At 40 C H2O2(aq) will decompose according to the following reaction:

2H2O2 (aq) -----> 2H2O (l) + O2 (g)

The following data were collected for the concentration of H2O2 at various times:
Time (s) H2O2 (mol/L)
1.000 2.16 x 10^4
0.500 4.32 x 10^4

(a) calculate the average rate of decomposition of H2O2 between0 and 2.16 x 104 s. Use this rate to calculate the averagerate of production of O2 (g) over the same time period.
(b) what are these rates for the time period 2.16 x 104 s to4.32 x 104 s ?

Answer:

The partial pressure of krypton in the flask is 0.59 atm and the total pressure in the flask is 1.39 atm

Explanation:

This must be solved with the Ideal Gas Law equation.

First of all we need the moles or Ar and Kr in the mixture

Moles = Mass / Molar mass

Molar mass Ar 39.95g/m

Moles Ar = 6.18 g/39.95 g/m → 0.154 moles

Molar mass Kr 83.8 g/m

Moles Kr = 9.66 g/ 83.8g/m → 0.115 moles

Total moles in the mixture: 0.154 moles + 0.115 moles = 0.269moles

Now, we have the total moles, we can calculate the total pressure.

P . V = n . R . T

(T° in K = T° in C + 273)

P. 5.38L = 0.269mol . 0.082 L.atm/mol.K . 340K

P = (0.269mol . 0.082 L.atm/mol.K . 340K) / 5.38 L

P = 1.39 atm

Now we have the total pressure, we can apply molar fraction so we can know the partial pressure of Kr.

Kr pressure / Total Pressure = Kr moles / Total moles

Kr pressure / 1.39 atm = 0.115 moles / 0.269 moles

Kr pressure = (0.115 moles / 0.269 moles) / 1.39atm

Kr pressure = 0.59 atm

Answer:

[H3O+] = 0.0117 M

[OH-] = 8.5 * 10^-13 M

Explanation:

Step 1: Data given

Concentration of HCl = 0.0117 M

Step 2:

HCl is a strong acid

pH of a strong acid = -log[H+] = - log[H3O+]

[H3O+] = 0.0117 M

pH = -log(0.0117)

pH = 1.93

pOH =14 - 1.93 = 12.07

pOH = -log[OH+] = 12.07

[OH-] = 10^-12.07 = 8.5 * 10^-13

Or

Kw / [H3O+] = [ OH-]

10^-14 / 0.0117 = 8.5*10^-13

Final answer:

The hydronium ion concentration [H3O+] in a 0.0117 M HCl solution is 0.0117 M, and the hydroxide ion concentration [OH-] is calculated to be 8.55 × 10^-13 M using the water dissociation constant.

Explanation:

To calculate the hydronium ion concentration, [H3O+], of a 0.0117 M HCl solution, we first need to understand that HCl is a strong acid which dissociates completely in water. This means that for every mole of HCl dissolved, there will be one mole of H3O+ ions in solution. Hence, the hydronium ion concentration in a 0.0117 M HCl solution is also 0.0117 M.

As for the hydroxide ion concentration, [OH-], we use the water dissociation constant (Kw), which is 1.0 × 10-14 at 25 °C. The product of the concentrations of the hydronium and hydroxide ions in any aqueous solution is equal to Kw. We can find [OH-] by rearranging the expression Kw = [H3O+][OH-] to solve for [OH-], giving us [OH-] = Kw / [H3O+]. Substituting in the values we get [OH-] = 1.0 × 10-14 M / 0.0117 M = 8.55 × 10-13 M.

Explanation:

The heat gained by the solution = q

[tex]q=mc\times (T_{final}-T_{initial})[/tex]

where,

q = heat gained = ?

c = specific heat of solution= [tex]4.18 J/^oC[/tex]

Mass of the solution(m) = mass of water + mass of calcium chloride

Mass of water = ?

Volume of water = 50.00 mL

Density of water = 1.00 g/mL

Mass = Density × Volume

m = 1.00 g/mL × 50.00 mL = 50.00 g

Mass of the solution (m) = 50.00 g + 4.51 g =54.51 g

[tex]T_{final}[/tex] = final temperature = [tex]25.8 ^oC[/tex]

[tex]T_{initial}[/tex] = initial temperature = [tex]22.6 ^oC[/tex]

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

[tex]q=54.51 g\times 4.18 J/g^oC\times (25.8-22.6)^oC[/tex]

[tex]q=729.126 J[/tex]

The heat gained by the solution is 729.126 J.

Heat energy released during the reaction  = q'

q' = -q  ( law of conservation of energy)

q' = -729.126 J

The heat energy released during the reaction is -729.126 J.

Moles of calcium chloride, n = [tex]\frac{4.51 g}{111 g/mol}=0.04063 mol[/tex]

[tex]\Delta H_{rxn}=\frac{q'}{n}=\frac{-729.126 J}{0.04063 mol}=-17,945.23 J/mol= -17.945 kJ/mol[/tex]

The ΔH of the reaction is -17.945 kJ/mol.

Answer:

- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5

- There are transferred 5 moles of e-

Explanation:

This is the reaction:

Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)

Let's think the oxidation numbers:

Fe2+ changed to Fe3+

It has increased the oxidation number → OXIDATION

Mn in MnO₄⁻ acts with +7 and it changed to Mn²⁺

It has decreased the oxidation number → REDUCTION

Let's make the half reactions:

Fe²⁺ → Fe³⁺  +  1e⁻    (it has lost 1 mol of e⁻)

MnO₄⁻ + 5e⁻ →  Mn²⁺  (it has gained 5 mol of e⁻)

Now we have to ballance the O. In acidic medium we complete with water as many oxygens we have, in the opposite side. We have 4 O in reactant side, so we fill up with 4 H2O in products side.

MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Now we have to ballance the H, so as we have 8 H in products side, we complete with 8H⁺ in reactants, this is the complete half reaction:

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Notice that have 1e⁻ in oxidation and 5e⁻ in reduction. We must multiply (x5) the half reaction of oxidation, so the electrons can be cancelled.

(Fe²⁺ → Fe³⁺  +  1e⁻ ) .5

5Fe²⁺ → 5Fe³⁺  +  5e⁻  

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

We sum both half reactions:

5Fe²⁺  + 8H⁺  + MnO₄⁻ + 5e⁻ →  5Fe³⁺  +  5e⁻   + Mn²⁺  + 4H₂O

The electrons are cancelled, so the ballanced reaction is this:

5Fe²⁺  + 8H⁺  + MnO₄⁻  →  5Fe³⁺  + Mn²⁺  + 4H₂O

The given redox reaction balances to 5Fe2+(aq) + MnO4⁻(aq) + 8H⁺(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l), with coefficients of 8 for H⁺ and 5 for Fe3+ in the balanced reaction. A total of 5 moles of electrons are transferred.

To balance the given redox reaction, you need to separate the two half-reactions, one for oxidation and one for reduction. In the case of Fe2+(aq) + MnO4⁻(aq) → Fe3+(aq) + Mn2+(aq), the oxidation half-reaction is Fe2+(aq) → Fe3+(aq) + e⁻, and the reduction half-reaction is MnO4⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn2+(aq) + 4H2O(l).

After balancing the half-reactions, you can add them together to get 5Fe2+(aq) + MnO4⁻(aq) + 8H⁺(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l).  This reaction shows that the coefficients in front of H⁺ and Fe3+ in the balanced reaction are 8 and 5, respectively. Also, the stoichiometric coefficients in front of the electrons in the two half-reactions indicate that a total of 5 moles of electrons are transferred during the reaction.

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Answer:

The correct answer is 1 the total number of proton plus neutrons in the products and reactants must be same.

Explanation:

During a radioactive reaction the nucleus of the radioactive atom undergo disintegration or breakdown to form new nucleus.

   Radioactive disintegration deals with the emission of α particle or β particle from a radioactive nucleus resulting in the formation of new nucleus.Although new nucleus is formed the total number of proton plus neutrons in product and reactant must be same.

A  rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.

A nuclear equation for a radioactive reaction, shows the reactants and product, where the proton number and atomic mass are conserved, unlike in chemical equations.

In a radioactive decay, there is emission of α particle or β particle from a radioactive nucleus to form a new nucleus.

Therefore, a rule for balancing a nuclear equation is: I. total number of protons + neutrons in the products and reactants must be the same.

Learn more about nuclear equation on:

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The theoretical yield of Li3N is 1.24 mol or 138.16 g. The percent yield of the reaction is 4.23%.

To find the theoretical yield of Li3N, we first need to determine the limiting reactant. This is done by comparing the moles of each reactant to the stoichiometric ratio of the balanced equation. The molar mass of Li is 6.94 g/mol and the molar mass of N2 is 28.02 g/mol. First, convert the given masses of Li and N2 to moles using their respective molar masses:

12.7 g Li X (1 mol Li / 6.94 g Li) = 1.83 mol Li

34.7 g N2 X (1 mol N2 / 28.02 g N2) = 1.24 mol N2

Then, divide the number of moles of each reactant by its stoichiometric coefficient to find the mole ratio:

1.83 mol Li / 6 = 0.305 mol Li3N

1.24 mol N2 / 1 = 1.24 mol Li3N

Since N2 gives a smaller value, it is the limiting reactant. Therefore, the theoretical yield of Li3N is 1.24 mol or 138.16 g.

To calculate the percent yield, divide the actual yield by the theoretical yield, then multiply by 100:

Percent yield = (Actual yield / Theoretical yield) X 100

Percent yield = (5.85 g / 138.16 g) X 100 = 4.23%

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Answer:

The correct option is: D) C (graphite)+ 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)

Explanation:

The standard enthalpy change of a given chemical reaction ([tex]\Delta H^{\circ }_{r}[/tex]) is equal to the difference of the sum of formation enthalpy ([tex]\Delta H^{\circ }_{f}[/tex]) of products and the sum of formation enthalpy of reactants.

[tex]\Delta H^{\circ }_{r} = \sum v. \Delta H^{\circ }_{f}(products) - \sum v. \Delta H^{\circ }_{f}(reactants)[/tex]

The standard enthalpy of formation is zero for elements that exist in their standard states.

The standard enthalpy change of a reaction is equal to the standard enthalpy of formation of methanol (l) [[tex]\Delta H^{\circ }_{r} = \Delta H^{\circ }_{f}(CH_{3}OH, l)[/tex]], only if:

1. The standard enthalpy of all the other reactants and products should be zero, i.e. all the other reactants and products should be in their standard states.

2. The stoichiometric coefficient of methanol (CH₃OH, l) should be 1.

Among the given options, only the option D satisfies both the conditions.

D) C (graphite)+ 2H₂ (g) + 1/2 O₂ (g)→ CH₃OH (l)

The standard enthalpy change for this chemical reaction:

[tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ] - \left [\Delta H^{\circ }_{f}(C, s, graphite) + \Delta H^{\circ }_{f}(H_{2}, g) + \Delta H^{\circ }_{f}(O_{2}, g)\right ][/tex]

The standard enthalpy of formation:

[tex]\Delta H^{\circ }_{f}(C, s, graphite) = 0 kJ/mol[/tex]

[tex]\Delta H^{\circ }_{f}(H_{2}, g) = 0 kJ/mol[/tex]

[tex]\Delta H^{\circ }_{f}(O_{2}, g) = 0 kJ/mol[/tex]

⇒ [tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ] - \left [0 kJ/mol + 0 kJ/mol + 0 kJ/mol\right ] [/tex]

⇒ [tex]\Delta H^{\circ }_{r} = \left [\Delta H^{\circ }_{f}(CH_{3}OH, l)\right ][/tex]

Therefore, the correct option is (D).

Answer:

-79.8 × 10⁴ J/mol

Explanation:

Arsine, AsH₃, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressure is 35 Torr at 111.95 °C and 253 Torr at 83.6 °C.

Then,

P₁ = 35 torr

T₁ = 111.95 + 273.15 = 385.10 K

P₂ = 253 torr

T₂ = 83.6 + 273.15 = 356.8 K

We can calculate the standard enthalpy of vaporization (ΔH°vap) using the two-point Clausius-Clapeyron equation.

[tex]ln(\frac{P_{2}}{P_{1}} )=\frac{-\Delta H\°_{vap}}{R} .(\frac{1}{T_{2}} -\frac{1}{T_{1}} )[/tex]

where,

R is the ideal gas constant

[tex]ln(\frac{253torr}{35torr})=\frac{-\Delta H\°_{vap}}{8.314J/K.mol} .(\frac{1}{356.8K}-\frac{1}{385.10K})\\ \Delta H\°_{vap}=-79.8 \times 10^{4} J/mol[/tex]

Answer:

To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.

Explanation:

A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:

2x*50% + x*30% + y*10% = 50L*32%

130x + 10y = 1600 (1)

-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-

Also, it is possible to write a formula using the total volume (50L), thus:

2x + x +y = 50L

3x + y = 50L (2)

If you replace (2) in (1):

130x + 10(50-3x) = 1600

100x + 500 = 1600

100x = 1100

x = 11L -Volume of 30% solution-

2x = 22L -Volume of 50% solution-

50L - 22L - 11L = 17 L -Volume of 10% solution-

I hope it helps!

Final answer:

To create a 50-liter mixture with 32% acid concentration, the chemist should use 20 liters of the 10% acid solution, 10 liters of the 30% acid solution, and 20 liters of the 50% acid solution.

Explanation:

The question involves solving a system of linear equations to determine the volume of each acid solution required to create a 50-liter mixture with a 32% acid concentration. Let's denote the amount of the 10% solution as x liters, the 30% solution as y liters, and the 50% solution as 2y liters (since it's twice the amount of the 30% solution).

The total amount of acid should be 32% of the 50-liter mixture: 0.10x + 0.30y + 0.50(2y) = 0.32 × 50.

We've also established the relationship between the 30% and 50% solutions: 2y is twice the amount of y.

x + 3y = 50

0.10x + 0.30y + 1.00y = 16

Combining and rearranging gives us x = 20 liters, y = 10 liters, and thus 2y = 20 liters. Therefore, to obtain the desired mixture, the chemist should use 20 liters of the 10% solution, 10 liters of the 30% solution, and 20 liters of the 50% solution.

Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)[/tex]  [tex]\Delta H=-2220.1kJ/mol[/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_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})][/tex]

where,

n = number of moles

[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero

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

[tex]-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})][/tex]

[tex]\Delta H_{C_3H_8}=72.4kJ/mol[/tex]

Therefore, the heat of formation of propane is 72.4 kJ/mol

The heat of formation of propane can be found using Hess's Law and the given enthalpy formations of CO2 and H2O, along with the enthalpy combustion of propane. Calculations reveal a heat of formation for propane of -103.85 kJ/mol.

The heat of formation of propane can be calculated using Hess's Law, which states that the enthalpy change in a chemical reaction is independent of the pathway between the initial and final states. In this case, we know the enthalpy of combustion for propane (C3H8) is -2220.1kJ/mol, and that the standard enthalpy of formation (ΔHf°) for CO2 and H2O are -393.5kJ/mol and -241.8kJ/mol, respectively.

Therefore, by rearranging the equation ΔH(combustion) = Σ(ΔHf° products) - Σ(ΔHf° reactants), we can solve for the ΔHf° of propane.

Inserting the given values into the equation, -2220.1 kJ/mol = [ 3(-393.5 kJ/mol) + 4(-241.8 kJ/mol)] - [x + 5(0)], where x represents the ΔHf° of propane, and '5(0)' is the contribution from the oxygen, which is zero because the change in enthalpy formation for an element in its stable form is zero. Thus, solving for x, you get a heat of formation for propane of -103.85 kJ/mol.

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Answer:

See the figure

Explanation:

As the first step, the Cl leaves the molecule and the double bond is moved to obtain a tertiary carbocation (the most stable one). Then the molecule of water attacks the primary carbon of the double bond, therefore the double bond moves again to the initial place. Finally, a hydronium ion is produced to remove the positive charge of the oxygen.

a. 2257.7 kW  b. 6.057kw/K

Steam existing at 100KPa with t = 20C, we can obtain values of enthalpy and entropy from table

h1 = hf = 417.4 kJ/kg

s1 = 1.302 KJ/kg.K

h2 = hg = 2675.1 kJ/kg

s2 = 7.359 kJ/K

a. Electrical Power consumption of system is given by Wel = mass x Change in Enthalpy

Wel = 1 x (h2 - h1) = 1 x (2675.1 - 417.4) = 2257.7 kw

b . Entropy Generation is given by Change in Entropy  = Entropy generation - heat dissipated/temperature, since we have a well insulated system with no losses, q =0, hence

ΔS = Sgen + 0/T = Sgen

m(S2 - S1) = Sgen

Sgen = 1 x (7.359 - 1.302) = 6.057 kW/K

All solutions A through E, being strong acids at the same concentration, will have virtually the same pH value under the same conditions. None will have a significantly higher pH. In general, the solution with the highest pH would be the one derived from the weakest acid or strongest base.

The pH scale is used to determine the acidity or basicity of a solution. The solutions given are all acidic, being solutions of hydrobromic acid (HBr), hydroiodic acid (HI), hydrofluoric acid (HF), hydrochloric acid (HCl), and perchloric acid (HClO4), respectively. Each of these acids dissociates in water to a different extent, affecting the pH of the solution. However, common to all of these acids is that they are strong acids and will fully dissociate in water.

In this case, all solutions have the same molarity (concentration) of 0.10 M. Because all the acids are strong acids and will fully dissociate, the pH of these solutions will be virtually the same. Therefore, none of the solutions will have a significantly higher pH than the others if measured under the same conditions.

In general, to identify the solution with the highest pH (or lowest acidity), you would look for the weakest acid or the strongest base out of the options provided.

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Among the given solutions, C) 0.10 M HF(aq) has the highest pH because HF is a weak acid and only partially dissociates in water, resulting in a higher pH compared to the strong acids HCl, HI, HBr, and HClO₄.

To determine which solution has the highest pH, we need to understand that pH is a measure of the hydrogen ion concentration in a solution. A lower pH signifies a higher hydrogen ion concentration. Since all the given solutions are acids, we must determine which one produces the least amount of hydrogen ions (H⁺).

HCl, HI, HBr, and HClO₄ are strong acids, meaning they fully dissociate in water. This full dissociation results in a high concentration of H⁺ ions, giving them all a pH of 1 for a 0.10 M solution:

HCl: pH = 1.0

HI: pH = 1.0

HBr: pH = 1.0

HClO4: pH = 1.0

HF, however, is a weak acid and only partially dissociates in water. This partial dissociation results in fewer H+ ions compared to the strong acids listed. Consequently, the pH of a 0.10 M HF solution will be higher (less acidic) than the pH of the other solutions.

Given this information, the solution that has the highest pH is:

C) 0.10 M HF(aq)

The product of the given reaction, an acid-catalyzed hydration of an alkene, would be 2-butanol, as per Markovnikov's Rule. This rule predicts the placement of the hydrogen and halide groups in a reaction.

The reaction given is an example of an acid-catalyzed hydration of an alkene. In such reactions, an alkene reacts with water in the presence of an acid (in this case, H3PO4) to form an alcohol. The prediction of the product involves recognizing the reaction type and the reagents involved.

For the given reaction: CH3CH=CHCH3 + H2OH3PO4⟶, the product would be 2-butanol.

But why does this process form 2-butanol? It all comes down to Markovnikov's rule, which predicts that in the addition of a protic acid HX to an alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents.

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NO2Cl is a polar molecule because the electronegativity difference between atoms of the molecule, being the chloride tho most electronegative. This fact and the asymmetry of the molecule makes a dipole moment.

In order to don’t include formal charges in the drawing you can use the image number 1. Nevertheless is more usual to use a lewis structure like image number 2.

Answer:

Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Explanation:

In the anode takes place the oxidation, in which the reducing agent loses electrons.

In the cathode takes place the reduction, in which the oxidizing agent gains electrons.

The corresponding half-reactions are:

Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻

Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)

Answer:

CCl₄, because it is the heaviest compound.

Explanation:

When a liquid is in a closed container, evaporation occurs, in what the molecules with the highest kinetic energy can scape of the liquid. The vapor that was formed does pressure on the liquid that remains, and when both phases stay in equilibrium the pressure is called vapor pressure.

We can notice that the vapor pressure is a measure of the volatility of a liquid. Substances with higher vapor pressure are more volatile. The volatility, however, depends on the nature of the forces in the compound and on the molar mass of it.

For the substances given, they are all covalent compounds and have dipole-induced-dipole-induced bonds between the molecules (because they are nonpolar). So, the lowest vapor pressure is in the heaviest compound, which is the most substituted: CCl₄.

Final answer:

Carbon tetrachloride (CCl4) has the lowest vapor pressure at room temperature due to its highest molecular weight and strongest intermolecular forces among the products of the reaction between methane and chlorine.

Explanation:

Methane, CH4, reacts with chlorine, Cl2, to produce chlorinated hydrocarbons such as methy chloride (CH3Cl), methylene chloride (CH2Cl2), chloroform (CHCl3), and carbon tetrachloride (CCl4). Looking at the structures of these molecules, we can see that carbon tetrachloride (CCl4) has the highest number of chlorine atoms, which are highly electronegative and contribute to more significant intermolecular forces like London dispersion forces. Consequently, the compound with the highest molecular weight and strongest intermolecular forces will exhibit the lowest vapor pressure at room temperature, which, in this case, is carbon tetrachloride (CCl4).

Answer:

The IUPAC of 1-pentanone is not correct.

Explanation:

The structures of each of the given compounds with correct IUPAC names are shown in the figure.

In case of ketones there must be two alkyl groups attached to the carbonyl carbon. In case of aldehydes there must be atleast one hydrogen attached to carbonyl carbon.

Now if we are saying that 1-pentanone, it means we are actually naming the compound with wrong functional group. This is infact an aldehyde with the correct name as "pentanal".

Rest of the given IUPAC names are correct.

Answer:

3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g)    ΔH° = -6.00 kJ

Explanation:

Let's consider the following balanced equation.

3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g)

When 1 mole of Fe₂O₃(s) reacts, 2.00 kJ of energy are evolved. Energy is an extensive property. In the balanced equation there are 3 moles of Fe₂O₃(s), so the evolved energy is:

[tex]3molFe_{2}O_{3}.\frac{2.00kJ}{1molFe_{2}O_{3}} =6.00kJ[/tex]

By convention, when energy is evolved it takes the negative sign. At constant pressure, the thermochemical equation is:

3 Fe₂O₃(s) + H₂(g) ⇒ 2 Fe₃O₄(s) + H₂O(g)    ΔH° = -6.00 kJ

where

ΔH° is the standard enthalpy of reaction (heat released at constant pressure)

Using the Combined Gas Law, the temperature is converted to Kelvin and pressure to atm, then the initial and final conditions are plugged into the formula to find that the final volume is approximately 0.808 liters.

The problem presented involves a gas law calculation where we need to find the final volume of the gas when measured at a different temperature and pressure while keeping the amount of gas constant. This is a straightforward application of the Combined Gas Law, which is stated as (P1 × V1) / T1 = (P2 × V2) / T2. Making sure to convert temperatures to Kelvin and pressures to atmospheres if they're not already can help in solving the problem accurately.

First, let's convert the temperatures from Celsius to Kelvin by adding 273.15. Therefore, 37 °C becomes 310.15 K and 18 °C becomes 291.15 K. Secondly, the pressure needs to be converted from mmHg to atm by dividing by 760. Thus, 695 mmHg becomes approximately 0.915 atm. Now we apply the Combined Gas Law:

P1 × (V1 / T1) = P2 × (V2 / T2)

(1.00 atm × 0.790 L) / 310.15 K = (0.915 atm × V2) / 291.15 K

After cross-multiplying and solving for V2, we find that the final volume of the gas when measured at 18°C and 695 mmHg, with a constant amount of gas, is approximately 0.808 liters.

Answer:

CH3CHCHCOOH : 11 sigma and 2 pi bonds

HCCCH2CHCHNH2: 12 sigma and 3 pi bonds

H2CCHCH2CCH: 10 sigma and 3 pi bonds

Explanation:

>In this carboxylic acid there are 11 sigma bonds between C-H, C-C ,C-O and  O-H atoms,while there are two pi bonds ,one is made by carbonyl carbon and other from alkene C=C carbons

> In this amine there are 12 sigma bonds made by C-C,C-H and N-H atoms,while 2 pi bonds made by alkyne carbons C≡C carbons and 1 pi bond is made by alkene C=C carbon atoms

>In this organic molecule there are 10 sigma bonds made by C-C and C-H bonds ,while  2 pi bonds made by alkyne carbons C≡C carbons and 1 pi bond is made by alkene C=C carbon atoms.

Answer:

The number of sigma and pi bonds in (a) 11 sigma, 2 pi, (b) 12 sigma, 3 pi, (c) 10 sigma and 3 pi bonds.

Explanation:

Sigma bonds are formed by the linear interactions of the two atoms. The pi bonds are covalent bonds formed at top and bottom of the sigma bonds.

(a) [tex]\rm CH_3CHCHCOOH[/tex]

The compound is bonded with 7 Hydrogen with sigma bond. There is 3 sigma bond in between the 4 carbon atoms. One carbon atom is attached with OH group of carboxylic acid with 1 sigma bond. So, total, 11 Sigma bonds are present in the structure.

The pi bonds are 2 in number, 1 in between C-C in backbone. The pi bond is in carboxylic group in between C-O.

(b) [tex]\rm HCCCH_2CHCHNH_2[/tex]

There are 12 sigma bonds in the backbone of given amine. The number of pi bonds are 3. There are two pi bonds in between the alkyene carbons and one pi bond is present in the alkene carbon-carbon bond.

(c) [tex]\rm H_2CCHCH_2CCH[/tex]

The structure comprises of 10 sigma bonds in the backbone. There is presence of 3 pi bonds in the given compound structure.

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Answer:

[tex]V_f=552 mL[/tex]

Explanation:

Initially:

Total pressure: 735 mmHg

Water vapor pressure: 21 mmHg

Hydrogen pressure: 714 mmHg

(This is because the total pressure is divided between both gases)

When water vapor is removed:

Total pressure: 735 mmHg

Hydrogen pressure: 735 mmHg

Assuming ideal gases:

Boyle-Mariotte's Law:

[tex]P_i*V_i = P_f * V_f[/tex]

[tex]V_f=\frac{P_i*V_i}{P_f}[/tex]

[tex]V_f=\frac{714 mmHg*568 mL}{735 mmHg}[/tex]

[tex]V_f=552 mL[/tex]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc T1= 98.37 °C

Initial temperature of water T1= 24.16 °C

Final temperature of water (and zinc) T2 = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

Qzinc = -Qwater

m(zinc)*c(zinc)* ΔT(zinc) = - m(water)*c(water)* ΔT(water)

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

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Answer:

a. 0.5 mol

b. 1.5 mol

c. 0.67

Explanation:

Fe3+ + SCN- -----> [FeSCN]2+

a. The ratio of the product to Fe3+ is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of Fe3+ was used. Leaving 0.5 mol remaining at equilibrium

b. The ratio of the product to SCN= is 1:1. Meaning that if 0.5 mol of product was produced up then 0.5 mol of SCN- was used. Leaving 1.5 mol remaining at equilibrium

c. KC =  0.5/(0.5*1.5) =  0.67

Answer:

[tex]K_{p}[/tex] of the reaction is  [tex]9.521 \times 10^{-3}[/tex].

[tex] K_{c}[/tex] of the reaction is  [tex]3.89 \times 10^{-4}[/tex].

Explanation:

Initial pressure of NOBr is "x" atm.

34% = 0.34

The equilibrium chemical reaction is a follows.

       [tex]2NOBr(g)\leftrightarrow 2NO(g)+Br_{2}(g)..............(1)[/tex]

Initial          x                                      0          0

Change    -0.34x                             +0.34x    +0.17x

Final        (x-0.34x)                          +0.34x     +0.17x

Total pressure = sum of the pressure at equilibrium.

[tex]0.25atm=x-0.34x+0.34x+0.17x[/tex]

[tex]0.25atm= 1.17x[/tex]

[tex]x= \frac{0.25}{1.17}=0.213[/tex]

Partial pressure of NOBr=  x-0.34x =x (1-0.34)

= 0.213(0.66) = 0.14 atm.

Hence, partial pressure of [tex][]P_{NOBr}][/tex] is 0.14 atm.

[tex]P_{NO}=0.34x = 0.34 \times 0.213 = 0.072[/tex]

[tex]P_{Br_{2}}=0.17x = 0.17 \times 0.213 = 0.036\,atm[/tex]

From equation (1)

[tex]K_{p}= \frac {P^{2}_{NO} P_{Br_{2}}}{P^{2}_{NOBr}}[/tex]

[tex]= \frac{(0.072)^{2} (0.036)}{(0.14)^{2}}= 9.521 \times 10^{-3}[/tex]

Therefore, [tex]K_{p}[/tex] of the reaction is  [tex]9.521 \times 10^{-3}[/tex].

[tex]K_{p}= K_{c}(RT)^{\Delta n}[/tex]

Rearrange the equation is as follows.

[tex]K_{c}= \frac{K_{p}}{(RT)^{\Delta n}}[/tex]

[tex]\Delta n[/tex] = number of moles of reactants - Number of moles of products

= 3-2 = 1

[tex]= \frac{9.521 \times 10^{-3}}{(0.0821 \times 298)^{1}}= 0.389 \times 10^{-3} = 3.89\times 10^{-4}[/tex]

Therefore,[tex] K_{c}[/tex] of the reaction is  [tex]3.89 \times 10^{-4}[/tex].

The [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.

[tex]2NOBr\rightleftharpoons 2NO+Br_2[/tex]

​When the initial pressure of [tex]NOBr[/tex] is p, then at equilibrium, therefore the dissociation percentage of [tex]NOBr \ \ is\ \ 34\%[/tex]

[tex]\to P(NOBr)=0.66\ p\\\\\to P(NO)=0.34\ p\\\\\to P(Br_2)= \frac{0.34p}{2}=0.17\ p\\\\[/tex]

Calculating the total pressure:

[tex]\to 0.66\ p+0.34\ p+0.17\ p=1.17\ p=0.25\ atm\\\\[/tex]

[tex]\to p=0.214 \ atm\\\\[/tex]

The equilibrium partial pressures:

[tex]\to P(NOBr)=0.66\ p=0.141\ atm\\\\\to P(NO)=0.34\ p=0.073\ atm\\\\\to P(Br_2)=0.17\ p=0.036\ atm\\\\[/tex]

Calculating the [tex]K_p[/tex]:

[tex]\to K_p =\frac{(P_{NO})^2 (P_{Br_2})}{(P_{NOBr})^2}\\\\[/tex]

          [tex]=\frac{(0.073)^2 \times (0.036)}{(0.141)^2}\\\\=\frac{(0.005329) \times (0.036)}{(0.019881)}\\\\=\frac{(0.000191844)}{(0.019881)}\\\\=0.009649615 \ atm[/tex]

Calculating the [tex]K_c[/tex]:

Using formula:

[tex]\to \Delta ng = \text{Number of moles of product - Number of moles of reactant}[/tex]

            [tex]= (2+1) - 2\\\\= 3 - 2\\\\=1 \\[/tex]

The relationship between [tex]K_p[/tex] and [tex]K_c[/tex] is:

[tex]\to K_p = K_C(RT)^{\Delta ng}[/tex]

Hence, by putting the values of  [tex]K_p[/tex], gas constant [tex]R, T,[/tex] and [tex]\Delta ng[/tex], we can calculate the value of [tex]K_c[/tex].

[tex]\to 0.0096\ atm = - K_c(0.0821 \ atm \ mol L^{-1}K^{-1} \times 298 K)^1[/tex]

[tex]\to K_c = \frac{0.0096 }{0.0821\times 298}[/tex]

         [tex]= \frac{0.0096}{ 24.46}\\\\ = 0.00039\ mol\ L^{-1}[/tex]

Therefore, the [tex]K_p[/tex] and [tex]K_c[/tex] is " [tex]0.009649615 \ atm[/tex] and [tex]0.00039\ mol\ L^{-1}[/tex]"for the dissociation at this temperature.

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Answer:

To calculate the molar mass add the atomic mass of each atom in the chemical formula.

Explanation:

For example Molar mass of NaHCO₃  calculated as

Molar mass is the sum of masses of all atom present in formula.

Atomic mass of sodium = 23 g/mol

Atomic mass of hydrogen = 1.008 g/mol

Atomic mass of carbon = 12 g/mol

Atomic mas of Oxygen = 16 g/mol

Molar mass of NaHCO₃ = 23 + 1.008 + 12 + 16× 3

Molar mass of NaHCO₃ = 23 + 1.008 + 12 + 48

Molar mass of NaHCO₃ = 84.008 g/mol

Answer:

Explanation: AH, AG and AS for the reaction at 873 K

delta G =-RT ln Kp

G= -8.314 x 873 ln 0.9

= 764.7J/mol

AG = AH-TAS

AH = 764.7 - 8314(0.9)

Kp =pCO/pCo2

Substitute the values at different temp. Solve simultaneously

Since mole fraction =kp/pCo

Mf =1.5

Answer:

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Answer:

The solution boils at 102.76 °C

Explanation:

Δt = m* Kb x i  

i = 1 with 1 particle in solution

m = molality

Kb = 0.512 °C/molal

Step 2: Calculate moles C2H6O2

molar mass of  C2H6O2 = 62.068 g/mol

Calculate number of moles:

Moles = Mass / molar mass

Moles = 1400 grams / 62.068 g/mol

Moles C2H6O2 = 22.56 moles

Step 3: Calculate molality

these are in 4192 g of water:

22.56 moles / 4.192 kg water

⇒ moles / kg of water = 5.38 moles / Kg = m olal

Step 4: Calculate  ΔT

ΔT= Kb * molal = 5.38 molal* 0.512 °C/m

ΔT = 2.76 °C

Boiling point = 100°C + 2.75 °C = 102.76 °C

the solution boils at 102.76 °C

The boiling point of the ethylene glycol solution is calculated by first finding the molality of the solution and then using this to find the boiling point elevation. The normal boiling point of water is thus raised by this amount, giving a final boiling point of the solution as 102.75 degrees Celsius.

To solve this problem, we need to calculate the molality of the ethylene glycol solution and then use this to determine the boiling point elevation of the solution.

Firstly, we need to convert the mass of ethylene glycol (C2H6O2) and water into moles. The molar mass of ethylene glycol is approximately 62.07 g/mol, so 1.4 kg (or 1400 g) of ethylene glycol is equivalent to approximately 22.54 moles. The mass of water is 4192 g, or roughly 4.192 kg.

The molality (m), is thus given by the formula m = moles of solute / kg of solvent, therefore the molality of the ethylene glycol solution is 22.54 moles / 4.192 kg = 5.38 m.

Next, we use the formula for boiling point elevation: ΔTb = Kb * m, where Kb is the ebullioscopic constant for water (given as 0.512 °C/m), and m is the molality calculated earlier. Therefore, the boiling point of the solution rises by ΔTb = 0.512 × 5.38 = 2.75 °C.

The normal boiling point of water is 100 °C, thus, the boiling point of the ethylene glycol solution is 100 °C + 2.75 °C = 102.75 °C.

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Answer:

V = 18.8 L

Explanation:

This question is solved by using the ideal gas law :

PV = nRT ∴  V= nRT/P

where n= 0.788 mmol = 0.788 mol (1 mmol is a thousandths of a mol)

R = R constant for ideal gases =  0.08206 Latm/Kmol

T = 17.0 ºC =  (17.0 + 273) K = 290 K

and V = volume in liters our unknown

Plugging our values and solving for V,

V = 0.788 mol x 0.08206 Latm/Kmol x 290 K / 1atm = 18.8 L

Applying the Ideal Gas Law with the given number of moles, temperature and pressure shows that the volume of dinitrogen difluoride gas collected is approximately 19.0 L.

The volume of a gas can be calculated using the Ideal Gas Law if the number of moles, temperature and pressure are known. In your question, we're given that 788mmol of dinitrogen difluoride gas is evolved at a temperature of 17.0 degree C and a pressure of 1 atm. Firstly, convert the temperature to Kelvins by adding 273.15 to the Celsius temperature, giving us 290.15K.

The Ideal Gas Law is expressed as PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L·atm/ K·mol for this situation) and T is temperature. We can rearrange the equation to solve for volume: V=nRT/P. Therefore, substituting in the known values: V = 0.788 mol * 0.0821 L·atm/ K·mol * 290.15K / 1 atm, we find that the volume of dinitrogen difluoride gas collected is approximately 19.0 L, given correct to the number of significant digits.

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