanol

[tex]Valence:\\ C=4,O=2,H=1[/tex]

2.4) [tex]HNO_{2}[/tex]

2.5) [tex]N_{2}[/tex] Nitrogen Gas

2.6) [tex]CH_{2}O[/tex] Formaldehyde

To draw the Lewis dot structures for the given molecules, we need to determine the total number of valence electrons for each molecule and then arrange the atoms and electrons to satisfy the octet rule. The Lewis dot structures for NH3, SO2, CH3OH, HNO2, N2, and CH2O are shown with the appropriate arrangements of atoms and their valence electrons.

To draw the Lewis dot structures for the given molecules, we first need to determine the total number of valence electrons for each molecule. For NH3, N has 5 valence electrons and each H has 1 valence electron, giving a total of 5 + 3 = 8 valence electrons. The Lewis dot structure for NH3 shows N as the central atom surrounded by three H atoms, each bonded by a single bond. Each H atom has two dots around it to represent its two valence electrons.

For SO2, S has 6 valence electrons and each O has 6 valence electrons, giving a total of 6 + 2(6) = 18 valence electrons. The Lewis dot structure for SO2 shows S as the central atom bonded to two O atoms by double bonds. Each O atom has six dots around it to represent its six valence electrons.

Continuing with CH3OH, C has 4 valence electrons, H has 1 valence electron, and O has 6 valence electrons, giving a total of 4 + 3(1) + 6 + 1 = 14 valence electrons. The Lewis dot structure for CH3OH shows C as the central atom bonded to three H atoms and one O atom. The O atom is bonded to the C atom by a single bond and has two dots around it to represent its two valence electrons.

For HNO2, H has 1 valence electron, N has 5 valence electrons, and each O has 6 valence electrons, giving a total of 1 + 5 + 2(6) = 18 valence electrons. The Lewis dot structure for HNO2 shows N as the central atom bonded to two O atoms by single bonds. Each O atom has six dots around it to represent its six valence electrons, and the H atom is bonded to one of the O atoms.

For N2, each N atom has 5 valence electrons, giving a total of 2(5) = 10 valence electrons. The Lewis dot structure for N2 shows two N atoms bonded by a triple bond, with each N atom having three dots around it to represent its three valence electrons.

Finally, for CH2O, C has 4 valence electrons, H has 1 valence electron, and O has 6 valence electrons, giving a total of 4 + 2(1) + 6 = 12 valence electrons. The Lewis dot structure for CH2O shows C as the central atom bonded to two H atoms and one O atom. The O atom is bonded to the C atom by a double bond and has four dots around it to represent its four valence electrons.

Answer: ionic solid

Explanation:

In an ionic solid, the ions are bound together by strong electrostatic attraction hence they are immobile and the solid is unable to conduct electricity. If this solid is dissolved in water, the ions move apart due to solvation and become mobile hence the solution conduts electricity. Similarly, when the solid melts, the ions also become free and the melt conduct electricity.

A solid that is hard, brittle, does not conduct electricity in solid form but does in liquid form or when dissolved in water, and has a high melting point is classified as an ionic solid.

The characteristics of the solid described are indicative of an ionic solid. These solids are typically hard and brittle, and they have high melting points. As a solid, ionic compounds do not conduct electricity because the ions are locked in place within the crystal lattice and thus cannot move freely. When these compounds melt, however, the ionic lattice breaks down and the ions are free to move, allowing the liquid to conduct electricity.

Similarly, when an ionic compound is dissolved in water, it dissociates into ions, which are free to move in the solution, making the solution an electrical conductor. This is because an electrolyte is present, which is a substance that contains free ions and can behave as an electrical conductor.

Answer:

In order to fully understand the issue of the ideal gas or perfect gas, we must pay attention to the following, an ideal or perfect gas does not really exist, it is a hypothetical gas whose sharing of the variables of pressure, volume and temperature can be fully described by the ideal gas equation.

The molecules that make up an ideal gas do not usually attract or repel each other, and their volume is negligible compared to the volume of the container that contains it. Although in our nature the case of an ideal gas does not exist, the differences between the behavior of a real gas in temperature and pressure margins do not substantially alter the calculations, so we can make use of the equation with all the security, to solve various gas exercises.

Explanation:

The collisions that occur between the molecules and with the molecules and with the walls is elastic because the moment is preserved, in addition to the kinetic energy.

It can be synthesized that a gas is ideal when all collisions that occur between atoms or molecules are completely elastic and there are no attractive forces that are intermolecular.

In ideal gases the kinetic energy is proportional to its temperature. The gases approach an ideal gas if they are mono atomic gases, if it is under pressure and also at room temperature.

The amount of gas in a body is measured in moles. One mole of any type of gas reaches 22.4 liters, in normal condition, 0 ° Celsius and 1 of the pressure atmosphere. That volume is called normal molar volume.

Ideal gases have an equation called the Ideal Gas Equation and is based on three main laws that are Boyle's law, Gay-Lussac's law, Charles's law and also Avogadro's law.

Answer:D

Explanation:

The high boiling point of HF is not attributable to the dispersion forces mentioned in the question. In HF, a stronger attraction is in operation, that is hydrogen bonding. This ultimately accounts for the high boiling point and not solely the dispersion model as in F2.

The unusually high boiling point of HF compared to F2 is due to the strong hydrogen bonding interactions between HF molecules.

The correct answer is D. Liquid F2 has weak dispersion force attractions between its molecules, whereas liquid HF has both weak dispersion force attractions and hydrogen bonding interactions between its molecules. Dispersion forces are a type of intermolecular force that occurs between all molecules, regardless of polarity.

However, these forces are generally weaker than other types of intermolecular forces such as hydrogen bonding. In HF, the significant electronegativity difference between hydrogen and fluorine leads to the formation of a polar covalent bond, which makes the HF molecules capable of hydrogen bonding, a stronger intermolecular force.

This hydrogen bonding results in a much higher boiling point for HF as compared to F2, which can only interact with other F2 molecules via relatively weaker dispersion forces.

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Al2O3, or aluminum oxide, is most likely to have low water solubility due to the dissolution process being highly endothermic. It requires more energy to break its intermolecular forces and disperse into water which results in a low solubility.

The solute most likely to have low water solubility due to the dissolution process being highly endothermic is Al2O3 (aluminum oxide). In chemical reactions, an endothermic process involves the absorption of heat. Solutes like Al2O3 need more energy to overcome intermolecular forces and disperse into the solvent, thus making the dissolution process highly endothermic and resulting in lower solubility in water.

On the other hand, solutes like RbF, CaF2, AgCl, and FeCl2 generally require less energy to dissolve in water, making their dissolution process less endothermic and more favorable under normal environmental conditions.

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The FM radio station broadcasting at a frequency of 103.4 MHz produces electromagnetic radiation with a wavelength of approximately 2.9 meters.

The wave equation, which is used to calculate the wavelength of electromagnetic radiation, is given as: c = fλ where c = 3.00 × 10^8 m/s is the speed of light in vacuum, f is the frequency of the electromagnetic wave in Hz (s⁻¹) and λ is its wavelength in m.

In this scenario, the radio station is broadcasting at a frequency of 103.4 MHz, which equals 103.4 x 10^6 Hz. From the wave equation, we can rearrange and solve for the wavelength: λ = c/f. Therefore the wavelength of the radio wave is approximately λ = 3.00 ×10^8 m/s / 103.4 ×10^6 s⁻¹ = 2.9 meters. So, the FM radio broadcast at a frequency of 103.4 MHz has a wavelength of approximately 2.9 meters in free space.

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The wavelength of an FM radio station broadcasting at a frequency of 103.4 MHz is calculated by using the formula c = λv, where c is the speed of light, λ is the wavelength and v is the frequency. The frequency is converted from megahertz to hertz, and the formula is rearranged to solve for wavelength. The calculated wavelength is approximately 2.913 m.

The question asks for the calculation of the wavelength of the radio waves being broadcasted by an FM station at 103.4 MHz. To find the wavelength, we can use the formula c = λv, where c is the speed of light, λ is the wavelength and v is the frequency.

The frequency (v) needs to be in Hz (hertz), so we must first convert the given frequency from megahertz (MHz) to hertz (Hz) - 1 MHz = 10⁸ Hz, therefore 103.4 MHz = 1.034 × 10⁸ Hz.

Then, using the speed of light, c = 2.998 × 10⁸ m/s and rearranging the formula to solve for λ (wavelength), we get λ = c/v.

Substituting the given values, λ = 2.998 × 10⁸ m/s / 1.03 × 10⁸ Hz = 2.913 m.

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Answer:

Sgen = 0.0366 W/K

Explanation:

for the body:

∴ Q = - 336 W...rate of heat loss

∴ T surface = 35°C ≅ 308 K

the rate of entropy transfer from the body:

⇒ ΔS = - Q/Ts

for the enviroment:

⇒ ΔS = Q/Te

∴ assuming: T = Tenv = 25°C ≅ 298 K

resulting in a net variation in the universe:

⇒ Sgen = ΔS = Q/Tenviroment - Q/Tsurface = Q(Ts - Te)/Ts*Te

⇒ Sgen = (336( 308-298))/(308×298) = 3360 WK/91784 K² = 0.0366 W/K

To calculate the equilibrium ratio of [I3-] to [I2], first calculate the moles of I3- formed from the moles of I2 added. Then, calculate the concentrations of I3- and I2 in the diluted solution. Finally, divide the concentration of I3- by the concentration of I2 to get the equilibrium ratio.

In this problem, we are given the concentration of a solution of KI and the amount of I2 added to the solution. We are asked to calculate the equilibrium ratio of [I3-] to [I2]. To solve this problem, we need to use the K value and the stoichiometry of the reaction.

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Final answer:

Calculating the equilibrium ratio of [I3-] to [I2] involves applying the equilibrium constant and concentration changes due to reaction, although the exact effect of dilution cannot be determined without the specified dilution factor.

Explanation:

To calculate the equilibrium ratio of [I3-] to [I2] when 6.00×[tex]10^{-2}[/tex] mol of I2 is added to 1.00 L of 6.00×[tex]10^{-1}[/tex] M KI solution, we first recognize the reaction I-(aq) + I2(aq) ⇒ I3-(aq) with a given equilibrium constant (K) of 710. Initially, [I-] is 0.6 M, and [I2] is 0.06 M. Let x be the change in concentration of I2 and I3- at equilibrium. Hence, [I2] at equilibrium becomes (0.06 - x) M, and [I3-] is x M.

The equilibrium expression for the reaction is K = [I3-]/([I-][I2]). Substituting the known values and solving for x will provide the equilibrium concentrations, allowing us to calculate the desired ratio of [I3-] to [I2]. Unfortunately, the exact dilution factor is not specified in the question, which would be necessary to calculate the effect of dilution accurately.

Explanation:

The given data is as follows.

     m = 0.5 kg,     [tex]T = 20^{o}C[/tex],     [tex]T_{2} = 90^{o}C[/tex]

It is known that specific heat of aluminium is 0.91 kJ/kg.

As we know that,   dQ = dU + dw

where,     dQ = heat transfer

                dU = change in internal energy

                dw = work transfer

For the given system, work transfer "w" is 0.

(a)    Hence, change in stored energy will be calculated as follows.

               Q = [tex]mC \Delta T[/tex]

                   = [tex]0.5 \times 0.91 \times (90 - 20)[/tex]

                   = 31.85 kJ

(b)    The amount of heat transferred will be equal to change in stored energy.

So,              dQ = Q = 31.85 kJ

(c)     Change in entropy will be calculated as follows.

                dS = [tex]mC ln \frac{T_{2}}{T_{1}}[/tex]          

                      = [tex]0.5 \times 0.91 \times ln \frac{90}{20}[/tex]

                      = 0.684 kJ/K

(d)     Entropy transfer by heat will be calculated as follows.

             [tex]\Delta S = \frac{dQ}{dT}[/tex]

                          = [tex]\frac{31.85}{(20 + 273)}[/tex]

                          = 0.1087 kJ/K

(e)    Entropy change will be calculated as follows.

              Entropy change = entropy transfer + entropy generation

           [tex]S_{2} - S_{1} = \frac{dQ}{T} + S^{o}_{gen}[/tex]

            0.684 kJ/K = 0.187 + [tex]S^{o}_{gen}[/tex]

                  [tex]S^{o}_{gen}[/tex] = 0.5752 kJ/K

Answer:

3.12 × 10⁻² atm

Explanation:

A student dissolves 20.0 g of glucose into 511 mL of water. At 25 °C, the vapor pressure of pure water at 25 C is 3.13 × 10⁻² atm. I think the question is: "What is the vapor pressure of the solution?"

According to Raoult's law, the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent present.

[tex]P_{solution}=P\°_{solvent}X_{solvent}[/tex]

The molar mass of glucose is 180.16 g/mol. The moles corresponding to 20.0 g of glucose are:

20.0 g × (1 mol/180.16 g) = 0.111 mol

The density of water at 25°C is 0.997 g/mL. The mass corresponding to 511 mL of water is:

511 mL × (0.997 g/mL) = 509 g

The molar mass of water is 18.02 g/mol. The moles corresponding to 509 g of water are:

509 g × (1 mol/18.02 g) = 28.2 mol

The total number of moles is 0.111 mol + 28.2 mol = 28.3 mol

The mole fraction of water is:

[tex]X_{solvent}=\frac{28.2mol}{28.3mol} =0.996[/tex]

The vapor pressure of a solvent above the solution is:

[tex]P_{solution}=3.13 \times 10^{-2} atm \times 0.996 = 3.12 \times 10^{-2} atm[/tex]

Answer:

The value of equilibrium constant for reverse reaction is [tex]7.7\times 10^{-4}[/tex]

Explanation:

The given chemical equation follows:

[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

The equilibrium constant for the above equation is [tex]1.7\times 10^6[/tex].

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

[tex]2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)[/tex]

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the square root of the equilibrium constant  of initial reaction.

So,

[tex]SO_3(g)\rightarrow SO_2(g)+\frac{1}{2} O_2(g)[/tex]

The value of equilibrium constant for half reverse reaction is:

[tex]K_{eq}'=(\frac{1}{1.7\times 10^6})^{\frac{1}{2}}=0.00077=7.7\times 10^{-4}[/tex]

Hence, the value of equilibrium constant for reverse reaction is [tex]7.7\times 10^{-4}[/tex]

Answer:

[tex]7.7\times10^{-4}[/tex]

Explanation:

The equation for which we have to find Kc is obtained by two - step transformation of the equation whose Kc is given.

1st step:

Reversing the reaction:

By reversing the reaction the reactants become products and vice-versa.

The new equilibrium constant will be:

[tex]Kc^{'}=\frac{1}{Kc}[/tex]

2nd step:

Dividing the equation throughout by 2:

New Kc becomes:

[tex]Kc^{''}=\sqrt{Kc^{'}}=\frac{1}{\sqrt{Kc} }[/tex]

[tex]=\frac{1}{\sqrt{1.7\times10^{6} } }=7.7\times10^{-4}[/tex]

Hence the equilibrium constant is [tex]7.7\times10^{-4}[/tex]

Answer: The temperature at which given potential between the two electrodes is attained is 331.13 K

Explanation:

The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction:  [tex]Fe(s)\rightarrow Fe^{2+}(0.1M)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V[/tex]

Reduction half reaction:  [tex]Ni^{2+}(3\times 10^{-3}M)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.25V[/tex]

Net reaction:  [tex]Fe(s)+Ni^{2+}(3\times 10^{-3}M)\rightarrow Fe^{2+}(0.1M)+Ni(s)[/tex]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:

[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]

Putting values in above equation, we get:

[tex]E^o_{cell}=-0.25-(-0.44)=0.19V[/tex]

To calculate the temperature at which the reaction is taking place, we use the Nernst equation, which is:

[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Fe^{2+}]}{[Ni^{2+}]}[/tex]

where,

[tex]E_{cell}[/tex] = electrode potential of the cell = +0.140 V

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.19 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol.K

F = Faraday's constant = 96500

T = temperature of the reaction

[tex][Fe^{2+}]=0.1M[/tex]

[tex][Ni^{2+}]=3\times 10^{-3}M[/tex]

Putting values in above equation, we get:

[tex]0.140=0.19-\frac{2.303\times 8.314\times T}{2\times 96500}\times \log(\frac{(0.1)}{(3\times 10^{-3})})\\\\T=331.13K[/tex]

Hence, the temperature at which given potential between the two electrodes is attained is 331.13 K

The temperature at which the potential between the two electrodes will be +0.140 V can be calculated using the Nernst equation. Rearranging the equation allows us to solve for temperature. Substituting the given values allows us to find the temperature.

First, we need to find the cell potential using the Nernst equation:

Ecell = E°cell - (0.0592 V / n) * log(Q)

In this case, since both sides of the cell are based on the same half-reaction with different concentrations, the number of electrons transferred (n) is 2. The equilibrium constant (Q) can be calculated using the concentrations of Ni2+ and Fe2+ ions in the half-cells. Rearranging the equation, we can solve for temperature (T):

T = (Ecell - E°cell) / ((0.0592 V / n) * log(Q))

Substituting the given values, we can solve for T.

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Answer:

The reaction will shift to the right in the direction of products.

Explanation:

According to Le Chatelier's Principle, the change in any state of the equilibrium say temperature, volume, pressure, or the concentration, the equilibrium will oppose these changes and will shift in such a way that the effect cause must be nullified.

For an endothermic reaction,

On increasing the temperature, reaction will go in forward direction (towards right), because according to Le Chatelier's principle as we increase the temperature,  the equilibrium is will be disturbed , so to again establish the equilibrium, the reaction will go in forward direction as it is endothermic in nature (towards right).

Thus, the [tex]\Delta H^0[/tex] of the given reaction is positive and thus, on increasing temperature, reaction will go in forward direction.

Answer:- The reaction will shift to the right in the direction of products.

The reaction will move to the left in the direction of the reactants as the temperature rises, favouring the endothermic path. The reaction will move left in the direction of the reactants, thus that is the right response.

A chemical process that takes heat from its surroundings is said to be going in an endothermic direction. Because the reactants and products of an endothermic reaction have different energies, heat energy is needed to help the reaction proceed and create the products. An endothermic reaction is one that absorbs heat while it is occurring. This can be felt as a drop in temperature in the immediate environment or shown by thermodynamic calculations. An endothermic process has a positive enthalpy change (H), which means energy is being absorbed. The reaction will move to the left in the direction of the reactants as the temperature rises, favouring the endothermic path. The reaction will move left in the direction of the reactants, thus that is the right response.

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Answer:

answer is a

Explanation:

Acidify and evaporate

Answer:

Protein's primary structure

Explanation:

The proteins have four different types of structures:

Primary: Describes the sequence of amino acids that is unique for each protein

Secondary: Makes reference to the proteins 3D geometry, it can be alpha-helix (product of the protein coiling) or beta-plated sheet (product of the protein folding).

Tertiary: Refers to the comprehensive 3D structure of the protein. There are many types of tertiary structures, due to hyfrophobic interactions, hydrogen unions, ionic unions.

Quaternary: Resambles the macromolecule formed by many protein molecules. Not all proteins have quaternary structure.

Answer:

5- N20 and N2O5

Explanation:

Full working is shown in the image attached. It is important to remember that NO2 dimerizes to N2O4 while N2PO occurs as monomers.

Answer:

Reactivity: Ethyl benzene > Toluene > Ethane

Explanation:

The order of reactivity depends on how easily the most reactive hydrogen can be abstracted.

The order of reactivity of hydrogen is : 1° < 2° < 3° < Benzyllic based on the stabilizing effects like inductive effect, hyperconjugation , resonance effect.

Hence the order of decreasing reactivity : Ethyl Benzene, Toluene , Ethane.

To sum up, ethylbenzene > toluene > ethane is the sequence in which reactivity decreases with bromine, and the previously described components are critical in dictating the rates of reaction.

Hydrocarbons containing bromine atoms exhibit varying degrees of reactivity, which can be attributed mainly to the stability of the free radical generated during hydrogen atom abstraction. When an unpaired electron delocalizes across the aromatic ring, a more stable benzyl radical is produced upon abstraction from a benzylic carbon in toluene and ethylbenzene. Compared to toluene, the benzyl radical generated from ethylbenzene has somewhat more reactivity due to this delocalization. Ethane has the lowest reactivity of the three compounds since it doesn't have an aromatic ring or benzylic hydrogens, which leads to the formation of a primary free radical that is much less stable.

Answer:

42,3g of H₂O

Explanation:

For the reaction:

4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(g)

62,8 g of NH₃ are:

62,8g×(1mol/17,031g) = 3,69 moles of NH₃

62,8 g of O₂ are:

62,8g×(1mol/32g) = 1,96 moles of O₂

For a complete reaction of 1,96 moles of O₂ you need:

1,96mol O₂×(4mol NH₃ / 5molO₂) = 1,57 moles NH₃. As you have 3,69 moles, limiting reactant is O₂.

Assuming a complete reaction, 1,96mol O₂ produce:

1,96mol O₂×(6mol H₂O / 5molO₂) = 2,35 moles of H₂O. In grams:

2,35 moles of H₂O×(18,01g/1mol) = 42,3g of H₂O

I hope it helps!

Answer:

a) 56 protons

b) 54 electrons

c) 81 neutrons

d) The sum of protons and neutrons is shown. The number of protons is always the same. So we can calculate the number of neutrons ( and also the isotopes)

e)137Ba (with 56 protons and 81 neutrons)

f) atomic mass is 136.9 u ; the mass number is the sum of protons and neutrons and is 137

Explanation:

Step 1: Data given

137 Ba2+ is an isotope of barium. The atomic number of barium( and its isotopes) is 56. This shows the number of protons.

For a neutral atom, the number of protons is equal to the number of electrons.

The different isotopes of an element have the same number of protons but a different number of neutrons.

137Ba2+ has 56 protons (this is the same as the atomic number)

137Ba2+ has 54 electrons ( since it's Ba2+, this means it has 2 electrons less than protons, that's why it's charged +2)

137Ba2+ has 81 neutrons ( 137 - 56 = 81)

In the symbol, the atomic number is not shown. The sum of the protons and neutrons is shown. (Since the number of protons is the same for every isotope, we can calculate the number of neutrons that way. By knowing the neutrons, we also know the isotope.

This isotope is 137Ba

Atomic mass is also known as atomic weight. The atomic mass is the weighted average mass of an atom of an element based on the relative natural abundance of that element's isotopes.

The atomic mass of 137Ba2+ is 136.9 u

The mass number is a count of the total number of protons and neutrons in an atom's nucleus.

The mass number of 137Ba2+ is 137

Answer:

1)There is 2.68 * 10^-11 J of energy needed

2) The nuclear binding energy for 1 mol of Ne is 1.6 *10^13 J/mol

Explanation:

Step 1: Data given

The nucleus of a21Ne atom has a amass of 20.98846 amu.

Step 2: Calculate number of protons and neutrons

The number of electrons and protons in an 21Ne atom = 10

The number of neutrons = 21 -10 =11

Step 3: mass of the atom

Mass of a proton = 1.00727647 u

Mass of a neutron = 1.0086649 u

The mass of the atom = mass of all neutrons + mass of protons

Mass of atom = 11*1.0086649 + 10*1.00727647  = 21.1680786 amu

Step 4: Calculate change of mass

The change in mass = Mass of atom - mass of neon

Δmass = 21.1680786 - 20.98846

Δmass = 0.1796186

Step 5: Calculate mass for a single nucleus

The change of mass for a single nucleus is = Δmass / number of avogadro

Δmass of nucleus = 0.1796186 / 6.022*10^23

Δmass of nucleus =2.98 * 10^-25 grams = 2.98 * 10^-28 kg

Step 6: Calculate energy to break a Ne nucleus

Calculate the amount of energy to break a Ne nucleus

ΔEnucleus = Δmass of nucleus * c²

⇒ with c = 2.9979 *10^8 m/s

ΔEnucleus = 2.98 * 10^-28 kg * (2.9979*10^8)² = 2.68 * 10^-11 J

What is the nuclear binding energy for 1 mol of Ne?

ΔE= ΔEnucleus * number of avogadro

ΔE= 2.68 * 10^-11 J * 6.022*10^23

ΔE= 1.6 *10^13 J/mol

Answer:

4 C3H5N3O9 ------> 6N2 + O2 + 10H2O + 12CO2

Explanation:

Nitroglycerin has a chemical formula C3H5N3O9. The balanced chemical equation is as follows:

4 C3H5N3O9 ------> 6N2 + O2 + 10H2O + 12CO2

We suppose that in a reaction, 44g of carbon dioxide is produced. The mass of nitroglycerin that must have reacted will be calculated as under:

Molecular mass of Nitroglycerin = 227g/mol

Molecular mass of Carbon dioxide = 44g/mol

No. of moles of carbon dioxide produced = 44/44 = 1 mole produced.

Now, from balanced chemical equation, we can see that

12 moles of carbon dioxide are produced by = 4 moles of nitroglycerin.

1 mole of carbon dioxide is produced by = 4/12 = 1/3 moles of nitroglycerin.

Mass of nitroglycerin which produced 1 mole of carbon dioxide =  1/3 x 227 = 75.666 grams.

Answer:

271g

Explanation:

The full explanation is seen in the image attached. See the solution below for details

Answer: The standard potential for this cell is +1.49 V at 25C.

Explanation:

[tex]E^0_{[Sn^{2+}/Sn]}=-0.14V[/tex]

[tex]E^0_{[Ti^{2+}/Ti]}=-1.63V[/tex]

As titanium has lower reduction potential, it will act as anode and tin will acts as cathode.

[tex]Ti+Sn^{2+}\rightarrow Ti^{2+}+Sn[/tex]

Using Nernst equation :

[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Ti^{2+}]}{[Sn^{2+}]}[/tex]

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = [tex]25^oC=273+25=298K[/tex]

n = number of electrons in oxidation-reduction reaction = 2

[tex]E^0=E^0_{cathode}- E^0_{anode}=-0.14-(-1.63)=1.49V[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = 1.49 V

[tex]E_{cell}[/tex] = emf of the cell = ?

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

[tex]E_{cell}=1.49-\frac{2.303\times (8.314)\times (298)}{1\times 96500}\log \frac{1}{1}[/tex]

[tex]E_{cell}=1.49V[/tex]

The voltage of the voltaic cell is 1.49 V.

A voltaic cell is a cell in which electrical energy is produced by a spontaneous chemical reaction.

The equation of the cell is; Ti(s) + Si^2+(aq) -----> Ti^2+(aq) + Si(s)

E°cathode = -0.14 V

E°anode = -1.63 V

E°cell = (-0.14 V) - (-1.63 V) = 1.49 V

Using the Nernst equation;

E = E°cell - 0.0592/n log[Ti^2+]/[Si^2+]

E = 1.49 V - 0.0592/2 log(1)/(1)

E = 1.49 V

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Answer:

The empirical formula is P2O5 (option B)

Explanation:

An empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers.

The actual numbers of atoms of each element that occur in the smallest freely existing unit or molecule of the compound is expressed by the molecular formula of the compound.

The molecular formula of a compound may be the empirical formula, or it may be a multiple of the empirical formula.

If the molecular formula is P4O10, this means for each for P-atoms we have 10-O atoms this is a ratio 4:10 or 1: 2.5

To find the empirical formula we divide the molecular formula by 2 what will give us P2O5

For each 2 P atoms we have 5 O-atoms. This is a ratio 1: 2.5

This is the simpliest form for the compound P4O10.

The empirical formula is P2O5 (option B)

An increase in temperature shifts an endothermic reaction to the right and an exothermic reaction to the left. For the reduced volume, it shifts the equilibriums to the side with the fewer moles of gas.

The effect of an increase in temperature on the given equilibrium systems depends on the sign of Delta H (ΔH), which represents the heat of reaction. For (a) 2NH3(g) ⇌ N2(g)+3H2(g) with ΔH = 92kJ and (b) N2(g) + O2(g) ⇌ 2NO(g) with ΔH = 181kJ, because these reactions are endothermic (ΔH is positive), an increase in temperature will shift the equilibrium towards the right to absorb the excess heat. On the other hand, reactions (c) 2O3(g) ⇌ 3O2(g) with ΔH = - 285kJ and (d) CaO(s) + CO2(g) ⇌ CaCO3(s) with ΔH = - 176kJ are exothermic (ΔH is negative), an increase in temperature will shift the equilibrium to the left to offset the increase in heat.

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Increasing temperature generally shifts endothermic reactions to the right and exothermic reactions to the left. Decreasing volume typically shifts equilibria towards the side with fewer moles of gas. Each reaction's equilibrium shift depends on its enthalpy change (ΔH) and the moles of gases involved.

(a) 2NH₃(g) ⇌ N₂(g) + 3H₂(g) ΔH = 92kJ

(b) N₂(g) + O₂(g) ⇌ 2NO(g) ΔH = 181kJ

(c) 2O₃(g) ⇌ 3O₂(g) ΔH = − 285kJ

(d) CaO(s) + CO₂(g) ⇌ CaCO₃(s) ΔH = −176kJ

Answer:

D

Explanation:

We know that the

reaction catalyzing power of a catalyst ∝ surface area exposed by it

Given

volume V1= 10 cm^3

⇒[tex]\frac{4}{3} \pi r^3= 10[/tex]

hence r= 1.545 cm

also, surface area S1= [tex]4\pi r^2[/tex]

now when the sphere is broken down into 8 smaller spheres

S2= 8×4πr'^2

now, equating V1 and V2 ( as the volume must remain same )

[tex]\frac{4}{3}\pi r^3=8\times\frac{4}{3} \pi r'^3[/tex]

and solving we get

r'= r/2

therefore, S2=[tex]8\times4\pi\frac{r}{2}^2[/tex]

S2=[tex]2\times4\pi r^2[/tex]

S2= 2S1

hence the correct answer is

. The second run has twice the surface area.

Answer:

0.00564 moles

Explanation:

Given that:

The rate constant, k = [tex]6.82\times 10^{-3}[/tex] s⁻¹

Initial concentration [A₀] = [tex]2.90\times 10^{-2}[/tex] mol

Time = 4.0 min = [tex]4.0\times 60[/tex] sec = 240 sec

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

So,  

[tex][A_t]=2.90\times 10^{-2}\times e^{-6.82\times 10^{-3}\times 240}=2.9\times \frac{1}{10^2}\times \frac{1}{e^{1.6368}}[/tex]

[tex][A_t]=0.00564\ moles[/tex]

The concentration after four minutes is 3.3 ×10−3.

Let us recall that for a first order reaction;

ln[A] = ln[A]o - kt

Where;

[A] = concentration at time t

[A]o = initial concentration

k = rate constant

t = time

[A]o = 2.90×10−2 mol/1.7 L = 0.0171 M

k = 6.82×10−3 s−1

t = 4 min or 240 s

Substituting values;

ln[A] = ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)

[A]  = e^ln[0.0171 M] - (6.82×10−3 s−1 × 240 s)

[A]  = 3.3 ×10−3.

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Final answer:

The problem involves physics principles, specifically the ideal gas law. To solve for the final temperature of a gas when pressure changes and the number of moles is halved from an initial condition, the relationship between pressure and temperature must be considered. However, the type of process (isothermal, isobaric, adiabatic) must be known for an accurate calculation.

Explanation:

The student's question involves finding the final temperature of a gas when the pressure changes and the number of moles is reduced by half, starting from an initial temperature (Ti) of 25.0°C. This problem can be solved by applying the ideal gas law and the concept that, for a given amount of gas, the pressure is directly proportional to the temperature (P ∝ T) when volume and the number of moles are constant. Given the initial conditions and the pressure change from 25.7 atm to 4.10 atm, the process is not specified as isothermal, isobaric, or adiabatic; therefore, additional details from the context of the part (a) of the experiment would be required to provide a comprehensive solution.

To find the final temperature based on the given information, one would have to assume the same type of process that occurred in part (a), where volume doubled and pressure got halved. If we assume a similar relationship between temperature and pressure as was demonstrated before, where if the pressure is halved from 2.50 atm, the temperature must also be halved from 303.15 K, we could calculate the final temperature for the new conditions by adjusting for the fact that the number of moles was halved. However, without explicit mention of whether this happens in an isothermal, isobaric, or adiabatic process, a direct calculation cannot be provided here.

Explanation:

Surface tension is defined as the attractive forces experienced by the surface molecules of a liquid by the molecules present beneath the surface layer of the liquid.

In a molecule, more is the number of hydroxyl groups present in it more will be the force of attraction faced by surface molecules towards the inner side of the liquid.

Hence, then the surface tension will increase.

Compoud (a) has only one -OH group, compound (b) has 3 -OH groups and compound (c) has 2 -OH groups.

Therefore, the given compounds are arranged on the basis of order of increasing surface tension at a given temperature are as follows.

    [tex]CH_3CH_2CH_2OH[/tex] < [tex]HOCH_2CH_2OH[/tex] < [tex]HOCH_2CH(OH)CH_2OH[/tex]

The surface tension in the following molecules in the increasing order is given as, [tex]\rm CH_2CH_2CH_2OH < HOCH_2CH_2OH < HOCH_2CH(OH)CH_2OH[/tex].

Surface tension can be given as the force applied by the molecules underneath the liquid surface to form the attraction and made the liquid occupy, the least surface area.

The compounds with a more number of attracting groups such as hydroxyl radicals tend to apply more attraction and have more surface tension.

The number of hydroxyl groups in the following compounds are:

The increasing order of the surface tension in the following molecules is, [tex]\rm CH_2CH_2CH_2OH < HOCH_2CH_2OH < HOCH_2CH(OH)CH_2OH[/tex].

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