Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties. Mg ( OH ) 2 ( s ) + 2 HCl ( g ) ⟶ MgCl 2 ( s ) + 2 H 2 O ( g )

Answer:

I) Solvent-solvent interactions involves B & F.

B. Interactions involving dipole-dipole attractions.

F. Interactions between the water molecules.

II) Solute-solute interactions involves A & D.

A. Interactions between the ions of sodium chloride.

D. Interactions involving ion-ion attractions.

III. Solute-solvent interactions involves G & C

G. Interactions formed between the sodium ions and the oxygen atoms of water molecules.

C. Interactions formed during hydration.

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

Answer:

25 moles

Explanation:

1 mole of nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia. 30 moles of hydrogen will react with 10 moles of nitrogen to give 20 moles of ammonia. As the actual yield is 50%, ammonia formed is 10 moles, the amount of nitrogen reacted is 5 moles, and the amount of hydrogen reacted is 15 moles. The mass of the remaining hydrogen is 15 moles and of the remaining nitrogen is 25 moles.

Answer:

4 MnO₄⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 1 OH⁻(aq) + 3 CH₃COO⁻(aq) + 4 H₂O(l)

Explanation:

To balance a redox reaction we use the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) ⟶ MnO₂(s)

Oxidation: CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq)

Step 2: Balance the mass adding H₂O and OH⁻ where necessary.

2 H₂O(l) + MnO₄⁻(aq) ⟶ MnO₂(s) + 4 OH⁻(aq)

5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l)

Step 3: Balance the charge adding eelctrons where necessary.

2 H₂O(l) + MnO₄⁻(aq) + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq)

5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l) + 4 e⁻

Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost are the same.

4 × (2 H₂O(l) + MnO₄⁻(aq) + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq))

3 × (5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l) + 4 e⁻)

Step 5: Add both half-reactions and cancel what is repeated.

8 H₂O(l) + 4 MnO₄⁻(aq) + 12 e⁻ + 15 OH⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 16 OH⁻(aq) + 3 CH₃COO⁻(aq) + 12 H₂O(l) + 12 e⁻

4 MnO₄⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 1 OH⁻(aq) + 3 CH₃COO⁻(aq) + 4 H₂O(l)

To balance the given reaction: CH3CH2OH(aq)+MnO−4(aq)⟶CH3COO−(aq)+MnO2(s), in a basic, aqueous solution, the balanced equation is 8CH3CH2OH(aq) + MnO−4(aq) + 8OH-(aq) ⟶ 8CH3COO-(aq) + MnO2(s) + 4H2O(l).

To complete and balance the given reaction: CH3CH2OH(aq)+MnO−4(aq)⟶CH3COO−(aq)+MnO2(s) in a basic, aqueous solution, we need to assign appropriate coefficients to each compound to ensure that the reaction is balanced. The balanced equation is:

8CH3CH2OH(aq) + MnO−4(aq) + 8OH-(aq) ⟶ 8CH3COO-(aq) + MnO2(s) + 4H2O(l)

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Answer:

1086.8 torr

Explanation:

Using Boyle's law  

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

Given ,  

V₁ = 0.671 L  

V₂ = 583 mL = 0.583 L ( 1 mL = 0.001 L )

P₁ = 1.24 atm

P₂ = ?

Using above equation as:

[tex]{P_1}\times {V_1}={P_2}\times {V_2}[/tex]

[tex]{1.24\ atm}\times {0.671\ L}={P_2}\times {0.583\ L}[/tex]

[tex]{P_2}=1.43\ atm[/tex]

The conversion of P(atm) to P(torr) is shown below:

[tex]P(atm)={760}\times P(torr)[/tex]

So,  

Pressure = 1.43*760 torr = 1086.8 torr

Answer: [tex]\Delta H=-16.5 kcal[/tex]

Explanation:

The balanced chemical reaction is,

[tex]C_6H_{12}O_6(s)\longrightarrow 2C_2H_5OH(l)+2CO_2(g)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{C_2H_5OH}\times \Delta H_{C_2H_5OH})+(n_{CO_2}\times \Delta H_{CO_2})]-[(n_{C_6H_{12}O_6}\times \Delta H_{C_6H_{12}O_6})][/tex]

where,

n = number of moles

Now put all the given values in this expression, we get

[tex]\Delta H=[(2\times -66.4)+(2\times -93.9)]-[(1\times -304.5)][/tex]

[tex]\Delta H=-16.5kcal[/tex]

Therefore, the enthalpy change for this reaction is -16.5 kcal

Answer:

+1, lose, 1, 4s, 4s and 3d

Explanation:

An element with the valence electron configuration 4s¹ would form a monatomic ion with a charge of +1. In order to form this ion, the element will lose 1 electron from the 4s subshell.

The corresponding oxidation reaction is:

K ⇒ K¹⁺ + 1 e⁻

[Ar] 4s¹ ⇒ [Ar]

If an element with the valence configuration 4s² 3d⁶ loses 3 electrons, these electrons would be removed from the 4s and 3d subshell(s).

The corresponding oxidation reaction is:

Fe ⇒ Fe³⁺ + 3 e⁻

[Ar] 4s² 3d⁶ ⇒ [Ar] 4s⁰ 3d⁵

An element with 4s1 valence configuration will form a +1 ion by losing 1 electron from 4s subshell. An element with 4s2 3d6 configuration will lose electrons firstly from 4s subshell and then from 3d, if it loses 3 electrons.

An element with the valence electron configuration 4s1 would form a monatomic ion with a charge of +1. In order to form this ion, the element will lose 1 electron from the 4s subshell.
Meanwhile, an element with the valence configuration 4s2 3d6 that loses 3 electrons will remove these electrons first from the 4s subshell (2 electrons), and then one from the 3d subshell. This is due to the fact that 4s electrons are generally removed before 3d electrons in transitional metals.

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Answer:

ΔH  = - 272 kJ

Explanation:

We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:

N2(g) +         3H2(g) →          2NH3(g)                 ΔH=−92.kJ  Multiplying by 2:      

2N2(g) +       6H2(g) →          4NH3(g)                        ΔH=− 184 kK

plus

4NH3(g) +     5O2(g) →          4NO(g) +6H2O(g)        ΔH=−905.kJ

__________________________________________________

2N2(g) +   6H2(g) + 5O2(g)→  4NO(g)  + 6H2O(g)      ΔH = (-184 +(-905 )) kJ

                                                                                     ΔH =    -1089 kJ

Notice how the intermediate NH3 cancels out.

As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH  for the formation  of one mol NO:

-1089 kJ/4 mol NO  x 1 mol NO =  -272 kJ (rounded to nearest kJ)

The net enthalpy change for the formation of one mole of nitric oxide from nitrogen, hydrogen, and oxygen is -318.25 kJ, which is obtained by summing up the enthalpy changes of the two relevant reactions and adjusting for the number of moles of NO produced.

To calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen, we necessitate to add the enthalpy changes of the two reactions. For the initial reaction, nitrogen and hydrogen react to produce ammonia, with ΔH = -92 kJ. In the second stage, ammonia reacts with oxygen to form nitric oxide and water, with ΔH = -905 kJ.

Since the second equation leads to formation of 4 moles of nitric oxide, we divide the enthalpy change for this reaction by 4 to get the enthalpy change for one mole of nitric oxide, which is -905/4 = -226.25 kJ.

The total enthalpy change for the formation of a mole of nitric oxide hence will be the sum of ΔH for the both reactions, i.e., -92 kJ + -226.25 kJ = -318.25 kJ.

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Answer: The molarity of KOH solution is 0.0663 M.

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of KHP = 0.4877 g

Molar mass of KHP = 204.22 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of KHP}=\frac{0.4877g}{204.22g/mol}=0.0024mol[/tex]

The chemical reaction for the formation of chromium oxide follows the equation:

[tex]KHC_8H_4O_4(aq.)+KOH\rightarrow K_2C_8H_4O_4(aq.)+H_2O(l)[/tex]

By Stoichiometry of the reaction:

1 mole of KHP reacts with 1 mole of KOH.

So, 0.0024 moles of KHP will react with = [tex]\frac{1}{1}\times 0.0024=0.0024mol[/tex] of KOH.

To calculate the molarity of KOH, we use the equation:

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

We are given:

Moles of KOH = 0.0024 moles

Volume of solution = 36.21 mL = 0.03621 L (Assuming)      (Conversion factor:  1L = 1000 mL)

Putting values in above equation, we get:

[tex]\text{Molarity of KOH }=\frac{0.0024mol}{0.03621L}=0.0663M[/tex]

Hence, the molarity of KOH solution is 0.0663 M.

Answer:

(a) The rate of disappearance of [tex]O_{2}[/tex] is: [tex]4.65*10^{-5}[/tex] M/s

(b) The value of rate constant is: 0.83036 [tex]M^{-2}s^{-1}[/tex]

(c) The units of rate constant is:  [tex]M^{-2}s^{-1}[/tex]

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

[tex]2NO(g)+O_{2}->2NO_{2}[/tex]

rate = [tex]-\frac{1}{2} \frac{d}{dt}[NO][/tex] = [tex]-\frac{d}{dt}[O_{2}][/tex] = [tex]\frac{1}{2}\frac{d}{dt}[NO_{2}][/tex] -----(1)

According to the question, the reaction is second order in NO and first order in  [tex]O_{2}[/tex].

Then we can say that, rate = k[tex][NO]^{2}[O_{2}][/tex] -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

[tex]-\frac{d}{dt}[NO][/tex] = [tex]9.3*10^{-5}[/tex] M/s.

(a) From (1), we can get the rate of disappearance of [tex]O_{2}[/tex].

    Rate of disappearance of  [tex]O_{2}[/tex] = [tex]-\frac{d}{dt}[O_{2}][/tex] = (0.5)*([tex]9.3*10^{-5}[/tex]) M/s = [tex]4.65*10^{-5}[/tex] M/s.

(b) The rate of the reaction can be obtained from (1).

    rate = [tex]-\frac{1}{2} \frac{d}{dt}[NO][/tex] = (0.5)*([tex]9.3*10^{-5}[/tex])

    rate = [tex]4.65*10^{-5}[/tex] M/s

   The value of rate constant can be obtained by using (2).

    rate constant = k = [tex]\frac{rate}{[NO]^{2}[O_{2}]}[/tex]

    k = [tex]\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}[/tex] = 0.83036 [tex]M^{-2}s^{-1}[/tex]

(c) The units of the rate constant can be obtained from (2).

    k = [tex]\frac{rate}{[NO]^{2}[O_{2}]}[/tex]

    Substituting the units of rate as M/s and concentrations as M, we get:

[tex]\frac{Ms^{-1} }{M^{3}}[/tex] = [tex]M^{-2}s^{-1}[/tex]

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

     [tex]rate\alpha [NO]^{2}[/tex]

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of [tex](1.8)^{2}[/tex] = 3.24

The rate of disappearance of O2 is 4.65×10−5 M/s, the rate constant is 0.19875 M^-1s^-1. The units of the rate constant for a third-order reaction are M^-1s^-1. If NO concentration were to increase by a factor 1.8, the reaction rate would increase by a factor of 3.24.

In a chemical reaction, the rate of disappearance of a reactant matches the stoichiometric ratio. Here, the ratio of NO to O2 is 2:1, therefore, the rate of disappearance of O2 is half that of NO. So, (a) the rate of disappearance of O2 is 9.3×10−5 M/s ÷ 2 = 4.65×10−5 M/s.

From the rate law, rate = k[NO]^2[O2], we can determine the rate constant (k) by substituting the known values into the equation. (b) the value of the rate constant k is rate ÷ ([NO]^2[O2]) = 9.3×10−5 M/s ÷ ((0.040 M)^2(0.035 M)) = 0.19875 M^-1s^-1.

As for (c) the units of the rate constant for a third-order reaction are M^-1s^-1. (d) If the concentration of NO were increased by a factor of 1.8, the rate would increase by a factor of (1.8)^2 due to the reaction being second order in NO. Hence the rate would increase by a factor of 3.24 times.

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Answer:

2.97 × 10¹³ g

Explanation:

First, we have to calculate the biomass the is burned. We can establish the following relations:

The biomass burned in the site of 400,000 acre is:

[tex]400,000acre\times\frac{10,000m^{2} }{2.47acre} \times \frac{10kgC}{m^{2} } \times 50\% = 8.10 \times 10^{9} kgC[/tex]

Let's consider the combustion of carbon.

C(s) + O₂(g) ⇒ CO₂(g)

We can establish the following relations:

The mass of  produced is CO₂:

[tex]8.10 \times 10^{12}gC \times \frac{1molC}{12.01gC} \times \frac{1molCO_{2}}{1molC} \times \frac{44.01gCO_{2}}{1molCO_{2}} =2.97 \times 10^{13} gCO_{2}[/tex]

Answer: [tex]CO_2(s)\rightarrow CO_2(g)[/tex]

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

[tex]\Delta S[/tex] is positive when randomness increases and [tex]\Delta S[/tex] is negative when randomness decreases.

a) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]

2 molesof gas are converting to 2 moles of another gas , thus [tex]\Delta S[/tex] is zero.

b) [tex]H_2O(s)\rightarrow H_2O(l)[/tex]

1 mole of solid is converting to 1 mole of liquid, the randomness increases and thus [tex]\Delta S[/tex] is positive.

b) [tex]Pb^{2+}(aq)+2Cl^-(aq)\rightarrow PbCl_2(s)[/tex]

2 moles of ions are converting to 1 mole of solid, the randomness decreases and thus [tex]\Delta S[/tex] is negative

d) [tex]CO_2(s)\rightarrow CO_2(g)[/tex]

1 mole of solid is converting to 1 mole of gas, the randomness increases drastically and thus [tex]\Delta S[/tex] is highly positive.

The answer is a. "The melting point of the fatty acid decreases as the number of double bonds increases" because when there are double bonded carbons it cancels the polarity of the carboxyl in the opposite direction, increasing that effect as double bonded carbons add to the molecule, giving the net dipole moment to the molecule. As the net dipole moment of the molecule decrease, the melting point of the fatty acid also decreases.

In addition, when the fatty acid is major than 8 carbons and have a double bond that prevents the fatty acid from crystal lattice formation, so the melting point will be lower also for this reason in this case.

Final answer:

The melting point of a fatty acid decreases as the number of double bonds increases because double bonds create kinks in the fatty acid chains, preventing tight packing and weakening intermolecular forces. The correct answer is option a.

Explanation:

For fatty acids with the same number of carbon atoms, the melting point changes as the number of double bonds in the fatty acid changes. The correct option is (a) - the melting point of the fatty acid decreases as the number of double bonds increases. This decrease in melting point occurs because unsaturated fatty acids, which have one or more double bonds, tend to have kinks or bends at the location of these bonds.

These kinks inhibit the fatty acids from packing closely together, resulting in weaker intermolecular Van der Waals forces (specifically London dispersion forces) and thus a lower melting point. In contrast, saturated fatty acids without double bonds have straight chains and can pack tightly together, leading to stronger intermolecular forces and higher melting points.

A good illustration of this concept is the substantial difference in melting points between palmitoleic acid, which contains a cis-double bond and melts over 60℃ lower than its saturated counterpart, palmitic acid. When considering the physical properties of fatty acids, it's important to note that more saturated fatty acids are typically more solid at room temperature due to their higher melting points, while unsaturated fatty acids are usually liquid.

The Lewis symbol for Ca is Ca²⁺, while F is F⁻. The Ca atom loses two electrons, and the F atom gains one electron. The most likely product is CaF₂.

(a) The Lewis symbol for Ca is Ca2+, while the Lewis symbol for F is F-. The action that occurs between Ca and F atoms is a redox reaction, where one atom loses electrons (oxidation) and the other gains electrons (reduction).
(b) The chemical formula of the most likely product is CaF2 since one Ca atom can bond with two F atoms.
(c) In the reaction, two electrons are transferred, with Ca losing two electrons and each F atom gaining one electron.
(d) Ca atom loses electrons in the reaction.

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Answer:

The final volume will be 24.7 cm³

Explanation:

Step 1: Data given:

Initial temperature = 180 °C

initial volume = 13 cm³ = 13 mL

The mixture is heated to a fina,l temperature of 587 °C

Pressure and amount = constant

Step 2: Calculate final volume

V1/T1 = V2/T2

with V1 = the initial volume V1 = 13 mL = 13*10^-3

with T1 = the initial temperature = 180 °C = 453 Kelvin

with V2 = the final volume = TO BE DETERMINED

with T2 = the final temperature = 587 °C = 860 Kelvin

V2 = (V1*T2)/T1

V2 = (13 mL *860 Kelvin) /453 Kelvin

V2 = 24.68 mL = 24.7 cm³

The final volume will be 24.7 cm³

Answer:

The correct option is: C.) Octahedral

Explanation:

Iodine pentafluoride (IF₅) is an inorganic interhalogen compound in which the oxidation state of iodine is +5 and the oxidation state of fluorine is (-1).

In this molecule, iodine is sp³d² hybridized and covalently bonded to five fluorine atoms. So there are 5 bond pair of electrons and 1 lone pair of electron around iodine.

Thus the steric number = 6

According to the VSEPR theory, the electron pair arrangement of a molecule with steric number 6 is octahedral.

Therefore, electron pair arrangement around iodine in IF₅ molecule is octahedral.

The molecule IF5 (Iodine Pentafluoride) has a square pyramidal electron pair arrangement around the central atom, Iodine, resulting from its electron configuration. In this arrangement, five fluorine atoms surround iodine in the shape of a pyramid with a square base. Therefore, the answer would be D) Square Pyramidal.

The molecule IF5 (Iodine Pentafluoride) follows the square pyramidal electron pair arrangement around the central atom (Iodine). In this arrangement, five Fluorine atoms are arranged around Iodine in the shape of a pyramid with a square base. This structure comes from the electron configuration of the central atom. Iodine has seven valence electrons, five of which are shared with the Fluorine atoms (forming covalent bonds), and the remaining two occupy a lone pair of non-bonding electrons. Therefore, the answer would be D) Square Pyramidal.

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Answer:

Negatively Charged: Beta - radiation

Neutral: Gamma radiation

Positively Charged: Beta + radiation.

Explanation:

The radioactive beam described is formed by three different types of particules:

Answer:

The heat of the given reaction is -23.0 kJ.

Explanation:

We are given with ;

[tex]NO(g) + NO_2(g)\rightarrow N_2O_3(g) \Delta H_{1,rxn}=-39.kJ[/tex]..[1]

[tex]NO(g) + NO_2(g) + O_2(g)\rightarrow N_2O_5(g), \Delta H_{2,rxn} = -112.5 kJ [/tex]..[2]

[tex]2NO_2(g) \rightarrow N_2O_4(g) ,\Delta H_{3,rxn} = -57.2 kJ [/tex]..[3]

[tex]2NO(g) + O_2(g)\rightarrow 2NO_2(g), \Delta H_{4,rxn} = -114.2 kJ[/tex]..[4]

[tex]N_2O_5(s)\rightarrow N_2O_5(g) ,\Delta H_{5,sub}= 54.1 kJ[/tex]..[5]

To find heat of reaction:

[tex]N_2O_3(g) + N_2O_5(s)\rightarrow 2N_2O_4(g),\Delta H_{6,rxn} = ?[/tex]..[6]

Using Hess's law:

[5] +2 × [3] + [4] - [2] - [1] = [6]

[tex]\Delta H_{6,rxn}=\Delta H_{5,sub}+2\times \Delta H_{3,rxn}+\Delta H_{4,rxn}-\Delta H_{2,rxn}-\Delta H_{1,rxn}[/tex]

[tex]\Delta H_{6,rxn}=54.1 kJ+(2\times (-57.2 kJ))+(-114.2 kJ)-(-112.5 kJ)-(-39.0 kJ)[/tex]

[tex]\Delta H_{6,rxn}=-23.0 kJ[/tex]

The heat of the given reaction is -23.0 kJ.

Answer:

2nd order reaction

Explanation:

Let us assume the reaction to be:

                         R → P

Where R is the reactant and P is the product.

So here, say initially we have "a" amount of reactant.

                        R → P

At t=0:             a      0       (initial condition)

At t=t:            a-x      x

Say x be the amount of reactant which forms the product in time t.

So from the rate law, we have

                 rate of decomposition = k (R)ⁿ

Where k is rate constant , R is amount of reactant at time t and n is the order of the reaction

From the question, at the instant when 5% and 20% have reacted, we will be left with 95% and 80% of the reactant respectively. So writing the rate law equation:

1.07 = k ( 95a / 100)ⁿ

0.76 = k ( 80a/100)ⁿ

Dividing these two equations, we get:

(1.07 / 0.76 ) = ( 95/80 )ⁿ

Taking logarithm on both sides we get

n = ( ㏒ (1.07 / 0.76) ) ÷ ㏒(95/80) = 2.0067 ≈ 2

Therefore the reaction is of order 2

“Vapor pressure increases with temperature” is the true statement from the given statements.

Option: A

Explanation:

Vapor pressure exerted in “thermodynamic equilibrium” with solid or liquid phase called “condensed phase” at a given temperature in packed or closed system. “Liquid particles” are arranged with more inter molecular space than solid and not as fixed as solid. Therefore when temperature is increased “kinetic energy” of the molecules also increases and thereby molecules transitioning into vapor also increases and in this way whole process is responsible for increase in “vapor pressure”.

Final answer:

The statement that vapor pressure increases with temperature is true because higher temperatures lead to increased kinetic energies, causing more molecules to escape into the vapor phase.

Explanation:

The correct statement among those listed is that vapor pressure increases with temperature. As the temperature rises, the kinetic energy of the molecules increases, leading to a greater number of molecules having enough energy to escape the liquid phase and enter the vapor phase, thereby increasing the vapor pressure. On the other hand, hydrogen bonds are not chemical bonds; they are strong intermolecular forces that occur between molecules, not within them as chemical bonds do.

Dispersion forces are generally weaker than dipole-dipole forces, and they increase with the mass and size of the molecules involved due to increased polarizability. Finally, intermolecular forces are responsible for the attractions between molecules, not for holding the atoms within a molecule together; that role is filled by intramolecular forces or chemical bonds.

Answer:

[tex]K_c=0.0867[/tex]

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{0.760}{1.50\ L}[/tex]

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

[tex]\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}[/tex]

Given:    

Equilibrium concentration of  O₂ = 0.130 mol

Volume = 1.50 L

[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]

[tex]Molarity=\frac{0.130}{1.50\ L}[/tex]

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

[tex]K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}[/tex]  

[tex]K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}[/tex]

[tex]K_c=0.0867[/tex]

Answer:

7 valence electrons

Explanation:

Astatine has the atomic number 85. Thus, its electron configuration is:

[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁵

As we can see, in the last level (6) it has 2 + 5 = 7 electrons, that is, astatine has 7 electrons in its valence shell. In the Lewis dot structure (attached) we write 3 pairs of electrons and 1 unpaired electron around the symbol of At.

Answer:

The answers are:

Purines:

C. contain four ring nitrogen atoms.

E. contain two heterocyclic rings.

Pyrimidines:

C. contain only two ring nitrogen atoms.

E. contain one heterocyclic ring.

Explanation:

Purines and Pyrimidines are nitrogenous bases which are the building blocks of nucleic acids (DNA and RNA).

Purines are composed by two fused heterocyclic rings, one of them is a 6-ring and the other is a 5-ring. Each ring contains two nitrogen atoms which form part of the ring. Thus, the nitrogen positions in purines are: 1', 3', 7' and 9'. Depending on the functional groups bonded to the two-ring structure, a purine base can be Guanidine (G) or Adenine (A).

The structure of Pyrimidines is a single heterocycle ring wich contains two nitrogen atoms in positions 1' and 3'. Depending of the functional groups, they can be: Cytosine (C), Thymidine (T) and Uracil (U, which is found in RNA).

Answer:

A. maintain electrical neutrality in the half-cells via migration of ions

Explanation:

Salt bridge -

For an electrochemical reaction , involving an anode and a cathode , both the electrodes are connect via a salt bridge to complete the circuit for the reaction .

One of the very important use of a salt bridge is to maintain the electrical neutrality of the respective half cells , which is achieved by the movement of ions .

Hence , from the given options , the correct option is ( a ) .

A salt bridge in an electrochemical cell helps maintain electrical neutrality. It does so by providing a pathway for ions to flow between the half-cells, balancing out the charge imbalance resulting from the reactions, thereby allowing the cell to continue to operate.

The purpose of the salt bridge in an electrochemical cell is to maintain electrical neutrality in the half-cells via the migration of ions. A salt bridge, often saturated with a salt solution, provides a pathway for ions to flow between the two half-cells. During the operation of the cell, the reactions produce or use ions in the solutions, which could result in a charge imbalance. The salt bridge helps to balance out these charges, keeping the solutions electrically neutral. This flow of ions helps complete the circuit, allowing the cell to continue to operate.

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Answer:

An organism that is completely composed D amino acids cannot survive.

Explanation:

Most of amino acids in all organism are present in L conformation.As result  in all organism the enzymes are specific for L amino acids but not for D amino acids.

   Bacterial cell wall contain some D amino acids such as D glutamate,D alanine but not entirely composed of D amino acids.

428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).

Let's consider the following thermochemical equation.

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(l)  ΔH° = 1168 kJ

The standard enthalpy of the reaction is positive, which means that the reaction is endothermic, that is, heat is absorbed.

We will convert 25.00 g of NH₃ to moles using its molar mass (17.03 g/mol).

[tex]25.00 g \times \frac{1mol}{17.03g} = 1.468 mol[/tex]

1168 kJ of heat are absorbed when 4 moles of NH₃ react. The heat absorbed when 1.468 moles of NH₃ react is:

[tex]1.468 mol \times \frac{1168kJ}{4mol} = 428.7 kJ[/tex]

428.7 kJ of heat are absorbed when 25.00 g of NH₃(g) reacts in the presence of excess O₂(g) to produce NO(g) and H₂O(l).

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Chemical reactions are defined as the reaction in which two or more reactants combine to form single or more products. The heat can be absorbed or evolved in the reaction. The heat absorbed is 428.7kJ when ammonia reacts with excess oxygen.

Given that,

The chemical equation between ammonia and oxygen is:

Now, the enthalpy of the reaction is positive, such that the reaction is endothermic. In endothermic reaction heat is absorbed.

Now, converting the mass of ammonia into moles, we get:

1168 kJ of heat is absorbed when 4 moles of ammonia react with oxygen. The heat absorbed in 1.468 moles, will be:

Therefore, when 25 grams of ammonia reacts in the presence of oxygen 428.7 kJ will be absorbed. Hence, it is an endothermic reaction.

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Answer:

All of these indicate plates exist and move.

Explanation:

Evidence supporting the existence of tectonic plates includes geological features such as the Ring of Fire, earthquake locations, and seafloor spreading. Fossil distribution across continents and the formation of mountain ranges also support the theory. Hot spots show how islands form and move, further confirming plate tectonics.

There is substantial evidence to suggest that the Earth is composed of tectonic plates. Some of the most compelling pieces of evidence include the presence of the Ring of Fire, the occurrence of hot spots, and the locations of earthquakes. These features highlight areas where plates interact. Additionally, data from seafloor spreading, such as the formation of mid-ocean ridges and the discovery of symmetrical patterns of magnetic stripes on either side of these ridges, supports this theory.

The presence of fossils provides further proof. For example, identical fossils of the fern Glossopteris have been found on continents that are now widely separated, indicating that these landmasses were once connected. Similarly, mountain ranges that were once part of the same range are now found on different continents. These phenomena can be explained by the movement of tectonic plates.

Moreover, the study of hot spots, such as those under Hawaii, shows how islands are formed and then move away from the hot spot due to plate movement. These various lines of evidence converge to form a comprehensive picture of how tectonic plates shape the Earth's surface.

Answer:

Longwave ridges = Global scale

High pressure system persists over the central plains for 4 days = Synoptic scale

Supercell thunderstorm that lasts for over an hour = mesoscale

A turbulent eddy = microscale

Explanation:

Global scale has a range of the entire earth

synoptic scale has a range of about 100-1000km and can last fro days to weeks

mesoscale has a range of 4-100km and lasts for a day maximum

microscale is the least of all atmospheric motions.

Answer:

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

Explanation:

Step 1: Data given

ΔH = −320.1 kJ/mol

ΔS = −86.00 J/K · mol.

Step 2: Calculate the temperature

ΔG<0 = spontaneous

ΔG= ΔH - TΔS

ΔH - TΔS  <0

-320100 - T*(-86) <0

-320100 +86T < 0

-320100 < -86T

320100/86 > T

3722.1 > T

The temperature should be lower than 3722.1 Kelvin (= 3448.9 °C)

We can prove this with Temperature T = 3730 K

-320100 -3730*(-86) <0

-320100 + 320780  = 680 this is greater than 0 so it's non spontaneous

T = 3700 K

-320100 -3700*(-86) <0

-320100 + 318200  = -1900 this is lower than 0 so it's spontaneous

The temperature is quite high because of the big difference between ΔH and ΔS.

This reaction is spontaneous for temperature lower than 3722.1 Kelvin or 3448.95 °C

Answer:

Ksp for PbCl2 is 1.7 *10^-5

Explanation:

Step 1: Data given

The concentration of Pb2+ ion in the solution was found to be 1.62 *10^−2 M

Step 2: The balanced equation

PbCl2 ⇔Pb^2+ + 2Cl -

Step 2: ICE chart

The initial concentration of Pb2+ is 0 M

There will react X M  and 2X of Cl-

At the equillibrium there is X M  of Pb^2+ and 2X M of Cl-

The concentration of Pb2+ ion in the solution was found to be  1.62 *10^−2  M

Step 3: Calculate Ksp

Since PbCl2 is solid, it doesn't aply for Ksp

Ksp = [Pb^2+][Cl-]²

Ksp = X*(2X)²

Ksp = 4X³

  ⇒ X = 1.62 *10^-2 M

Ksp = 4*( 1.62 *10^-2)³

Ksp =1.7 *10^-5

Ksp for PbCl2 is 1.7 *10^-5

The change in entropy, given the 10 kJ of energy needed to melt the ice sculpture at room temperature, works out to be 1.22x10³ J/K. This underscores the substantial amounts of energy required for phase changes.

Given the problem at hand, the energy required to melt the ice sculpture is given by Q = 10 kJ, and the melting temperature of ice (T) is 273 K. The change in entropy can be calculated using the formula ΔS = Q/T. Substituting the provided values, the change in entropy works out to be 1.22x10³ J/K. This calculation emphasizes the concept that phase changes require significant energy. For instance, the energy needed to turn ice into liquid water, as in this example, is immense compared to the energy associated with mere temperature changes.

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The change in entropy for the melting ice sculpture at room temperature is calculated to be 34.1 J/K. This represents an increase in disorder due to the melting, an endothermic process that breaks bonds in the ice, allowing the molecules to move freely.

The calculation of the change in entropy for the melting ice sculpture is based on the formula ΔS = Q/T, where Q is the heat required for the phase change (melting), and T is the absolute temperature in Kelvin during the process. Given that Q is 10 kJ (or 10,000 J) and T is room temperature, which is approximately 293 K, we substitute these values into the formula.

So, ΔS = Q/T = 10000 J / 293 K = 34.1 J/K.

This represents the change in entropy of the surroundings when the ice sculpture melts at room temperature. It's important to note that the process of melting is an endothermic process requiring energy to break the bonds in the ice, which contributes to this increase in entropy: the system becomes more disordered as the structured ice lattice becomes freely moving water molecules.

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