Sodium (Na) reacts with chlorine gas to form the ionic compound NaCl. Which of the following statements is true? A The reaction leaves each atom more unstable. B Sodium gains electrons from the chlorine to form the sodium ion. C Sodium ions are anions in the ionic bond. D By reacting, each atom fills its outer energy level with electrons.

Answer:

I may not be correct but i think its pH = Kg

Explanation:

Answer:

B. in such a way that pH = pKa

Explanation:

Ideally, the pH of the desired solution should have the same pKa as the pH, making the ratio 1:1.

Answer:

[HI] = 0.5239M

[H₂] = 7.05x10⁻²

[I₂] = 7.05x10⁻²

Explanation:

The reaction of HI to produce H₂ and I₂ is:

2HI → H₂ + I₂

Where K of reaction is defined as:

K = [H₂] [I₂] / [HI]²

Replacing with the concentrations of the gases in equilibrium:

K = [4.34x10⁻²] [4.34x10⁻²] / [0.323] ²

K = 0.0181

If you add 0.228 mol = 0.228M (Because volume of the flask is 1.0L), the concentration when the system reaches the equilibrium are:

[HI] = 0.228M + 0.323M - X = 0.551M - X

[H₂] = 4.34x10⁻² + X

[I₂] = 4.34x10⁻² + X

Where X is reaction coordinate.

Replacing in K formula:

K = 0.0181 = [4.34x10⁻²+ X] [4.34x10⁻²+ X] / [0.551 - X] ²

0.0181 =  0.00188356 + 0.0868 X + X² / 0.303601 - 1.102 X + X²

0.005495 - 0.01995 + 0.0181X² = 0.00188356 + 0.0868 X + X²

0 = -0.003611 + 0.10675X + 0.9819X²

Solving for X:

X = -0.136 → False solution, there is no negative concentrations.

X = 0.0271 → Right solution.

Replacing for concentrations of each species:

Answer:

Oxygen and Magnesium.

Based on the difference in electronegativity, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, whereas lithium and calcium, sulfur and bromine, and manganese and chlorine cannot.

Ionic compounds are formed when there is a high electronegativity difference between two elements.

To determine whether the following pairs of elements can form ionic compounds, we need to look at their electronegativity difference:

1. Lithium and calcium: The electronegativity difference between Lithium and calcium is only 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.

2. Sulfur and bromine: The electronegativity difference between sulfur and bromine is 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.

3. Manganese and chlorine: The electronegativity difference between manganese and chlorine is 1.5, which is less than 1.7. Therefore, they cannot form an ionic compound.

4. Potassium and sulfur: The electronegativity difference between potassium and sulfur is 2.4, which is greater than 1.7. Therefore, they can form an ionic compound.

5. Chlorine and lithium: The electronegativity difference between chlorine and lithium is 2.8, which is greater than 1.7. Therefore, they can form an ionic compound.

6. Oxygen and magnesium: The electronegativity difference between oxygen and magnesium is 1.7, which is equal to 1.7. Therefore, they can form an ionic compound.

Therefore, based on the electronegativity difference between the elements, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, while lithium and calcium, sulfur and bromine, and manganese and chlorine cannot form ionic compounds.

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Answer:

0.27 g

Explanation:

The reaction equation:

[tex]Na_{2} CO_{3} + CaCl_{2}[/tex] → [tex]2NaCl + CaCO_{3}[/tex]

106g of Na2CO3 - 1 mole

0.5g of Na2CO3 = 0.5 ÷ 106

= 0.0047 moles.

1 mole of NaCl - 58.5

⇒ 0.0047 moles = 0.0047 × 58.5

= 0.27g.

When 0.5 grams of Na₂CO₃ react with excess CaCl₂, 0.6 g of NaCl are formed.

We combine 0.5 grams of Na₂CO₃ with excess CaCl₂ and we want to know the mass of NaCl produced. This is a stoichiometry problem.

Stoichiometry refers to the relationship between the quantities of reactants and products before, during, and following chemical reactions.

First, we will write the balanced chemical equation.

Na₂CO₃ + CaCl₂ ⇒ 2 NaCl + CaCO₃

We will consider the following relationships.

[tex]0.5 g Na_2CO_3 \times \frac{1molNa_2CO_3}{105.99gNa_2CO_3} \times \frac{2molNaCl}{1molNa_2CO_3} \times \frac{58.44gNaCl}{1molNaCl} = 0.6gNaCl[/tex]

When 0.5 grams of Na₂CO₃ react with excess CaCl₂, 0.6 g of NaCl are formed.

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Answer:

[tex]P_{H_2}=1.9kPa[/tex]

Explanation:

Hello,

Here, by using the Dalton's law, we can quantify the total pressure of a gaseous mixture by knowing the partial pressure of each gas, in case, hydrogen, helium and argon:

[tex]P_T=P_{H_2}+P_{He}+P_{Ar}[/tex]

In such a way, since we actually know the partial pressure of helium and argon, and the total pressure, we can compute the partial pressure of hydrogen as shown below:

[tex]P_{H_2}=P_T-P_{He}+P_{Ar}=99.3kPa-42.7kPa-54.7kPa\\\\P_{H_2}=1.9kPa[/tex]

Best regards.

Answer:

Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

Explanation:

The pH of the buffer solution before the addition of HCl is:

[tex]pH = pKa + log(\frac{[KF]}{[HF]})[/tex]

[tex]pH = -log(6.8 \cdot 10^{-4}) + log(\frac{0.371}{0.494}) = 3.04[/tex]  

The hydrochloric acid added will react with the potassium fluoride as follows:

H₃O⁺(aq)  +  F⁻(aq) ⇄   HF(aq) + H₂O(l)

The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:

[tex] \eta_{KF}_{i} = C_{KF}*V = 0.371 M*1 L = 0.371 mol [/tex]

[tex] \eta_{HF}_{i} = C_{HF}*V = 0.494 M*1 L = 0.494 moles [/tex]

The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:

[tex] \eta_{HCl} = \eta_{HCl} - \eta_{KF}_{i} = 0.408 moles - 0.371 moles = 0.037 moles [/tex]  

Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.  

We can calculate the pH after the addition of HCl:

HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)    (1)

The number of moles of HF after the reaction of KF with HCl is:

[tex] \eta_{HF} = 0.494 moles + (0.408 moles - 0.371 moles) = 0.531 moles [/tex]

And the concentration of HF after the reaction of KF with HCl is is:

[tex] C_{HF} = \frac{\eta_{HF}}{V} = \frac{0.531 moles}{1 L} = 0.531 moles/L [/tex]

Now, from the equilibrium of equation (1) we have:

[tex] Ka = \frac{[H_{3}O^{+}][F^{-}]}{[HF]} [/tex]

[tex] Ka = \frac{x^{2}}{0.531 - x} [/tex]  (2)

By solving equation (2) for x we have:

x = 0.0187

Finally, the pH after the addition of HCl is:

[tex] pH = -log (H_{3}O^{+}) = -log (0.0187) = 1.73 [/tex]

Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.

I hope it helps you!

Answer:

This is because Hydrogen peroxide must be acidic, thereby making hydrogen ions taking part in the reaction which act as reactant. It is important also to use H2SO4 because permanganate is a strong oxidizing agent and it cannot oxidize sulphuric acid if enough is added but it can oxidize chloride ions to chlorine if hydrochloric acid is used.

Explanation:

Hydrogen peroxide is used as a chemical agents in cleaning processes for chemical industries and semiconductor plants. In the hydrogen peroxide assay, sulphuric acid is used and the reaction is an exothermic reaction. This produce strong peroxysulphuric acid

Answer:

30.0g/mol

Explanation:

Step 1: Given data

Step 2: Calculate the moles of the gas

We will use the ideal gas equation.

[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1atm \times 0.0933L}{\frac{0.0821atm.L}{mol.K} \times 273.15K} = 4.16 \times 10^{-3} mol[/tex]

Step 3: Calculate the molar mass of the gas

4.16 × 10⁻³ moles correspond to a mass of 0.125 g. The molar mass of the gas is:

[tex]\frac{0.125g}{4.16 \times 10^{-3} mol} =30.0g/mol[/tex]

The molar mass of the 0.125 g  of the gas occupies 93.3 mL at STP is 30.0 g/mol.

Number of moles of Gas at STP,

[tex]\bold{n =\dfrac {PV}{RT}}[/tex]

where,

P - pressure

V- volume

R- gas constant

T - temperature

Put the values in the formula,

[tex]\bold{n =\dfrac {1 \times 0.0933} {0.082 \times 273.15 }}\\\\\bold{n =4.16 \timesw 10^-^3}[/tex]

The molar mass of the gas can be calculated using formula,

[tex]\bold {m = \dfrac {w}{n}}\\\\\bold {m = \dfrac {0.125} {4.16 \times 10^-^3}}\\\\\bold {m = 30g/mol}[/tex]

The molar mass of the 0.125 g  of the gas occupies 93.3 mL at STP is 30.0 g/mol.

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Answer:

Percent Yield = 94.237%

Explanation:

CO = Carbon Dioxide = Molar Mass 28g/mol

C = Carbon = 12g/mol

O = Oxygen = 16g/mol

Theoretical yield = 93.7 grams

Actual yield = 88.3 grams

Percent yield  = (actual yield /theoretical yield )x100

Percent Yield = (88.3/93.7)x100

Percent Yield = 94.237%

Answer:

The correct answer is 28.2 %.

Explanation:

Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,  

Ptotal = PO2 + PN2 + PHe

= 55 + 90 + 50

= 195 atm

The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100

Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,  

%VO2 = (55/195) * 100 = 28.2%

%VN2 = (90/195) * 100 = 46%

%VHe = (50/195) * 100 = 25.64%

Hence, the percent oxygen by volume present in the blend is 28.2 %.  

Answer:it has fixed quantity

Explanation:energy between chemical bonds is hard to measure

Answer:

It has a fixed quantity.

Explanation:

Answer:

1.02 × 10³ g

Explanation:

Step 1: Write the balanced equation

2 CaCO₃ + H₂SO₄ ⇒ CO₂ + H₂O + CaSO₄

Step 2: Calculate the moles corresponding to 500 g of sulfuric acid

The molar mass of sulfuric acid is 98.08 g/mol.

[tex]500g \times \frac{1mol}{98.08g} = 5.10mol[/tex]

Step 3: Calculate the moles of calcium carbonate that react with 5.10 moles of sulfuric acid

The molar ratio of CaCO₃ to H₂SO₄ is 2:1. The moles that react of calcium carbonate are (2/1) × 5.10 mol = 10.2 mol

Step 4: Calculate the mass corresponding to 10.2 moles of calcium carbonate

The molar mass of calcium carbonate is 100.09 g/mol.

[tex]10.2mol \times \frac{100.09 g}{mol} =1.02 \times 10^{3} g[/tex]

Answer:

884.8 mL

Explanation:

If we have STP condition (Standard temperature and pressure), we will have the following relationship:

[tex]1~mol=22.4~L[/tex] or [tex]1~mol=22400~mL[/tex]. With this in mind we can do the conversion:

[tex]0.0395~mol\frac{22400~mL}{1~mol}=884.8~mL[/tex]

In the 0.0395 mol of fluorine gas, we will have 884.8 mL of gas at STP conditions.

Answer:

15g

Explanation:

Molar mass of acetic acid = 60g/mol

Molarity = mass ÷ molar mass

mass = 0.25 × 60

= 15g

∴ 15g of acetic acid weighed into 1L volumetric flask would give 0.25M  of acetic acid.

Alternatively, if the solution is being prepared from a stock solution of known concentration, (say 500mL of 0.5M solution), the formular C2V1 =C2V2 can be used to find the dilution volume.

Answer:

33Cl

Explanation:

The atomic mass of an element is made up of the proton and neutron.

chlorine has a constant number of 17 protons.

33-17 = 16

Answer:

1) Human activities such as the combustion of fossil fuels, has increased the levels of greenhouse gases----  Chemical reaction.

2) At dew point, the water vapor begins to condense out of air----  Physical changes.

3)  Human activities, such as the combustion of fossil fuels, generate the aerosols----  Chemical reaction.

4) You observe hail when the temperatures are below the freezing point----  Physical changes.

5) Carbon dioxide is produced during respiration of plants, animals, fungi, and microorganisms-----  Chemical changes.

A soluble base is formed when sodium reacts with water.

When sodium reacts with water, it produces strongly alkalic sodium hydroxide which is also called caustic soda and hydrogen gas. In this chemical reaction, energy is absorbed which means it is an exothermic reaction so we can conclude that a soluble base is formed when sodium reacts with water.

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Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will  B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: The value of Kc remains the same.

As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, The reaction must: A. run in the forward direction to reestablish equilibrium.

By decreasing the volume, and so that Kc remains constant, being:

[tex]Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2} ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume} } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume[/tex]

 where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so,  the number of moles of Cl₂ should decrease.The number of moles of Cl2 will  B. decrease.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, the value of Qc C. is less than Kc.

Answer:

q = - 3.569KJ 0r 3.569KJ Liberated heat (signifying the change in heat is negative)

Explanation:

liberated heat implies that change in heat is negative , therefore

q = -m c ΔT

where, m = mass of the Silver = 249 g

c = specific heat capacity of Silver = 0.235 Jg^{-1} °C^{-1

 ΔT = change in temperature = 87°C- 26 °C= 61°C

q = -m x c x ΔT

= - 249 x 0.235 x 61 = - 3569.415J  rounded to -3569J

Changing to KJ becomes= -3569/1000= - 3.569 KJ

q = - 3.569KJ 0r 3.569KJ liberated heat.

Answer:

C. The energy of motion

Explanation:

Kinetic energy is the energy associated with the movement of objects.

The kinetic energy of an object depends on both its mass and velocity, with its velocity playing a much greater role.

Example of Kinetic Energy:

1. An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.

// have a great day //

Kinetic energy is the energy of motion.

So kinetic energy is present when objects move.

For example, a snowball rolling down a mountain.

Answer:

B, C

Explanation:

The atoms or ions with the valid Lewis dot structures are B and C.

In A;

The Lewis structure of the carbon is correct. Each of the four dots represent the four valence electrons.

The  nitrogen with one dot on top, left and to the bottom and has a charge of minus 3 is wrong. For it to have a charge of -3 it must have 8 lewis dots ( two on the top, right, bottom and to the left)

The nitrogen with four dots (on top, right, bottom and to the left) is wrong.

In B;

An oxygen has two dots on top and bottom and one dot to the left and to the right. This is correct , the 6 dots represent the valence electrons of oxygen.

In C;

A carbon has two dots on top, right, bottom and to the left and a charge of plus four. This is correct because the charge indicates that it has gained four extra electrons so its valence electrons is now 8.

In D;

An oxygen has two dots on top, left and to the bottom and a charge of minus 2. This is wrong because the lewis dots are incomplete. Two dots are missing.

Answer: 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.

Explanation:

According to the dilution law,

[tex]C_1V_1=C_2V_2[/tex]

where,

[tex]C_1[/tex] = concentration of stock solution = 12.0 M

[tex]V_1[/tex] = volume of stock solution = ?

[tex]C_2[/tex] = concentration of diluted solution= 6.00 M

[tex]V_2[/tex] = volume of diluted acid solution = 500 ml

Putting in the values we get:

[tex]12.0\times V_1=6.00\times 500[/tex]

[tex]V_1=250ml[/tex]

Thus 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.

Answer:

0.0586 kg

Explanation:

From the question,

Heat lost by the aluminium = heat gain by nitrogen.

CM(t₁-t₂) = cm................... Equation 1

Where C = Specific heat capacity of aluminum, M = mass of aluminum, t₂ = Final temperature, t₁ = initial temperature, c = latent heat of vaporization of nitrogen, m = mass of nitrogen.

make m the subject of the equation

m = CM(t₁-t₂)/c................ Equation 2

Given: C = 900 J/kg.K, M = 60 g = 0.06 kg, t₁ = 20 °C, t₂ = -196 °C

Constant: c = 199200 J/kg

Substitute these values into equation 2

m = 900×0.06×[20-(-196)]/199200

m = 900×0.06×216/199200

m = 0.0586 kg.

Answer:

[tex]m_{N_2}=58.6gN_2[/tex]

Explanation:

Hello,

In this case, for an average temperature of -176 °C, the vaporization enthalpy of liquid nitrogen is 199.2 J/g, thus, we first compute the heat lost by the aluminium by considering it cooled mass, specific heat and temperature change:

[tex]Q=mCp(T_2-T_1)=60g*0.90\frac{J}{g\°C}*(-196-20)\°C\\ \\Q=-11664J[/tex]

Next, heat lost by the aluminium is gained by the nitrogen:

[tex]-Q_{Al}=Q_{N_2}=11664J[/tex]

Therefore, the vaporized nitrogen is:

[tex]m_{N_2}=\frac{Q_{N_2}}{\Delta H_v}=\frac{11664J}{199.2J/g}\\\\m_{N_2}=58.6gN_2[/tex]

Best regards.

Answer:

Volume, V2 at STP = 818.61ml.

Explanation:

Given the following parameters;

Volume, V1 = 1140ml

Volume, V2 =?

Temperature, T1 = 37ºC to Kelvin = 273+37 = 310K

Temperature, T2 = 273K is the standard temperature.

Pressure, P1 = 82.6kPa

Pressure, P2 = 101.3kPa is the standard pressure.

To solve for volume at standard temperature and pressure (STP), we would use the combined gas law;

The combined gas law is a combination of the other three gas laws, namely Gay Lusac's law, Boyle's law and Charles's law.

Combined gas law states that the ratio of the product or multiplication of volume and pressure to the temperature of a gas is equal to a constant.

Mathematically, [tex]PV/T = K[/tex]

P1V1/T1 = P2V2/T2

Let's make V2 to the subject formula;

Cross multiplying gives,

P1V1T2 = P2V2T1

Hence, [tex]V2 = (P1V1T2)/P2T1[/tex]

Substituting the parameters;

V2 = (82.6*1140*273)/101.3*310

V2 = 25706772/31403

V2 = 818.61ml.

Answer:

1. The correct option is;

c. maintains charge balance in the cell

2. The correct option is;

c. +3.272 V

Explanation:

The aqueous solution in a galvanic cell is the electrolyte which is a ionic solution containing that permits the transfer of ions between the separated compartment of the galvanic cell such that the overall system is electrically neutral

Therefore, the aqueous solution maintains the charge balance in the cell

2. Here we have;

B₂ + 2e⁻ → 2B⁻ Ecell = 0.662 V

A⁺ + 1e⁻ → A Ecell = -1.305 V

Hence for the overall reaction, we have;

2A + B₂ → 2AB gives;

(0.662) - 2×(-1.305)  = +3.272 V.

Answer:65N

Explanation:75-10=65

100-100=0

Therefore=65n

Answer:

the answer is 65N

Answer:

32 K is approx. -241.15° C

Answer:

ΔH = 130.5 kJ

Explanation:

Hello,

In this case, by using the Hess law, we compute the enthalpy of the required reaction:

C(s) + H2O(g) --> CO(g) + H2(g)

Thus, the first step is to keep the following reaction unchanged:

C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ

Next, we invert and halve this reaction:

2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ

So the enthalpy of reaction is inverted and halved:

CO2 (g) → CO (g) + 1/2 O2 (g) ΔH= 283 kJ

Then, we also invert and halve this reaction:

2 H2 (g) + O2 (g) → 2 H2O ΔH= -483.6 kJ

So the enthalpy of reaction is inverted and halved as well:

H2O → H2 (g) + 1/2 O2 (g) ΔH= 241.8 kJ

Finally, we add the three reactions to obtain the required reaction:

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + 1/2 O2 (g) + CO (g) + 1/2 O2 (g) + CO2 (g)

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + O2 (g) + CO (g) + CO2 (g)

= C (s) + H2O → H2 (g) CO (g)

So enthalpy is computed by:

ΔH = -393.5 kJ + 283 kJ + 241.8 kJ

ΔH = 130.5 kJ

Best regards.

Considering the Hess's Law, the enthalpy change for the reaction is 131.3 kJ.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C(s) + H₂O(g) → CO(g) + H₂(g)

You know the following reactions, with their corresponding enthalpies:

Equation 1: C (s) + O₂(g) → CO₂ (g)     ΔH = -393.5 kJ

Equation 2:  2 CO (g) + O₂ (g) → 2 CO₂ (g)     ΔH= -566.0 kJ

Equation 3: 2 H₂ (g) + O₂ (g) → 2 H₂O (g)     ΔH= -483.6 kJ  

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C(s) on reactant side and it is present in first equation so let's write this as such.

Now, 1 mole of CO(g) must be a product and is present in the second equation. Since this equation has 2 moles of CO(g) on the reactant side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of CO(g).

When an equation is inverted, the sign of ΔH also changes.

And since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is divided by 2, the variation of enthalpy also is divided by 2.

Finally, 1 mole of H₂O(g) must be a reactant and is present in the third equation. Since this equation has 2 moles of CO(g) on the product side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of H₂O(g).

So, the sign of ΔH also changes and the variation of enthalpy is divided by 2.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: C (s) + O₂(g) → CO₂ (g)     ΔH = -393.5 kJ

Equation 2: CO₂ (g) → CO (g) + [tex]\frac{1}{2}[/tex] O₂ (g)     ΔH= 283 kJ

Equation 3: H₂O (g) → H₂ (g) + [tex]\frac{1}{2}[/tex] O₂ (g)    ΔH= 241.8 kJ  

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C(s) + H₂O(g) → CO(g) + H₂(g)  ΔH= 131.3 kJ

Finally, the enthalpy change for the reaction is 131.3 kJ.

Learn more:

Answer:

E = 1.47

Explanation:

To do this, you need to apply the Nerst equation which is the following:

E = E° - RT/nF lnQ (1)

Where:

E: cell voltage

E°: Standard potential reduction

R: universal constant

T: temperature of the system

n: number of electrons transfered during the reaction

F: Faraday constant.

Q: Equilibrium constant

However, as the reaction is taking place at 25 °C, and R and F have constant values, we can reduce the above expression to the following:

E = E° - 0.05916/n lnQ  (2)

We can get the value of Q because it has to do with the reaction which is the following:

2IO₃⁻(aq) + 12H⁺(aq) + 5Co(s) ----------> I₂(s) + 5Co²⁺(aq) + 6H₂O(l)

Now, using only the aqueous state the expression of Q will be:

Q = [Co²⁺]⁵ / [H⁺]¹² [IO₃⁻]²

Replacing the values we have:

Q = (7.82)⁵ / (1.54)¹² * (6.64)²

Q = 3.728

Knowing this, all we need to know now is the standard potential reduction of the reaction. To do so, we need to write the two semi equations of reduction and oxidation:

2IO₃⁻ + 12H⁺ + 10e⁻ ---------> I₂ + 6H₂O       E₁° = 1.20 V

5Co ---------> 5Co²⁺ + 10e⁻                          E₂° = 0.28 V

E° = 1.2 + 0.28 = 1.48 V

Now that we have all the values (n = 10) we can write now the nernst equation to calculate the cell voltage:

E = 1.48 - 0.05916/10 ln (3.728)

E = 1.48 - 0.005916 (1.315872)

E = 1.47 V

This will be the cell voltage