The combustion of ethane (C2H6)(C2H6) produces carbon dioxide and steam. 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) 2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g) How many moles of CO2CO2 are produced when 5.95 mol5.95 mol of ethane is burned in an excess of oxygen? moles of CO2:CO2:

Answer: Option (e) is the correct answer.

Explanation:

A bond that is formed when an electron is transferred from one atom to another results in the formation of an ionic bond.

For example, NaBr will be an ionic compound as there is transfer of electron from Na to Br.

Whereas a bond that is formed by sharing of electrons is known as a covalent bond.

For example, [tex]CBr_{4}[/tex] will be a covalent compound as there is sharing of electron between carbon and bromine atom.

Also, when electrons are shared between the combining atoms and there is large difference in electronegativity of these atoms then partial charges develop on these atoms. As a result, it forms a polar covalent bond.

For example, in a HBr compound there is sharing of electrons between H and Br. Also, due to difference in electronegativity there will be partial positive charge on H and partial negative charge on Br.  

Thus, we can conclude that out of the given options HBr is the only compound that has polar covalent bonds.

HBr is the compound with polar covalent bonds.

The compound that has polar covalent bonds among the given options is HBr only. A polar covalent bond occurs when there is a difference in electronegativity between the two atoms involved. In HBr, hydrogen (H) is less electronegative than bromine (Br), resulting in a partial positive charge on hydrogen and a partial negative charge on bromine.

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Answer : The work done on the surroundings is, 709.1 Joules.

Explanation :

The formula used for isothermally irreversible expansion is :

[tex]w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)[/tex]

where,

w = work done

[tex]p_{ext}[/tex] = external pressure = 1.00 atm

[tex]V_1[/tex] = initial volume of gas = 1.00 L

[tex]V_2[/tex] = final volume of gas = 8.00 L

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

[tex]w=-p_{ext}(V_2-V_1)[/tex]

[tex]w=-(1.00atm)\times (8.00-1.00)L[/tex]

[tex]w=-7L.atm=-7\times 101.3J=-709.1J[/tex]

The work done by the system on the surroundings are, 709.1 Joules. In this, the negative sign indicates the work is done by the system on the surroundings.

Therefore, the work done on the surroundings is, 709.1 Joules.

The work done by the gas on the surroundings when it expands is -709.1 J.

The subject of this question is the work done by a gas during isothermal expansion. In physics, the work done by a gas is given by the formula W = -PΔV, where P is the pressure and ΔV is the change in volume. For this question, the external pressure P is 1.00 atm (which is equivalent to 101.3 J/L), the initial volume V₁ is 1.00 L, and the final volume V₂ is 8.00 L.

Thus, the change in volume ΔV is V₂ - V₁ = 8.00 L - 1.00 L = 7.00 L. Substituting into the formula for work, we find W = -(1.00 atm)(7.00 L) = -7.00 L⋅atm.

One last step needed is to convert from liters-atmospheres to joules. As given in the question, 1 L⋅atm is approximately 101.3 J, so W = -7.00 L⋅atm * 101.3 J/L⋅atm = -709.1 J. The negative sign indicates work was done by the gas on the surroundings.

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Explanation:

[tex]2 N_2O_5(g)\rightarrow 4 NO_2(g)+O_2(g)[/tex]

Rate of the reaction ,k= [tex]3.38\times 10^{-5} s^{-1}[/tex]

Half life of the [tex]N_2O_5=t_{\frac{1}{2}}[/tex]

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{3.38\times 10^{-5} s^{-1}}[/tex](first order kinetics)

[tex]t_{\frac{1}{2}}=20,502.958 seconds[/tex]

Half life of the [tex]N_2O_5[/tex] is 20,502.958 seconds.

Integrated rate equation for first order kinetics in gas phase is given as:

[tex]k=\frac{2.303}{t}\log\frac{p_o}{2p_o-p}[/tex]

p= pressure of the gas at given time t.

[tex]p_o[/tex] = Initial pressure of the gas

(i) When, t = 50 sec

[tex]p_o=500 torr[/tex]

[tex]3.38\times 10^{-5} s^{-1}=\frac{2.303}{50 s}\log\frac{500 Torr}{2(500 Torr)-p}[/tex]

p = 500.49 Torr

(ii)When, t = 20 min = 1200 sec

[tex]p_o=500 torr[/tex]

[tex]3.38\times 10^{-5} s^{-1}=\frac{2.303}{1200 s}\log\frac{500 Torr}{2(500 Torr)-p}[/tex]

p = 519.83 Torr

Final answer:

The half-life of N2O5 is 20443 seconds. The pressure after 50 seconds is 476.83 Torr and the pressure after 20 minutes is also 476.83 Torr.

Explanation:

The rate constant for the first-order decomposition of N2O5 in the reaction 2 N2O5(g) → 4 NO2(g) + O2(g) is kr=3.38×10−5 s−1 at 25 °C. To find the half-life of N2O5, we can use the formula for first-order reactions:

t1/2 = ln(2) / k,

where t1/2 is the half-life and k is the rate constant. Plugging in the values, we get:

t1/2 = ln(2) / (3.38×10−5 s−1) = 20443 seconds.

To calculate the pressure after a certain time, we can use the ideal gas law:

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. Since the volume and number of moles are constant, we can rearrange the equation to:

P = (nRT) / V.

To find the pressure after a certain time, we need to know the number of moles at that time. Using the initial pressure, we can calculate the number of moles using the ideal gas law:

n = PV / RT.

Let's calculate the pressure at different times:

(i) After 50 seconds:

n = (500 Torr) / (0.0821 L·atm/(mol·K) × 298 K) = 19.39 moles

P = (19.39 moles × 0.0821 L·atm/(mol·K) × 298 K) / 1 L = 476.83 Torr

(ii) After 20 minutes (1200 seconds):

n = (500 Torr) / (0.0821 L·atm/(mol·K) × 298 K) = 19.39 moles

P = (19.39 moles × 0.0821 L·atm/(mol·K) × 298 K) / 1 L = 476.83 Torr

Answer : The net ionic equation will be,

[tex]3Cu^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)[/tex]

Explanation :

First we have to balance the chemical reaction.

The given balanced ionic equation will be,

[tex]3CuCl_2(aq)+2K_3PO_4(aq)\rightarrow Cu_3(PO_4)_2(s)+6KCl(aq)[/tex]

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The ionic equation in separated aqueous solution will be,

[tex]3Cu^{2+}(aq)+6Cl^-(aq)+6K^+(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)+6K^+(aq)+6Cl^-(aq)[/tex]

In this equation, [tex]K^+\text{ and }Cl^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]3Cu^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)[/tex]

Answer:

0.04 m³.

Explanation:

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

(P₁V₁) = (P₂V₂).

V₁ = 0.025 m³, P₁ = 1.5 x 10⁶ Pa,

V₂ = ??? m³, ​P₂ = 0.95 x 10⁶ Pa.

∴ V₂ = (P₁V₁)/(P₂) = (0.025 m³)(1.5 x 10⁶ Pa)/(0.95 x 10⁶ Pa) = 0.04 m³.

Answer: The final volume of the container is [tex]0.040m^3[/tex]

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

The equation given by this law is:

[tex]P_1V_1=P_2V_2[/tex]

where,

[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.

[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.

We are given:

[tex]P_1=1.50\times 10^6Pa\\V_1=0.0250m^3\\P_2=0.950\times 10^6\\V_2=?m^3[/tex]

Putting values in above equation, we get:

[tex]1.50\times 10^6Pa\times 0.0250m^3=0.950\times 10^6Pa\times V_2\\\\V_2=\frac{1.50\times 10^6\times 0.0250}{0.950\times 10^6}=0.040m^3[/tex]

Hence, the final volume of container is [tex]0.040m^3[/tex]

Answer & Explanation:

1- The sulfur present in the crude oil can poison the catalysts that are used in the refining of crude oil into the many useful products obtained from crude oil.

2- The crude oil, containing sulfur compounds, introduced into a car or truck engine, it will again interfere with the catalytic converter installed in the exhaust system of the vehicle.

3-  if sulfur is present in fuel, when the fuel is burned it will be converted to SO₂ which is a known precursor to acid rain.

4- Sulfur compounds can lead to increased corrosion in distillation equipment at the high temperatures used.

The correct answer is that H2S, mercaptans, sulphur compounds, and salt are removed from crude oils to prevent corrosion, environmental pollution, and to meet product specifications.

H2S (hydrogen sulphide), mercaptans (thiols), and other sulphur compounds are removed from crude oil for several reasons:

1. Corrosion Prevention: H2S is a highly corrosive gas that can cause significant damage to pipelines, storage tanks, and refinery equipment. It can lead to the formation of sulphuric acid when it reacts with moisture, which accelerates corrosion. Removing H2S and other sulphur compounds helps to mitigate this risk.

2. Environmental Concerns: Sulphur compounds in fuels contribute to the formation of acid rain when burned, as they can react with moisture and oxygen in the atmosphere to form sulphuric acid. Additionally, H2S is toxic and can be harmful to human health and the environment if released into the atmosphere.

3. Product Specifications: Many refined products, such as gasoline, diesel, and jet fuel, have strict sulphur content specifications. Low sulphur fuels burn cleaner and are less harmful to the environment. The removal of sulphur compounds is necessary to meet these specifications.

4. Safety: H2S is also a toxic gas that can be lethal at high concentrations. Its removal is essential for the safety of workers and for the safe operation of refineries and transportation systems.

5. Catalyst Poisoning: Sulphur compounds can act as poisons for catalysts used in various refining processes, reducing their effectiveness and lifetime. Removing sulphur before processing increases the efficiency and longevity of these catalysts.

Salt is removed from crude oil primarily because it can cause corrosion and scaling in refinery equipment, which can lead to reduced efficiency and increased maintenance costs. Additionally, salt can contaminate the final products, affecting their quality and usability.

The process of removing these impurities typically involves a combination of physical and chemical treatments, such as distillation, adsorption, hydrodesulphurization, and caustic washing, depending on the nature and concentration of the contaminants.

Answer:

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

Explanation:

Density of the gasoline ,d= 0.692 g/mL

Volume of gasoline in an tanks,V = 22.0 gallons = 83,279.02 mL

Let mass of the gasoline be M

[tex]Density= \frac{Mass}{Volume}[/tex]

M = V × d = 83,279.02 mL × 0.692 g/mL=57,629.081 g

Given that gasoline is primarily octane.

[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]

Mass of octane burnt in the tank = M = 57,629.081 g

Moles of octane =[tex]\frac{57,629.081 g}{114.08g/mol}=505.1637 mol[/tex]

According to reaction, 2 moles of octane gives 16 moles of carbon-dioxide.

Then 505.1637 mol of octane will give:

[tex]\frac{16}{2}\times 505.1637 mol=4,041.3100 mol[/tex] of carbon-dioxide

Mass of 4,041.3100 mol of carbon-dioxide:

4,041.3100 mol × 44.01 g/mol = 177,858.05 g

Mass of carbon-dioxide produced in pounds = 391.28771 pounds

391.28771 pounds of carbon-dioxide was released into the atmosphere when 22.0 gallon tank of gasoline is burned in an automobile engine.

Answer:

[tex]\boxed{\text{0.200 mol}}[/tex]

Explanation:

The balanced equation is

2CO + O₂ ⇌ 2CO₂

Step 1. Calculate the known concentrations.

[tex]\text{[O$_{2}$]} = \dfrac{\text{0.100 mol}}{\text{3.50 L}} = \text{0.02857 mol/L}\\\\\text{[CO$_{2}$]} = \dfrac{\text{0.400 mol}}{\text{3.50 L}} = \text{0.1143 mol/L}[/tex]

Step 2. Calculate the concentration of CO

[tex]K_{\text{eq}} = \dfrac{\text{[CO$_{2}$]$^{2}$}}{\text{[CO]$^{2}$[O$_{2}$]}} = 1.4 \times10^{2}[/tex]

[tex]\begin{array}{rcl}\\\\\dfrac{0.1143^{2}}{\text{[CO]$^{2}$} \times 0.02857} & = & 1.4 \times10^{2}\\\\0.01306 & = & 4.000\text{[CO]$^{2}$}\\\\\text{[CO]$^{2}$} & = &\dfrac{0.01306}{4.000}\\\\\text{[CO]$^{2}$} & = & 0.003265\\\text{[CO]} & = & \mathbf{0.05714}\\\end{array}[/tex]

3. Calculate the moles of CO

n = 3.50 L × 0.057 14 mol·L⁻¹ = 0.200 mol

[tex]\text{At equilibrium, there are }\boxed{\textbf{0.200 mol}} \text{ of CO in the flask}[/tex]

Kc is the equilibrium constant and depicts the reactants or the product concentration. At equilibrium, the moles of the carbon monoxide is 0.200 moles.

Moles is the ratio of the mass of the substance to that of the molar mass of the substance.

The balanced chemical reaction is given as,

[tex]\rm 2CO + O_{2} \rightleftharpoons 2CO_{2}[/tex]

The known concentration of oxygen is calculated as:

[tex]\begin{aligned}\rm [O_{2}] &= \dfrac{0.100\;\rm mol}{3.50\;\rm L}\\\\&= 0.0285\;\rm mol/L\end{aligned}[/tex]

The known concentration of carbon dioxide is calculated as:

[tex]\begin{aligned}\rm [CO_{2}] &= \dfrac{0.400\;\rm mol}{3.50\;\rm L}\\\\&= 0.1143\;\rm mol/L\end{aligned}[/tex]

The concentration of the carbon monoxide is calculated as:

[tex]\begin{aligned}\rm k_{eq} = \rm \dfrac{[CO_{2}]^{2}}{[CO]^{2}[O_{2}]} &= 1.4 \times 10^{2}\\\\\rm \dfrac{(0.1143)^{2}}{[CO]^{2}\times 0.0285} &= 1.4 \times 10^{2}\\\\\rm [CO]^{2}& = 0.00326\\\\&= 0.0571\end{aligned}[/tex]

Moles of the carbon monoxide is calculated as:

[tex]\begin{aligned}\rm n &= \rm mass \times concentration\\\\&= 3.50 \;\rm L \times 0.057 14 \;\rm mol\; L^{-1}\\\\&= 0.200\;\rm mol\end{aligned}[/tex]

Therefore, at equilibrium, the moles of the carbon monoxide is 0.200 mol.

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Answer:

[tex]6.19\times 10^{-3}[/tex] is the mole fraction of potassium dichromate.

Explanation:

Mass of potassium dichromate = 24.42 g

Moles of  potassium dichromate =[tex]n_1=\frac{24.42 g}{294.185 g/mol}=0.0830 mol[/tex]

Mass of water = 240.0 g

Moles of water =[tex]n_2=\frac{240.0 g}{18.015 g/mol}=13.3222 mol[/tex]

Mole fraction is calculated by:

[tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]

[tex]\chi_1=\frac{0.0830 mol}{0.0830 mol+13.3222 mol}=0.00619=6.19\times 10^{-3}[/tex]

[tex]6.19\times 10^{-3}[/tex] is the mole fraction of potassium dichromate.

Answer:

True => ΔH°f for C₆H₆ = 49 Kj/mole

Explanation:

See Thermodynamic Properties Table in appendix of most college level general chemistry texts. The values shown are for the standard heat of formation of substances at 25°C. The Standard Heat of Formation of a substance - by definition - is the amount of heat energy gained or lost on formation of the substance from its basic elements in their standard state. C₆H₆(l) is formed from Carbon and Hydrogen in their basic standard states. All elements in their basic standard states have ΔH°f values equal to zero Kj/mole.

Answer:

Explanation:

Describing a solution as concentrated tells that the solution has a relative large concentration, but it is a qualitative description, not a quantitative one, so this does not tell really how concentrated the solution is. This is, the term concentrated is a kind of vague; it just lets you know that the solution is not very diluted, but, as said initially, that there is a relative large amount (concentration) of solute.

One conclusion, of course, is that the solute is soluble: else the solution were not concentrated.

On the other hand, the terms saturated and supersaturated to define a solution are specific.

A saturated solution has all the solute that certain amount of solvent can contain, at a given temperature. A supersaturated solution has more solute dissolved than the saturated solution at the same temperature; superstaturation is a very unstable condition.

From above, there is no way that you can conclude whether a solution is supersaturated or not from the statement that a solution is concentrated, so the answer is none of the above.

Answer:

A

Explanation:

A linear correlation means the increase in one variable cause an increase in the other variable. In a graph, the linear correlation can be demonstrated by a right-slanted straight diagonal line. Therefore if an increase in carbon dioxide causes a directly proportional increase in global temperatures then the two are correlated.

A linear correlation between two variables, such as the concentration of carbon dioxide in our atmosphere and the global temperature, does not necessarily imply that changes in one cause changes in the other. Correlation does not imply causation.

The correct answer to your question is B: No. The presence of a linear correlation between two variables does not imply that one of the variables is the cause of the other variable. In statistics, correlation describes the degree to which two variables move in relation to each other, but it does not imply causation. In other words, just because the concentration of carbon dioxide in our atmosphere and the global temperature move together (e.g., they both increase or decrease at the same time), it does not necessarily mean that changes in the concentration of carbon dioxide cause changes in the global temperature. Other factors might be influencing both, or the relationship might be coincidental.

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Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

[tex]0.0811 M=\frac{n}{0.0017 L}[/tex]

n = 0.0001378 mol

[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

[tex]\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol[/tex] of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

[tex]x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L[/tex]

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

To find the concentration of sulfuric acid in the given sample of rain, we can use the concept of titration. By calculating the moles of NaOH used in the titration and using the stoichiometry of the reaction, we can determine the concentration of sulfuric acid. The concentration of sulfuric acid in this sample of rain is 0.00685 M.

To calculate the concentration of sulfuric acid in the given sample of rain, we can use the concept of titration. From the given information, we know that 1.7 mL of 0.0811 M NaOH was required to reach the end point. Since the titration reaction is between NaOH and HCl, we can use the stoichiometry of the reaction to calculate the concentration of sulfuric acid.

First, we need to find the moles of NaOH used in the titration:

Moles of NaOH = volume (L) × Molarity

Moles of NaOH = 0.0017 L × 0.0811 M

Moles of NaOH = 0.000137 mol

Since the ratio of HCl to NaOH in the titration reaction is 1:1, the moles of HCl (sulfuric acid) in the rain is also 0.000137 mol.

Finally, we can calculate the concentration of sulfuric acid in the sample:

Concentration of sulfuric acid = moles/volume

Concentration of sulfuric acid = 0.000137 mol/0.020 L

Concentration of sulfuric acid = 0.00685 M

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Answer: [tex]1.109\times 10^3[/tex] with four significant digits.

Explanation:

Given,

[tex]\Delta H[/tex] = 178.5 KJ/mole = 178500 J/mole    (1kJ=1000J)

[tex]\Delta S[/tex] = 161.0 J/mole.K

According to Gibbs–Helmholtz equation:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G[/tex] = Gibbs free energy  

[tex]\Delta H[/tex] = enthalpy change  

[tex]\Delta S[/tex] = entropy change  

T = temperature in Kelvin

[tex]\Delta G[/tex]= +ve, reaction is non spontaneous

[tex]\Delta G[/tex]= -ve, reaction is spontaneous

[tex]\Delta G[/tex]= 0, reaction is in equilibrium

As per question the reaction is spontaneous that means the value of  [tex]\Delta G[/tex] is negative or we can say that the value is less than zero.

Thus

[tex]T\Delta S>\Delta H[/tex]

[tex]T\times 161J/Kmol> 178500J/mol[/tex]

[tex]T>1109K[/tex]

Significant figures are the figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Thus the temperature is [tex]1.109\times 10^3[/tex] in kelvins above which this reaction is spontaneous.

Answer:

x = 10.

Explanation:

M = (no. of moles of Na₂CO₃·xH₂O) / (Volume of the solution (L)).

∴ no. of moles of Na₂CO₃·xH₂O = (M of Na₂CO₃·xH₂O)*(Volume of the solution (L)) = (0.0366 M)*(5.0 L) = 0.183 mol.

∵ no. of moles = mass/molar mass

∴ Molar mass of Na₂CO₃·xH₂O = (mass of Na₂CO₃·xH₂O)/(no. of moles of Na₂CO₃·xH₂O) = (52.2 g)/(0.183 mol) = 285.245 g/mol.

∵ The molar mass of Na₂CO₃·xH₂O = Molar mass of Na₂CO₃ + x (molar mass of water).

The molar mass of Na₂CO₃·xH₂O = 285.245 g/mol.

Molar mass of Na₂CO₃ = 105.9888 g/mol.

molar mass of water = 18.0 g/mol.

∴ 285.245 g/mol = 105.9888 g/mol + x (18.0 g/mol).

x (18.0 g/mol) = 285.245 g/mol - 105.9888 g/mol = 179.257 g/mol.

∴ x = (179.257 g/mol)/(18.0 g/mol) = 9.958 ≅ 10.

Answer: Correct answer is a - bromo-3-isopropylcyclodecane

Answer: The molecular formula for the given organic compound is [tex]C_2H_2O_4[/tex]

Explanation:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=21.33g[/tex]

Mass of [tex]H_2O=4.366g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.33 g of carbon dioxide, [tex]\frac{12}{44}\times 21.33=5.82g[/tex] of carbon will be contained.

In 18g of water, 2 g of hydrogen is contained.

So, in 4.366 g of water, [tex]\frac{2}{18}\times 4.366=0.485g[/tex] of hydrogen will be contained.

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{5.82g}{12g/mole}=0.485moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.485 moles.

For Carbon = [tex]\frac{0.485}{0.485}=1[/tex]

For Hydrogen  = [tex]\frac{0.485}{0.485}=1[/tex]

For Oxygen  = [tex]\frac{0.969}{0.485}=1.99\approx 2[/tex]

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is [tex]C_1H_{1}O_2=CHO_2[/tex]

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

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

[tex]n=\frac{90.04g/mol}{45g/mol}=2[/tex]

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

[tex]C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4[/tex]

Thus, the molecular formula for the given organic compound is [tex]C_2H_2O_4[/tex]

Answer:

537.68 torr.

Explanation:

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

(P₁T₂) = (P₂T₁).

P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,

P₂ = ??? torr, ​T₂ = 74°C + 273 = 347 K.

∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.

By using Gay-Lussac's Law and converting Celsius to Kelvin, we find that the new pressure of the gas is approximately 537.8 torr when the temperature increases to 74°C.

To find the new pressure (P₂) of the gas when its temperature changes, we use the ideal gas law, specifically Gay-Lussac's Law, which states that for a given amount of gas at constant volume, the pressure of the gas is directly proportional to its absolute temperature.

First, we need to convert the temperatures from Celsius to Kelvin:

Gay-Lussac's Law formula is:

P₁/T₁ = P₂/T₂

We rearrange the formula to solve for P2:

P₂ = P₁ * (T₂/T₁)

Substitute the known values:

P₂ = 485 torr * (347.15 K / 313.15 K) ≈ 537.8 torr

The new pressure of the gas inside the container is approximately 537.8 torr.

Answer: The molarity of barium hydroxide solution is 0.118 M.

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HClO_4[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]

We are given:

[tex]n_1=1\\M_1=0.102M\\V_1=24.0mL\\n_2=2\\M_2=?M\\V_2=10.3mL[/tex]

Putting values in above equation, we get:

[tex]1\times 0.102\times 24.0=2\times M_2\times 10.3\\\\M_2=0.118M[/tex]

Hence, the molarity of [tex]Ba(OH)_2[/tex] solution will be 0.118 M.

Using the volume and molarity of the perchloric acid provided, the stoichiometry of the reaction with barium hydroxide, and the volume of the barium hydroxide solution used in the titration, the molarity of the barium hydroxide solution is calculated to be 0.1189 M.

To calculate the molarity of the barium hydroxide solution, we will need to use the titration data provided and the stoichiometry of the reaction between barium hydroxide (Ba(OH)2) and perchloric acid (HClO4). The balanced chemical equation for the reaction is:

Ba(OH)2 (aq) + 2 HClO4 (aq) → Ba(ClO4)2 (aq) + 2 H2O (l)

From the equation, we see that one mole of barium hydroxide reacts with two moles of perchloric acid. We can use the volume and molarity of the perchloric acid to find the moles of perchloric acid, and then use the mole ratio to find the moles of barium hydroxide. After that, we can calculate the molarity of the barium hydroxide solution.

Moles of HClO4 = Molarity × Volume = 0.102 M × 0.024 L = 0.002448 moles

From the 1:2 mole ratio, we get:

Moles of Ba(OH)2 = 0.002448 moles HClO4 / 2 = 0.001224 moles

Now, we calculate the molarity of Ba(OH)2 using the volume of the Ba(OH)2 solution:

Molarity of Ba(OH)2 = Moles of Ba(OH)2 / Volume in liters = 0.001224 moles / 0.0103 L = 0.1189 M

The molarity of the barium hydroxide solution is therefore 0.1189 M.

Answer:

liquid phase

Explanation:

see the non-graph solution (used only formulas of the thermodynamic); the final temperature of the mixture is ~1139.701 (°K). The details are in the attachment.

Note:

[tex]c_l \ is \ for \ liquid \ gold;\\ c_s \ is \ for \ solid \ gold.[/tex]

The edge length of a unit cell of a face-centered cubic structure is equal to the cube root of the volume of the unit cell. Hence, the edge length of the unit cell of Iron oxide (FeO) is [tex]\rm 4.3745 \ nm[/tex].

The rock salt crystal structure is a face-centered cubic structure. This means that there are atoms at each corner of the cube, and there is also an atom in the center of each face of the cube.

To find the unit cell edge length:

n = number of atoms

= 4

Atomic weight of [tex]\rm Fe = A_f[/tex]

= 55.85 g/mol

Atomic weight of [tex]\rm O = A_0[/tex]

= 16 g/mol

Density [tex]\rm = 5.70 g/cm^3[/tex]

Avogadro's number (NA) [tex]= 6.023\times10^2^3[/tex]

Volume of crystal,

[tex]\rm V = n \left(\frac{A_f + A_0}{{density} \times NA \right)}[/tex]

[tex]\frac{4\left((55.85 + 16)}{5.70 \times 6.023 \times 10^{23}}\\\right)}=8.3714 \times 10^{-23} \text{ cm}^3[/tex]

Unit cell edge length,

[tex]\rm a = \left( V \right)^{1/3}\\\left\\=( 8.3714 \times 10^{-23} \text{ cm}^3 \right)^{1/3}\\\\=4.3745 \times 10^{-8} \text{ cm} \times 10^{7} \frac{\text{nm}}{\text{cm}}\\= 0.43745\ nm[/tex]

Thus, the unit cell edge length is [tex]\rm 0.43745\ nm[/tex].

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The unit cell edge length of iron oxide (FeO) can be calculated using its density and the formula for mass and volume of a unit cell in the rock salt crystal structure.

The rock salt crystal structure is a type of cubic lattice structure. In this type of structure, the atoms are arranged in a face-centered cubic (FCC) unit cell. The unit cell edge length can be calculated using the formula:

Edge length = (4 * volume)^(1/3)

Given that the density of FeO is 5.70 g/cm3, the mass of the unit cell can be determined using the formula:

Mass = density * volume

By substituting the values of density and volume, the mass of the unit cell can be calculated. Finally, by rearranging the equation for mass to solve for volume, the unit cell edge length can be determined.

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The equilibrium constant (Kc) for the reaction CO(g) + Cl2(g) ⇆ COCl2(g) is calculated as 204 based on the given equilibrium concentrations.

The question is asking us to calculate the equilibrium constant, Kc, for the reaction of carbon monoxide and molecular chlorine to form carbonyl chloride (also known as phosgene). This reaction is represented by the equation CO(g) + Cl2(g) ⇆ COCl2(g). The equilibrium constant is calculated using the concentrations of the reactants and products at equilibrium.

In this case, we have [CO] = 0.0210 M, [Cl2] = 0.0450 M, and [COCl2] = 0.204 M. To find Kc, we use the formula:

Kc = [COCl2] / ([CO] * [Cl2])

Substituting the given concentrations into the formula, we get:

Kc = 0.204 / (0.0210 * 0.0450) = 204

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The equilibrium constant at the given temperature is  [tex]215.74.[/tex]

The equilibrium constant [tex](K_c)[/tex] for the reaction between carbon monoxide [tex](CO)[/tex] and molecular chlorine [tex](Cl_2)[/tex] to form carbonyl chloride [tex](COCl_2)[/tex] can be calculated using the equilibrium concentrations provided:

[tex]\[ [CO] = 0.0210 \text{ M}, \quad [Cl_2] = 0.0450 \text{ M}, \quad [COCl_2] = 0.204 \text{ M} \][/tex]

The balanced chemical equation is:

[tex]\[ CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g) \][/tex]

The equilibrium constant expression (K_c) for this reaction is:

[tex]\[ K_c = \frac{[COCl_2]}{[CO][Cl_2]} \][/tex]

Substituting the given equilibrium concentrations into the expression:

[tex]\[ K_c = \frac{[0.204]}{[0.0210][0.0450]} \][/tex]

 [tex]\[ K_c = \frac{0.204}{0.0210 \times 0.0450} \][/tex]

[tex]\[ K_c = \frac{0.204}{0.000945} \][/tex]

[tex]\[ K_c \approx 215.74 \][/tex]

 Therefore, the equilibrium constant at the given temperature is approximately 215.74.

The correct format for the answer is:

[tex]\[ \boxed{K_c \approx 215.74} \][/tex]

This value indicates that at equilibrium, the concentration of carbonyl chloride is much higher than the concentrations of carbon monoxide and chlorine gas, suggesting that the reaction strongly favors the formation of [tex]COCl_2[/tex] under these conditions.

The reaction of lead(II) nitrate with potassium chloride in water produces a precipitate of lead(II) chloride and a solution of potassium nitrate. The limiting reactant, which determines the maximum quantity of product, is currently unknown without further calculation, as is the precise quantity of precipitate recovered and excess reactant remaining, given the 89.3% reaction yield.

The balanced chemical equation for this reaction is: 2KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2KNO3(aq). In this reaction, the lead(II) nitrate reacts with potassium chloride to create a precipitate of lead(II) chloride and a solution of potassium nitrate.

To identify the limiting reactant, we would have to compare the mole ratio of the reactants. However, without the calculation, I am unable to specify which is the limiting reactant. The limiting reactant is the reactant that gets completely consumed during the reaction and determines the maximum amount of product that can be formed.

If the percent yield is 89.3%, it means that 89.3% of the theoretical maximum amount of product (predicted by stoichiometry from the limiting reactant) was actually made. Again, without the calculation, I can't provide the exact number for the grams of precipitate recovered and the grams of the excess reactant remaining.

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Answer: The expression for equilibrium constant is [tex]K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}[/tex]

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 the general chemical equation:

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

The expression for [tex]K_c[/tex] is given as:

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

For the given chemical reaction:

[tex]2HgO(s)+H_2O(l)+2Cl_2(g)\rightleftharpoons 2HOCl(aq.)+HgO.HgCl_2(s)[/tex]

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

[tex]K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}[/tex]

The concentration of solid is taken to be 0.

So, the expression for [tex]K_{eq}[/tex] is given as:

[tex]K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}[/tex]

Final answer:

To find the arsenic concentration in ppb, we convert the volume of the sample to liters and then use the formula Concentration (ppb) = (Mass of As (ng) / Volume of water (L)) x 1 billion. For a 14.3 cm3 sample containing 159.5 ng As, the arsenic concentration is 11,153.85 ppb.

Explanation:

To find the concentration of arsenic (As) in the lake water sample in parts per billion (ppb), we need to use the given information about the mass of arsenic and the volume of the water sample. Since we are provided with a 14.3 cm3 sample containing 159.5 ng As and assuming that the density of the lake water is 1.00 g/cm3, we can calculate the concentration.

First, we need to convert the volume of the lake water from cm3 to liters since 1 cm3 is equal to 1 mL and there are 1,000 mL in a liter. For the 14.3 cm3 sample, this converts to:

14.3 cm3 × (1 mL/1 cm3) = 14.3 mL

14.3 mL × (1 L/1,000 mL) = 0.0143 L

Next, we can calculate the concentration of arsenic in the water sample in ppb using the following equation:

Concentration (ppb) = (Mass of As (ng) / Volume of water (L)) × 1 billion

Filling in the values we have:

Concentration (ppb) = (159.5 ng / 0.0143 L) × 1,000,000,000 ng/billion

Concentration (ppb) = 11,153.85 ppb

The arsenic concentration in the sample is therefore 11,153.85 ppb.

Answer:

The correct answer is C: hydrogen bonds

Explanation:

Water's surface tension and heat storage capacity are accounted for by its hydrogen bonds. Molecules of water are very strongly attracted to each other through hydrogen bonding.

Hydrogen bonds are constantly formed and broken in water molecules. Surface tension results from this hydrogen bonding which means that Water has a higher storage capacity for heat.

That is why at night, the Earth gets colder must faster than water.

Water takes time to slowly release heat to keep the atmospheric temperature moderate at night.

Final answer:

Hydrogen bonds are responsible for water's high surface tension and heat storage capacity. They result in strong cohesion between water molecules and a significant amount of energy is required to change water's state due to these bonds.

Explanation:

Water's surface tension and heat storage capacity are accounted for by its hydrogen bonds. The correct answer to the student's question is C) hydrogen bonds. These bonds occur because the hydrogen atoms in a water molecule have a slight positive charge, while the oxygen atom has a slight negative charge. Each water molecule can form four hydrogen bonds with surrounding molecules, leading to a high degree of cohesion. This cohesion is the main reason behind water's significant surface tension.

Additionally, because of the strength of these bonds, water has a high heat capacity and high heat of vaporization. It requires a large amount of heat energy to break the hydrogen bonds for water to change state, which explains why water is able to absorb and store a lot of heat without undergoing a significant increase in temperature.

Answer:

#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.

Explanation:

Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆

1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.

=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP

2. Calculate moles of F₂ used

=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used

3. Calculate moles of XeF₆ produced from reaction ratios …

Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆

4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number  (6.02 x 10²³ molecules/mole)

=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)

                                  = 2.75 x 10²³ molecules XeF₆.

To determine the number of molecules of XeF₆,

First, we will determine the number of moles of F₂ present in the 12.9L of F₂

From the ideal gas equation

PV = nRT

Where P is the pressure

V is the volume

n is the amount of substance ( number of moles)

R is ideal gas constant

T is the temperature

From the question,

P = 2.6 atm

V = 12.9 L

R = 0.082057 L atm mol⁻¹ K⁻¹

T = 298 K

Putting these values into the equation

PV = nRT

2.6 × 12.9 = n × 0.082057 × 298

33.54 = n × 24.452986

∴ n = 33.54 ÷ 24.452986

n = 1.3716116 moles

The number of moles of F₂ present is 1.3716116 moles

From the given equation of reaction

Xe(g) +3F₂(g)→XeF₆(g)

1 mole of Xe reacts with 3 moles of F₂ to produce 1 mole of XeF₆

∴ 1.3716116 moles of F₂ will give (1.3716116/3) moles of XeF₆

Hence, the number of moles of XeF₆ produced = 0.457204 moles

To determine the number of molecules formed,

From the formula

Number of molecules = number of moles × Avogadro's number

∴ Number of molecules of XeF₆ formed = 0.457204 × 6.02214 ×10²³

Number of molecules of XeF₆ formed = 2.7533 × 10²³ molecules

Number of molecules of XeF₆ formed ≅ 2.75 × 10²³ molecules

Hence, the number of molecules of XeF₆ that are formed from 12.9 L of F₂ (at 298 K and 2.6 atm) is 2.75 × 10²³ molecules.

The correct option is E) 2.75 × 1023 molecules XeF6

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Answer: The molarity of potassium hydroxide is 0.186 M

Explanation:

To calculate the molarity of base, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HBr

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.

We are given:

[tex]n_1=1\\M_1=0.194M\\V_1=24.2mL\\n_2=1\\M_2=?M\\V_2=25.2mL[/tex]

Putting values in above equation, we get:

[tex]1\times 0.194\times 24.2=1\times M_2\times 25.2\\\\M_2=0.186M[/tex]

Hence, the concentration of potassium hydroxide is 0.186 M.

Answer:

The polymerization of amino acids produces polypeptides.

Explanation:

The polymerization of amino acids producing polypeptide is known as translation. mRNA produces amino acids.

The polymerization of amino acids produces polypeptides. Hence, option D is correct.

Polymerization is defined as "A chemical process that combines several monomers to form a polymer or polymeric compound."

Polymerization of amino acids is the formation of proteins. Amino acids are small molecules consisting of an amino group, a carboxyl group, a hydrogen atom, and an ‘R’ group.  

In polymerization, two amino acids make a bond with each other with a peptide bond, which is formed between the amino group of one amino acid and the carboxyl group of another amino acid.

Hence, the polymerization of amino acids produces polypeptides which are also known as translation. mRNA produces amino acids.

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