The manufacturer specs for a particular balloon indicate the maximum inflated volume is 3.00 L. The balloon is filled with 2.50 L of helium at sea level (assume P= 1.00 atm) and released. When the balloon rises to a higher altitude where the pressure is 450.0 mm Hg, will the balloon burst? Show a calculation to support your answer. Assume constant temperature.

Final answer:

C2H4OH is most likely to form intermolecular hydrogen bonds due to its hydroxyl (-OH) group, which is highly polar and able to engage in hydrogen bonding.

Explanation:

The compound most likely to form intermolecular hydrogen bonds is C2H4OH. This is because C2H4OH has an -OH (hydroxyl) group, which is capable of forming hydrogen bonds due to the high polarity of the O-H bond. Let's analyze why the other options are less likely:

Therefore, the answer is C2H4OH, as it is the only one among the given options with the ability to form significant hydrogen bonds due to its -OH group.

Answer :

The rate of the reaction at t=0 is, [tex]9.0\times 10^{-13}M.s^{-1}[/tex]

The overall order of reaction is, second order reaction.

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The general reaction is:

[tex]A+B\rightarrow C+D[/tex]

The general rate law expression for the reaction is:

[tex]\text{Rate}=k[A]^a[B]^b[/tex]

where,

a = order with respect to A

b = order with respect to B

R = rate  law

k = rate constant

[tex][A][/tex] and [tex][B][/tex] = concentration of A and B reactant

Now we have to determine the rate law for the given reaction.

The given balanced equations is:

[tex]O_3(g)+NO(g)\rightarrow O_2(g)+NO_2(g)[/tex]

In this reaction, [tex]O_3[/tex] and [tex]NO[/tex] are the reactants.

The rate law expression for the reaction is:

[tex]\text{Rate}=k[O_2][NO][/tex]    ..........(1)

Given :

Rate constant = [tex]k=3.0\times 10^{-5}M^{-1}s^{-1}[/tex]

Concentration of [tex]O_3[/tex] = [tex]5.0\times 10^{-4}M[/tex]

Concentration of [tex]NO[/tex] = [tex]6.0\times 10^{-5}M[/tex]

Now put all the given values in equation 1, we get:

[tex]\text{Rate}=(3.0\times 10^{-5}M^{-1}s^{-1})\times (5.0\times 10^{-4}M)\times (6.0\times 10^{-5}M)[/tex]

[tex]\text{Rate}=9.0\times 10^{-13}M.s^{-1}[/tex]

Thus, the rate of the reaction at t=0 is, [tex]9.0\times 10^{-13}M.s^{-1}[/tex]

Now we have to determine the overall order of reaction.

From the given rate law expression we conclude that,

The order of reaction with respect to [tex]O_3[/tex] is, first order reaction.

The order of reaction with respect to [tex]NO[/tex] is, first order reaction.

Thus, overall order of reaction will be:

Overall order of reaction = 1 + 1 = 2

Thus, the overall order of reaction is second order reaction.

The calculated rate of the reaction at t=0 is 9.0×10−1 M×s−1, and the overall order of the reaction is second order, resulting from being first order with respect to each reactant.

The rate of the reaction O3(g) + NO(g) → O2(g) + NO2(g) at t=0 can be calculated using the given rate law and concentrations. The rate law is rate = k[O3][NO], where k is the rate constant, and the brackets denote the concentration of the reactants O3 and NO respectively. To calculate the reaction rate at t=0, we plug in the given values: rate = (3.0×106 M−1×s−1)(5.0×10−4 M)(6.0×10−5 M), which gives a rate of 9.0×10−1 M×s−1.

The overall order of the reaction can be determined by looking at the exponents of the reactant concentrations in the rate law. Since the rate law is rate = k[O3]1[NO]1, it is first order with respect to both O3 and NO. This means that the overall order of the reaction is the sum of the individual orders, which is 1 + 1 = 2, making the reaction second order overall.

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Answer:

The correct answer is:

The pH of pure water decreases as the temperature is increased.

Explanation:

Ionization constant of water :

[tex]2H_2O(l)\rightleftharpoons H_3O^+(aq) + OH^-(aq)[/tex]

Water will dissociate into equal amount of hydronium and hydroxide ions

[tex]K_w=[H_3O^+]\times [OH^-][/tex]

As we can see that from the given information that ionic product of water increase with increase in temperature which means concentration of hydronium ion concentration increases.

And according to definition of pH of the solution is a negative logarithm of hydronium ions concentration.

[tex]pH=-\log [H_3O^+][/tex]

The pH of the solution of solution is inversely proportional to the concentration of hydronium ions.

As the hydronium ion concentration of water increases with increase in temperature , the pH of the water will decrease.

Answer:

4 shared pairs

Explanation:

According to the VESPR, a molecule with a central atom C linked to 4 Cl would form a tetrahedral, whose angles are 109.5°. When more lone pairs are added, the repulsion is greater due to the unshared electrons leading to a distorted tetrahedral. The more lone pairs, the lower the bond angle.

Answer:

60.0mL

Explanation:

For the process of titration , the formula we use is -

M₁V₁ = M₂V₂

Where ,

M₁ = initial concentration

V₁ = initial volume

M₂ = final concentration

V₂ = final volume .

Hence , from the question ,

The given data is ,

M₁ = 0.050 M

V₁ = 30.0mL

M₂ = 0.025 M

V₂ = ?

Now, to determine the unknown quantity , the formula can be applied ,

Hence ,

M₁V₁ = M₂V₂

Putting the respective values ,

0.050 M * 30.0mL = 0.025 M * ?

solving the above equation ,

V₂ = ? = 60.0mL

This titration reaction occurs between a weak base (NH3) and a strong acid (HCl). The distinct pH stages refer to the volume of HCl added. Initially, the pH is determined by the NH3. However, as more HCl is added and reacts with NH3, the pH is determined by the balance of NH3 and NH4+ (from NH3 + HCl -> NH4Cl). If there’s an excess of HCl, the pH drops becoming more acidic.

This question relates to the titration process in chemistry. Let's start by identifying the reactants: NH3 (a weak base) and HCl (a strong acid). The chemical reaction for this titration can be written as NH3 + HCl -> NH4Cl.

To calculate after specific volumes of titrant have been added, we must first understand the principle of titration. In titration, the reaction proceeds until one of the reactants is completely used up. The point at which this happens is known as the equivalence point.

a) At 0 mL HCl, no reaction has occurred so the pH is determined by the NH3 present.

b) At 20.0 mL of HCl, a partial reaction has occurred. The moles of NH3 originally present were 30.0 mL * 0.050 M = 1.5 mmol. After 20 mL of HCl added (0.5 mmol), we have 1.0 mmol NH3 left.

c) At 59.2 mL of HCl added, all the NH3 has reacted because the moles of HCl added are equal to the moles of NH3 initially present.

d) At 60.0 mL of HCl, an excess of HCl has been added and we have a solution of NH4Cl plus excess HCl.

e) At 71.3 mL and f) at 73.3 mL, we still have an excess of HCl, leading to a more acidic solution.

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Answer:

Molar mass of silane = 32.15 g/mol

Explanation:

Given:  

Pressure = 2.35 atm

Temperature = 22 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (22 + 273.15) K = 295.15 K  

Volume = 42.9 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

2.35 atm × 4.81 L = n × 0.0821 L.atm/K.mol × 295.15 K  

⇒n = 0.4665 moles

Given, Mass = 15.0 g

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]0.4665\ mole= \frac{15.0\ g}{Molar\ mass}[/tex]

Molar mass of silane = 32.15 g/mol

Answer:

ΔH° = 206.1 kJ

ΔG° = 142.1 kJ

Explanation:

Let's consider the first step in the synthesis of methanol.

Step 1: CH₄(g) + H₂O(g) ⟶ CO(g) + 3 H₂(g) ΔS° = 214.7 J / K

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(p) are the standard enthalpies of formation of reactants and products

ΔH° = [1 mol × ΔH°f(CO(g)) + 3 mol × ΔH°f(H₂(g))] - [1 mol × ΔH°f(CH₄(g)) + 1 mol × ΔH°f(H₂O(g))]

ΔH° = [1 mol × (-110.5 kJ/mol) + 3 mol × (0 kJ/mol)] - [1 mol × (-74.81 kJ/mol) + 1 mol × (-241.8 kJ/mol)]

ΔH° = 206.1 kJ

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

ΔG° = 206.1 kJ - 298 K × (214.7 × 10⁻³ kJ/K)

ΔG° = 142.1 kJ

The values of ΔH∘ and ΔG∘ for the reaction can be calculated using their respective formulas and knowing the standard enthalpy and free energy of formation of all the reactants and products. The data is usually found in Chemistry textbooks, specifically in Appendix G.

To calculate ΔH∘ and ΔG∘ at 298K for step 1, we use the Gibb's Helmholtz equation, ΔG = ΔH - TΔS. However, the problem doesn't provide enough information to calculate these directly. Let's presume, however, that we know the values of ΔH∘ for methane and water. Using Appendix G data (usually found in Chemistry textbooks), we could apply the formula ΔH = [ΔH products] - [ΔH reactants] to get ΔH∘.

Similarly, for ΔG∘, if the standard free energy of formation of the compounds is known, we could apply the formula ΔG = [ΔG products] - [ΔG reactants].

Since these values are usually tabulated, this could give us the values of ΔH∘ and ΔG∘ for step 1 of the reaction.

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Answer: The pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.

Explanation:

We are given:

Equilibrium partial pressure of [tex]O_2[/tex] = 0.29 atm

For the given chemical equation:

                   [tex]2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)[/tex]

Initial:              a

At eqllm:        a-2x          2x          x

Calculating for the value of 'x'

[tex]\Rightarrow x=0.29[/tex]

Equilibrium partial pressure of NO = 2x = 2(0.29) = 0.58 atm

Equilibrium partial pressure of [tex]NO_2[/tex] = a - 2x = a - 2(0.29) = a - 0.58

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{O_2}\times (p_{NO})^2}{(p_{NO_2})^2}[/tex]

We are given:

[tex]K_p=158[/tex]

Putting values in above expression, we get:

[tex]158=\frac{0.29\times (0.58)^2}{(a-0.58)^2}\\\\a=0.555,0.604[/tex]

Neglecting the value of a = 0.555 because it cannot be less than the equilibrium concentration.

So, [tex]a=0.604[/tex]

Equilibrium partial pressure of [tex]NO_2[/tex] = (a - 0.58) = (0.604 - 0.58) = 0.024 atm

Hence, the pressure of NO and [tex]NO_2[/tex] in the mixture is 0.58 atm and 0.024 atm respectively.

Answer: High P and Low T

Where P is the Pressure and T is the Temperature.

Explanation:

1. High Pressure : The above chemical equation has the reactant and product sides.

The product side has total moles of 3 and reactant side has total moles of 4.

To obtain maximum yield in the production of CH4 , pressure must be high because it will favour the side the less number of moles(volume) which is the product side.

2.Low Temperature : The enthaply change indicated as negative shows it is an exothermic reaction. And for an exothermic reaction, the temperature must be lower so as to favor the forward reaction and to make up for the heat/energy loss.

Answer: The number of moles of gas remaining in the lungs is 0.063 moles

Explanation:

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where,

[tex]V_1\text{ and }n_1[/tex] are the initial volume and number of moles

[tex]V_2\text{ and }n_2[/tex] are the final volume and number of moles

We are given:

[tex]V-1=1.9L\\n_1=0.080mol\\V_2=1.5L\\n_2=?[/tex]

Putting values in above equation, we get:

[tex]\frac{1.9}{0.080}=\frac{1.5}{n_2}\\\\n_2=\frac{1.5\times 0.080}{1.9}=0.063[/tex]

Hence, the number of moles of gas remaining in the lungs is 0.063 moles

Answer: The specific heat of iridium is 0.130 J/g°C

Explanation:

When iridium is dipped in water, the amount of heat released by iridium will be equal to the amount of heat absorbed by water.

[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]

The equation used to calculate heat released or absorbed follows:

[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]

[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex]      ......(1)

where,

q = heat absorbed or released

[tex]m_1[/tex] = mass of iridium = 23.9 g

[tex]m_2[/tex] = mass of water = 20.0 g

[tex]T_{final}[/tex] = final temperature = 22.6°C

[tex]T_1[/tex] = initial temperature of iridium = 89.7°C

[tex]T_2[/tex] = initial temperature of water = 20.1°C

[tex]c_1[/tex] = specific heat of iridium = ?

[tex]c_2[/tex] = specific heat of water = 4.18 J/g°C

Putting values in equation 1, we get:

[tex]23.9\times c_1\times (22.6-89.7)=-[20\times 4.18\times (22.6-20.1)][/tex]

[tex]c_1=0.130J/g^oC[/tex]

Hence, the specific heat of iridium is 0.130 J/g°C

Taking into account the definition of calorimetry, the specific heat of iridium is 0.13 [tex]\frac{J}{gC}[/tex].

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

In this case, you know:

Replacing in the expression to calculate heat exchanges:

For iridium: Qiridium= c × 23.9 g× (22.6 C - 89.7 C)

For water: Qwater= 4.186 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qiridium = + Qwater

- c × 23.9 g× (22.6 C - 89.7 C)= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 C)

Solving:

c × 23.9 g× ( 89.7 C - 22.6 C)= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× (22.6 C - 20.1 C)

c × 23.9 g× 67.1 C= 4.18 [tex]\frac{J}{gC}[/tex]× 20 g× 2.5 C

c × 1603.69 g×C= 209 J

[tex]c=\frac{209 J}{1603.69 gC}[/tex]

c= 0.13 [tex]\frac{J}{gC}[/tex]

Finally, the specific heat of iridium is 0.13 [tex]\frac{J}{gC}[/tex].

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Answer:

70.8 mmHg

Explanation:

To find partial pressure of carbon dioxide we first find the mole fraction of carbon dioxide

n (carbon dioxide) = 5/44.01 g/mol = 0.11361 mol

n (Helium) = 3.75 g/4 g/mol = 0.9375 mol

The partial pressure of carbon dioxide will be the mole fraction of carbon dioxide multiplied by the total pressure

Partial pressure = 0.11361g/(0.11361g + 0.9375 g)* 655 mmHg

                          = 70.8 mmHg

The standard enthalpy of formation, which has been established for a huge variety of compounds, is enthalpy change. The reactants undergo chemical changes and combine to produce products in any general chemical reaction. Here the enthalpy change is -297 kJ.

The reaction enthalpy is the change in enthalpy, denoted by the symbol ΔrH.  By deducting the total enthalpies of all the reactants from the total enthalpies of the products, the reaction enthalpy is determined.

Here we will multiply the equation (1) with "2" and then will subtract it from equation "2".

N₂(g) + 2O₂ (g)→ 2NO₂(g) ...... ΔH° = 66.4 kJ  (1) × (2)

2N₂O(g) → 2N₂(g) + O₂(g)  .......ΔH° = -164.2 kJ (2)

2N₂(g) + 4O₂ (g)→ 4NO₂(g) .......  ΔH° = 132.8 kJ

2N₂O(g) + 3O₂(g) → 4NO2(g)

ΔH° = ΔH°(eq 2) - 2ΔH°(eq 1) = -164.2 -(2X66.4) = -297 kJ

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Final answer:

Using Hess's Law and manipulating the given reactions to achieve the target reaction, the standard enthalpy change for the reaction 2N2O(g) + 3O2(g) -> 4NO2(g) is found to be -31.4 kJ.

Explanation:

To find the standard enthalpy change ΔH° for the reaction 2N2O(g) + 3O2(g) → 4NO2(g), we need to apply Hess's Law, which states that the total enthalpy change of a reaction is the sum of the enthalpy changes of individual steps that lead to the overall reaction.

The reactions given are:

(1) N2(g) + 2O2(g) → 2NO2(g)...... ΔH° = 66.4 kJ

(2) 2N2O(g) → 2N2(g) + O2(g)......ΔH° = -164.2 kJ

To use these reactions to find the enthalpy change for (3), we need to manipulate them so that when added, they result in the target reaction. We notice that reaction (1) must be doubled to provide the 4NO2(g) as in reaction (3). This also doubles its enthalpy change. Then, we can simply add the first and second reactions together to get the desired reaction.

So, we have:

(1) x 2: 2N2(g) + 4O2(g) → 4NO2(g)...... ΔH° = 2 x 66.4 kJ = 132.8 kJ

(2):  2N2O(g) → 2N2(g) + O2(g)......ΔH° = -164.2 kJ

To find the ΔH° for (3), we add the ΔH° from the doubled reaction (1) to the ΔH° from reaction (2):

132.8 kJ + (-164.2 kJ) = -31.4 kJ

Hence, the standard enthalpy change for the reaction 2N2O(g) + 3O2(g) → 4NO2(g) is -31.4 kJ.

To calculate the pKa for the acid-base indicator using the given absorbance and molar absorptivity values, apply Beer's Law to determine the concentrations of HIn and In-, then use the Henderson-Hasselbalch equation.

To calculate the pKa for the acid-base indicator HIn, we must use the absorbance data provided and apply Beer's Law, which relates the absorbance to the concentrations of the protonated form (HIn) and deprotonated form (In-) of the indicator.

We are given that at pH 6.12, the absorbance (A) is 0.818, the molar absorptivity (ε) of HIn is 2929 M-1 cm-1 and that of In- is 20060 M-1 cm-1, and the total concentration of the indicator (C) is 0.000127 M.

From Beer's Law we know that:

A = εHInb[HIn] + εnb[In-]

where b is the path length of the cuvette used, which is 1 cm. The concentration of HIn and In- sum up to the total concentration of the indicator:

[HIn] + [In-] = C

We must also consider the Henderson-Hasselbalch equation, which relates the pKa to the pH and the ratio of deprotonated to protonated forms:

pH = pKa + log([In-]/[HIn])

Using the absorbance at pH 6.12 and the total indicator concentration, we can calculate the fractions of HIn and In- and then use the Henderson-Hasselbalch equation to solve for the pKa.

Answer:

m H2O = 56 g

Explanation:

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT)Al = (mCΔT)H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C

⇒ mH2O = 55.722 g ≅ 56 g

To calculate the mass of water in the calorimeter, use the conservation of energy principle and the specific heat capacity formula. The mass of water in the calorimeter is 61 g.

To calculate the mass of water in the calorimeter, you can use the conservation of energy principle. The heat lost by the aluminum equals the heat gained by the water. The heat gained by water can be calculated using the specific heat capacity formula: Q = mcΔT.

From the given information, we know the initial temperature of the water (21.1°C), the final temperature of the water (26.8°C), and its specific heat capacity (4.18 J/g°C). Considering the amount of heat lost by the aluminum and the heat gained by the water, you can calculate the mass of water in grams to be 61 g.

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Answer:

The molecular shape and ideal bond angle of the [tex]PH_{3}[/tex] is trigonalbipyramidal and [tex]109.5^{o}[/tex] respectively.

Explanation:

The structure of  [tex]PH_{3}[/tex]  is as follows.(in attachment)

From the structure,

Phosphor atom has one lone pair and three hydrogens are bonded by six electrons.

Therefore, total electrons invovled in the formation [tex]PH_{3}[/tex]  is eight.

Hence, four electron groups which indicate the tetrahedral shape. But one pair is lone pair i.e, present on the phosphor atom.

Therefore, ideal geometry of the [tex]PH_{3}[/tex]  molecule is Trigonalbipyramidal.

The ideal angle of trigonalbipyrmidal is [tex]109.5^{o}[/tex].

All three bonds of P-H has [tex]109.5^{o}[/tex].

Therefore, ideal bond angle is [tex]109.5^{o}[/tex].

The electron-group arrangement of PH3 is tetrahedral, the molecular shape is trigonal pyramidal, and the ideal bond angle is 104.5°.

In the case of PH3, the electron-group arrangement is tetrahedral and the molecular shape is trigonal pyramidal. The ideal bond angle for PH3 is 104.5°.

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You may find the structure of compounds as well the oxidation number of sulfur in each compound written with blue color. The nonbonded electrons are represented by black dots.

Explanation:

In the methyl sulfenic acid sulfur have a oxidation number of 0, because we have +1 from carbon and -1 from hydroxil group.

In the methyl sulfinic acid sulfur have a oxidation number of +2 , because we have +1 from carbon, -1 from hydroxil group and -2 from oxygen.

In the  methyl sulfonic acid sulfur have a oxidation number of +4 , because we have +1 from carbon, -1 from hydroxil group, -2 from one oxygen and -2 form the other oxygen.

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Final answer:

The structures and oxidation states of the three sulfur containing acids are: I. Methyl sulfenic acid with sulfur's oxidation state of +2, II. Methyl sulfinic acid with sulfur's oxidation state of +4, and III. Methyl sulfonic acid with sulfur's oxidation state of +6.

Explanation:

I. Methyl sulfenic acid:
Structure: CH3-S(=O)(=O)H
Oxidation state of sulfur: +2

II. Methyl sulfinic acid:
Structure: CH3-S(=O)(=O)-H
Oxidation state of sulfur: +4

III. Methyl sulfonic acid:
Structure: CH3-S(=O)(=O)-OH
Oxidation state of sulfur: +6

Answer:

Lewis structure is shown in the image below.

Explanation:

Acetate ion (CH₃COO⁻)

Valence electrons of carbon = 4

Valence electrons of oxygen = 6

Valence electrons of hydrogen = 1

Charge = 1 (Negative which means that the electrons are being added)

The total number of the valence electrons  = 2(4) + 2(6) + 3(1) + 1 = 24

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,  

The Lewis structure is shown in the image below.

The Lewis structure of the acetate ion consists of a carbon atom bonded to two oxygen atoms. There are also resonance structures with the double bond and a negative charge on different atoms. The Lewis dot structure includes lone pairs of electrons for each atom.

The Lewis structure of the acetate ion (CH3COO-) can be represented by drawing the individual Lewis structures for each resonance form. The main structure consists of a carbon atom single-bonded to two oxygen atoms, with one oxygen atom also bonded to a hydrogen atom. The remaining carbon atom is double-bonded to one of the oxygen atoms and also bonded to three hydrogen atoms. Each O atom has three lone pairs of electrons.

The resonance structures can be depicted by moving the double bond between the carbon atoms and by moving the lone pairs of electrons around. One of the resonance structures places a negative charge on the single-bonded oxygen, while the other resonance structure places the negative charge on the double-bonded oxygen.

The Lewis dot structure for the acetate ion includes lone pairs of electrons for all the atoms, as well as the bonding pairs of electrons representing the chemical bonds between the atoms. The oxygen atoms each have three lone pairs of electrons, the carbon atoms each have one lone pair of electrons, and the hydrogen atoms each have no lone pairs of electrons.

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Answer:

The anode half reaction is : [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

Explanation:

In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.

In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.

We can split the given cell reaction into two half-cell reaction such as-

Oxidation (anode): [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

Reduction (cathode): [tex]Sn^{4+}(aq.)+2e^{-}\rightarrow Sn^{2+}(aq.)[/tex]

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overall: [tex]Fe(s)+Sn^{4+}(aq.)\rightarrow Fe^{2+}(aq.)+Sn^{2+}(aq.)[/tex]

So the anode half reaction is : [tex]Fe(s)\rightarrow Fe^{2+}(aq.)+2e^{-}[/tex]

In the electrochemical cell with redox reaction, Sn4+ (aq) + Fe (s) → Sn2+ (aq) + Fe2+ (aq), the anode half-reaction is Fe (s) → Fe2+ (aq) + 2e-, as Fe is oxidized from 0 to +2.

In the given chemical reaction, Sn4+ (aq) + Fe (s) → Sn2+ (aq) + Fe2+ (aq), we see that tin (Sn) and iron (Fe) change their oxidation states. Tin is being reduced from an oxidation state of +4 to +2, and iron is being oxidized from 0 to +2.

Type of reaction happening at anode is always an oxidation. Oxidation is defined as a reaction where a substance loses electrons. Since Fe goes from Fe to Fe2+, it loses 2 electrons in the process. Therefore, the anode half-reaction for this electrochemical cell is: Fe (s) → Fe2+ (aq) + 2e-

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N-terminal Arg-Lys-Phe C-terminal

The universal codon code is attached for reference. The central dogma (DNA --> RNA--> Proteins) concept is used to translate the information in DNA into proteins.

Explanation:

We shall begin with the transcription of the DNA strand to form an mRNA. Remember that RNA has no Thiamine base but instead in its place is Uracil;

3' GCATTCAAG 5'       DNA strand

5’ CGUAAGUUC 3’  RNA strand

We can then begin translating the mRNA into protein using the universal codon code (attached). Remember a codon is a 3 nucleotide sequence that codes for a specific amino acid;

5’ CGUAAGUUC 3’     RNA strand

N-terminal Arg-Lys-Phe C-terminal   Peptide strand

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The DNA template strand 3' GCATTCAAG 5' can be transcribed into complementary DNA sequence 5' CGTAAGTTC 3'. This is then transcribed into mRNA sequence 5' CGUAAGUUC 3'. Using the genetic code, this is translated into the amino acid sequence: Arg‑Lys‑Phe.

To determine the amino acid sequence, we first need to establish the complementary DNA sequence for the provided DNA template strand 3' GCATTCAAG 5'. DNA complementary sequence pairs adenine (A) with thymine (T) and cytosine (C) with guanine (G) so we get the complementary DNA sequence (5' to 3' direction) as 5' CGTAAGTTC 3'.

Next, this DNA sequence is transcribed into mRNA using the similar pairing rules but thymine is replaced by uracil (U) in RNA. Thus, the mRNA sequence is 5' CGUAAGUUC 3'. Then we translate this mRNA into amino acids. The genetic code translates each mRNA codon into an amino acid. Therefore, we get the following amino acid sequence: Arg‑Lys‑Phe. Arg stands for Arginine, Lys for Lysine and Phe for Phenylalanine. These abbreviations are internationally accepted standard to represent the corresponding amino acids.

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Answer:

-105 kJ

Explanation:

The enthalpy change of a reaction is the sum of the energy of the bonds of the reactants and the products. The bonds at the reactants are being broken, so it's an endothermic reaction, so the bond energy must be positive.

The bonds at the products are being formed, so the process is exothermic, and the bond energy must be negative. There are being broken 1 N≡N bond and 3 H-H bonds, and are being formed 6 N-H bonds:

Reactants: 945 + 3*432 = 2241 kJ

Products: 6*(-391) = -2346 kJ

ΔH = 2241 - 2346

ΔH = -105 kJ

Explanation:

The given data is as follows.

        Space craft fuel rate = 353 L/min

As 1 liter equals 1000 ml and 1 min equals 60 seconds.

So,     [tex]353 \times \frac{1000 ml}{60 sec}[/tex]

           = 5883.33 ml/sec

It is also given that density of the fuel is 0.7 g/ml and standard enthalpy of combustion of fuel is -57.9 kJ/g.

Fuel rate per second is 5883.33 ml.

             [tex]5883.33 ml \times 0.7 g/ml[/tex]

               = 4118.33 g

Hence, calculate the maximum power as follows.

          Power = Fuel consumption rate × (-enthalpy of combustion)

                      = 4118.33 g/s \times 57.9 kJ/g

                      = 238451.36 kJ/s

or,                  = 238451.36 kW

Thus, we can conclude that maximum power produced by given spacecraft is 238451.36 kW.

Answer : The correct option is, (C) spontaneous only at low temperatures.

Explanation :

According to Gibb's 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

As we know that:

[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

The given chemical reaction is:

[tex]2S(s)+3O_2(g)\rightarrow 2SO_3(g)[/tex]

As we are given that, the given reaction is exothermic that means the enthalpy change is negative.

In this reaction, the randomness of reactant molecules are more and as we move towards the formation of product the randomness become less that means the degree of disorderedness decreases. So, the entropy will also decreases that means the change in entropy is negative.

Now we have to determine the spontaneity of this reaction when ΔH is negative and ΔS is negative.

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

[tex]\Delta G=(-ve)-T(-ve)[/tex]

[tex]\Delta G=(+ve)[/tex]   (at high temperature) (non-spontaneous)

[tex]\Delta G=(-ve)[/tex]   (at low temperature) (spontaneous)

Thus, the reaction is spontaneous only at low temperatures.

Answer:

A.) Rate=k[A]

Explanation:

The rate law has the following general form:

Rate = k . [A]ᵃ . [B]ᵇ

where,

k is the rate constant

[A] and [B] are the molar concentrations of the reactants

a and b are the reaction orders with respect to A and B

a + b is the overall order of reaction

The rate expression for a first order reaction could be

A.) Rate=k[A]          YES. The reaction order is 1.

B.) Rate=k[A]²[B]     NO. The reaction order is 2 + 1 = 3.

C.) Rate=k[A][B]      NO. The reaction order is 1 + 1 = 2.

D.) Rate=k[A]²[B]²  NO. The reaction order is 2 + 2 = 4

41Ca (Calcium 41) undergoes electron capture to become 41K (Potassium 41). This 41K then undergoes alpha decay to become 37Cl (Chlorine 37).

The starting nuclide is 41Ca (Calcium 41), which undergoes electron capture (EC). This process involves the nucleus of an atom drawing in a nearby electron and combining it with a proton to form a neutron. The result is a nuclide of an element with atomic number one less than the original. Therefore, 41Ca becomes 41K (Potassium 41).

Next, 41K undergoes alpha decay, a type of radioactive decay where an atomic nucleus emits an alpha particle (consisting of two protons and two neutrons - essentially a helium nucleus). This results in a decrease of both atomic and mass number (by 2 and 4 respectively). Applying this to 41K, we get 37Cl (Chlorine 37) as the product of the alpha decay.

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Answer:

[tex]HOCl[/tex], [tex]K_a=2.0\times 10^{-8}[/tex]

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,  

pH = - log [H⁺]

Strong acids dissociate completely and thus, 0.5 M of a solution of a strong acid yields a pH of 1.0 .

The expression of the pH of the calculation of weak acid is:-

[tex]pH=-log(\sqrt{k_a\times C})[/tex]

Where, C is the concentration = 0.5 M

Given, pH = 4.0

So, for [tex]HOCl[/tex], [tex]K_a=2.0\times 10^{-8}[/tex]

[tex]pH=-log(\sqrt{2.0\times 10^{-8}\times 0.5})[/tex]

pH = 4.0

Hence, the acid is HOCl.

Final answer:

The acid with a 0.50 M solution resulting in a pH of 4.0 is likely hydrofluoric acid (HF), due to its intermediate acid strength indicated by its Ka value of 6.8 x 10^-4.

Explanation:

The question involves determining the identity of an unknown acid based on its molarity and pH balance. A 0.50 M solution of the acid has a pH of 4.0. The pH is a measure of the hydrogen ion concentration in a solution, and the pKa value is the negative logarithm of the acid dissociation constant (Ka), which provides insight into the strength of an acid. Strong acids fully dissociate in water, resulting in a lower pH for a given concentration, while weak acids do not completely dissociate, leading to higher pH values.

Given the pH of 4.0 for a 0.5 M solution, we are looking for an acid with a pKa close to 4, since this would indicate a weak acid that only partially dissociates. Among the given options, HF with a Ka value of 6.8 x 10-4 would likely be the correct acid since it has an adequate strength to result in a pH of 4 for a 0.5 M solution. On the other hand, HBr is a strong acid and would yield a much lower pH for the same concentration, while HOCl and C6H5OH have much lower Ka values, meaning they are weaker acids and would result in a higher pH than 4.

0.2577 moles of ions

Explanation:

The dissociation of Rb₂SO₄ in water:

Rb₂SO₄ (s) + H₂O (l) → 2 Rb²⁺ (aq) + SO₄²⁻ (aq)

where:

(s) - solid

(l) - liquid

(aq) - aqueous

(aq) - aqueous

Knowing the dissociation of Rb₂SO₄ in water, we devise the following reasoning:

if         1 mol of Rb₂SO₄ dissociate in 2 moles of Rb²⁺ and 1 mole of SO₄²⁻

then   0.0859 moles of Rb₂SO₄ dissociate in X moles of Rb²⁺ and Y moles of SO₄²⁻

X = (0.0859 × 2) / 1 = 0.1718 moles of Rb²⁺

Y = (0.0859 × 1) / 1 = 0.0859 moles of SO₄²⁻

total moles of ions = moles of Rb²⁺ ions + moles of SO₄²⁻ ions

total moles of ions = 0.1718 + 0.0859 = 0.2577 moles

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Total moles of ions released when 0.0859 mol of Rb2SO4 dissolves completely in water are 0.2577 mol.

In the International System of Units, Mole is the base unit of the amount of any substance.

The dissociation of [tex]Rb_2SO_4[/tex] in water:

[tex]\rm Rb_2SO_4 (s) + H_2O (l) = 2 Rb^2^+ (aq) + SO_4^\;^2^- (aq)[/tex]

If 1 mol of [tex]Rb_2SO_4[/tex] dissociating into 2 moles of Rb²⁺ and 1 mole of SO₄²⁻

Then, 0.0859 mol of [tex]Rb_2SO_4[/tex] will dissociate into?

[tex]X = \dfrac{(0.0859 \times 2) }{1 } = 0.1718 moles\; of\; Rb^2^+[/tex]

[tex]Y= \dfrac{(0.0859 \times 2) }{1 } = 0.0859\; moles\; of\; So_4\;^2^-[/tex]

Total moles of ions  = moles of Rb²⁺ ions + moles of SO₄²⁻ ions

Total moles of ions = 0.1718 + 0.0859 = 0.2577 moles

Thus, the moles of ions are 0.2577 mol.

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Answer: - 521.6kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

[tex]H_2(g)+F_2(g)\rightarrow 2HF(g)[/tex]    [tex]\Delta H^0_1=-546.6kJ[/tex]   (1)

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(l)[/tex] [tex]\Delta H^0_2=-571.6kJ[/tex]  (2)

The final reaction is:

[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex]  [tex]\Delta H^0_3=?[/tex]   (3)

By multipling (1) by 2

[tex]2H_2(g)+2F_2(g)\rightarrow 4HF(g)[/tex]    [tex]\Delta H^0_1'=-2\times 546.6kJ=-1093.2kJ[/tex]  (1')

Subtracting (2) from (1')

[tex]2F_2(g)+2H_2O(l)\rightarrow 4HF(g)+O_2(g)[/tex]  [tex]\Delta H^0_3=?[/tex]   (3)

Hence [tex]\Delta H^0_3=\Delta H^0_1'-\Delta H^0_2=-1093.2-(-571.6)kJ=-521.6kJ[/tex].

The enthalpy for the reaction is -521.6kJ

Explanation:

1). The given data is as follows.

       [tex]T_{i} = 25.87^{o}C[/tex],      [tex]T_{f} = 38.13^{o}C[/tex]

               C = 5.73 [tex]kJ/^{o}C[/tex]

Hence, calculate the change in enthalpy of the reaction as follows.

      [tex]\Delta E_{rxn} = -C \times \Delta T[/tex]

                     = [tex]-5.73 \times (38.13 - 25.87)^{o}C[/tex]

                     = -70.25 KJ

As,  number of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                    = [tex]\frac{1.55}{(6 \times 12 + 14 \times 1)}[/tex]

                                    = 0.018 mol

Therefore, enthalpy of reaction in kJ/mol hexane is as follows.

            [tex]\Delta E_{rxn} = \frac{-70.25 KJ}{0.018 mol}[/tex]

                          = [tex]-3.90 \times 10^{3}[/tex] kJ/mol

Thus, we can conclude that [tex]\Delta E_{rxn}[/tex] for the reaction in kJ/mol hexane is [tex]-3.90 \times 10^{3}[/tex] kJ/mol .

2).  As we know that,

     Number of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]

                                  = [tex]\frac{1.55}{(7 \times 12 + 8 \times 1)}[/tex]

                                   = 0.017 mol

       [tex]\Delta E_{rxn} = \E_{rxn} per mol \times \text{number of moles}[/tex]

                      = [tex]-3.91 \times 10^{3} \times 0.017[/tex]

                      = -65.875 kJ

As,    [tex]\Delta E_{rxn} = C \times \Delta T [/tex]

                    -65.875 kJ = [tex]-C \times (37.57 - 23.12)^{o}C[/tex]

                           C = 4.56 [tex]kJ/^{o}C[/tex]

Hence, heat capacity of the bomb calorimeter is 4.56 [tex]kJ/^{o}C[/tex].

When 1.550 g of liquid hexane (C₆H₁₄) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 °C to 38.13 °C. The change in the internal energy of the reaction is -3.90 × 10³ kJ/mol.

When 1.55 g of toluene (C₇H₈) undergoes combustion in a bomb calorimeter, the temperature rises from 23.12 °C to 37.57 °C. The heat capacity of the bomb calorimeter is 4.55 kJ/°C.

First, we will calculate the moles corresponding to 1.550 g of hexane using its molar mass (86.18 g/mol).

[tex]n = 1.550 g \times \frac{1mol}{86.18 g} = 0.01799mol[/tex]

Then, we will calculate the heat (Q) absorbed by the bomb calorimeter using the following expression.

[tex]Q = C \times \Delta T = \frac{5.73kJ}{\° C} \times (38.13\° C - 25.87\° C) = 70.2 kJ[/tex]

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qb) and the heat released by the combustion (Qc) is zero.

[tex]Qb + Qc = 0\\\\Qc = -Qb = -70.2 kJ[/tex]

Finally, we will calculate the change in the internal energy of the reaction (ΔErxn) using the following expression.

[tex]\Delta Erxn = \frac{Qc}{n} = \frac{-70.2kJ}{0.01799mol} = -3.90 \times 10^{3} kJ/mol[/tex]

First, we will calculate the moles corresponding to 1.55 g of toluene using its molar mass (92.14 g/mol).

[tex]1.55 g \times \frac{1mol}{92.14g} = 0.0168 mol[/tex]

The combustion of toluene has a ΔErxn of –3.91 × 10³ kJ/mol. The heat released by the combustion of 0.0168 moles of toluene is:

[tex]0.0168 mol \times \frac{-3.91\times 10^{3}kJ }{mol} = -65.7 kJ[/tex]

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qb) and the heat released by the combustion (Qc) is zero.

[tex]Qb + Qc = 0\\\\Qb = -Qc = 65.7 kJ[/tex]

We can calculate the heat capacity of the bomb calorimeter (C) using the following expression.

[tex]C = \frac{Qb}{\Delta T } = \frac{65.7 kJ}{37.57 \° C - 23.12 \° C } = 4.55 kJ/ \° C[/tex]

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Explanation:

In the Figure can be seen the structure of a soap surfactant. It has two key parts: the hydrophobic tail and the hydrophilic head.

Surfactant molecules bond with each other forming miscelles, expossing the part that is compatible with the solvent and protecting the other one.

This means thar the polar head enables the surfactant to dissolve in water and the non-polar tail enables it to dissolve in non polar solvents such as oil mixtures.