It is proposed to use Liquid Petroleum Gas (LPG) to fuel spark-ignition engines. A typical sample of the fuel on a volume basis consists of: 70% propane C3H8; 5% butane C4H10; and 25% propene C3H6 The higher heating values of the fuels are 50.38 MJ/kg for propane, 49.56 MJ/kg for butane, and 48.95 MJ/kg for propene. a) Work out the overall combustion reaction for stoichiometric combustion of 1 mole of LPG with air, and determine the stoichiometric F/A and A/F ratios. b) What are the higher and lower heating values per unit mass of LPG?

Answer:

The product is triphenylmethane dye

Explanation:

The H2SO4 removes the OH

Leaving triphenylmethane dye

Find attached the suggested structure

Final answer:

To calculate the cell potential for the given galvanic cell reaction, you can use the Nernst equation and the standard cell potential. Plug in the concentrations of Fe3+ and Au3+ and calculate the cell potential using the Nernst equation at 25°C.

Explanation:

The cell potential for the galvanic cell can be calculated using the Nernst equation, which is given by: Ecell = E°cell - (0.0592/n) * log(Q)

where E°cell is the standard cell potential, n is the number of electrons transferred in the reaction, and Q is the reaction quotient.

In this case, the reaction is: Fe(s) + Au3+(aq) -> Fe3+(aq) + Au(s)

The standard cell potential can be found in Appendix L. Plugging in the concentrations of Fe3+ and Au3+, as well as the standard cell potential, into the Nernst equation will give the cell potential at 25°C.

The cell potential for the given galvanic cell at 25°C is approximately 1.589 V.

This calculation involves determining standard reduction potentials, using the Nernst equation, and considering the reaction quotient.

To calculate the cell potential for the galvanic cell with the reaction:
Fe(s) + Au³⁺ (aq) → Fe³⁺ (aq) + Au (s) at 25°C with [Fe³⁺] = 0.00150 M and [Au³⁺] = 0.797 M, we can follow these steps:

Determine the standard reduction potentials:

Calculate the standard cell potential (E°cell) using the formula:

E°cell = E°cathode - E°anode

Use the Nernst equation to find the cell potential under non-standard conditions:

Plugging in the values:

Therefore, the cell potential for the given galvanic cell at 25°C is approximately 1.589 V.

The molar mass of the metal M was calculated using Faraday's laws of electrolysis and found to be 63.65 g/mol, which closely corresponds to copper (Cu). Therefore, the identity of metal M is likely to be copper.

To determine the identity of the metal M, we will need to use Faraday's laws of electrolysis to calculate the molar mass of the metal M.

The charge (Q) that passed through the electrode can be calculated by the formula Q = It, where I is the current in amperes and t is the time in seconds. In this case, I = 6.13 A and t = 73.1 s, so Q = 6.13 × 73.1 = 448.203 C (coulombs).

Next, we will use the equivalent weight of metal M, which is the molar mass of M divided by the number of electrons transferred per ion of M during the electrolysis. For MBr₂, there are 2 moles of electrons transferred per mole of M (since the valency is 2).

Using Faraday's constant (approximately 96500 C/mol), we can determine the number of moles of M deposited by dividing the charge by Faraday's constant multiplied by the valency, which is moles of M = Q / (2 ×96500).

From this, we can calculate the molar mass: Molar mass of M = mass of M deposited / moles of M.

Using our given data, the moles of M deposited would be 448.203 C / (2 × 96500 C/mol) = 0.002325 moles. The molar mass would then be 0.148 g / 0.002325 mol = 63.65 g/mol. This molar mass is very close to that of copper (Cu), which is 63.55 g/mol. Thus, the correct answer is Cu.

The identity of the metal M is copper, Cu.

To determine the identity of metal M, we need to follow these steps:

1. Calculate the total charge (in coulombs) that passed through the molten salt during electrolysis using the equation [tex]\( Q = I \times t \)[/tex], where [tex]\( I \)[/tex] is the current in amperes and [tex]\( t \)[/tex] is the time in seconds.

2. Use Faraday's constant to find the moles of electrons that passed through the solution. Faraday's constant is [tex]\( 96485 \text{ C/mol} \).[/tex]

3. Relate the moles of electrons to the moles of metal M deposited, since the electrolysis of [tex]\( \text{MBr}_2 \)[/tex] involves the transfer of two moles of electrons for each mole of metal M produced.

4. Calculate the molar mass of metal M using the mass of metal M deposited and the moles of metal M calculated in the previous step.

5. Identify the metal M based on its molar mass.

Let's perform the calculations:

 1. Total charge [tex]\( Q \)[/tex] is calculated as:

[tex]\[ Q = I \times t = 6.13 \text{ A} \times 73.1 \text{ s} = 448.563 \text{ C} \][/tex]

2. Moles of electrons [tex]\( n(e^-) \)[/tex] is given by:

[tex]\[ n(e^-) = \frac{Q}{F} = \frac{448.563 \text{ C}}{96485 \text{ C/mol}} \approx 4.65 \times 10^{-3} \text{ mol} \][/tex]

3. Since two moles of electrons are required to deposit one mole of metal M, the moles of metal M [tex]\( n(\text{M}) \)[/tex] is half the moles of electrons:

[tex]\[ n(\text{M}) = \frac{n(e^-)}{2} = \frac{4.65 \times 10^{-3} \text{ mol}}{2} \approx 2.325 \times 10^{-3} \text{ mol} \][/tex]

4. Molar mass [tex]\( M \)[/tex]of metal M is calculated as:

[tex]\[ M = \frac{m}{n} = \frac{0.148 \text{ g}}{2.325 \times 10^{-3} \text{ mol}} \approx 63.66 \text{ g/mol} \][/tex]

5. Based on the calculated molar mass, we can identify the metal M. The metal with a molar mass closest to [tex]\( 63.66 \text{ g/mol} \)[/tex] is copper, Cu, which has a molar mass of approximately[tex]\( 63.55 \text{ g/mol} \)[/tex].

Therefore, the identity of metal M is copper, Cu.

Answer:

distance with a negative value just means that it's going in the opposite direction.

Explanation:

example: -50m East is the just 50m west

In physics, negative values in a calculation often refer to displacement, indicating movement in the opposite direction of the chosen positive direction.

Distance itself remains a positive value as it does not account for direction.

On the other hand, distance refers to the total path length traveled and is always a positive value, since it does not take direction into account.

Answer:

Al2(SO4)3 + 6(NH3•H2O) → 2Al(OH)3 + 3(NH4)2SO

Explanation:

Aluminium sulfate react with ammonium hydroxide to produce aluminium hydroxide and ammonium sulfate.

If you are looking for the equation

Answer:

See explaination for detailed answer

Explanation:

In the IR spectrum, the broad peak at 3322 cm-1 corresponds to OH stretching while the peaks at 2929-2961 cm-1 correspond to C-H stretching. Thus the presence of alcohol is evident in the IR spectrum.

The 13C NMR suggests the presence of seven C-atoms in this ester. The peak corresponding to carbonyl carbon appear most downfield at ~172 ppm. The other six peaks are in the aliphatic region suggesting an aliphatic ester.

In the 1H NMR, we see a singlet at 2.0 ppm with the integral value of 3. This singlet is characteristic to the protons of the acetate (CH3CO) group as seen in ethyl acetate. This suggests that the acetic acid was employed in this esterification reaction. Using this information along with what we know from 13C NMR we can be certain that the given alcohol contains 5 carbons (total 7 carbon – 2 carbon from acetate group). Therefore the starting material must be pentyl alcohol.

The 1H NMR peaks for the pentyl group are

The most downfield triplet at 4.1 ppm corresponding to OCH2. This is due to deshielding of the CH2 by the electronegative O-atom

The most upfield triplet at 0.9 ppm corresponding to CH3.

A multiplet at 1.3 ppm corresponding to CH2 CH2 which is attached toCH3 moiety

A pentate at 1.6 ppm corresponding to CH2 which is attached toOCH3 moiety

Therefore, the given alcohol is n-pentyl alcohol and the ester is pentyl acetate (Molar mass 130.0994)

See attachment for diagram

Answer:

9.28 × 10⁻¹¹ mol/L

Explanation:

Let's consider the solution of iron(III) hydroxide.

Fe(OH)₃(s) ⇄ Fe³⁺(aq) + 3 OH⁻(aq)

We can relate the solubility (S) of the hydroxide with the solubility product (Ksp) using an ICE chart.

      Fe(OH)₃(s) ⇄ Fe³⁺(aq) + 3 OH⁻(aq)

I                              0                  0

C                            +S               +3S

E                              S                 3S

The solubility product is:

Ksp = [Fe³⁺] × [OH⁻]³ = S × (3S)³ = 27 S⁴

[tex]S=\sqrt[4]{\frac{Ksp}{27} } = \sqrt[4]{\frac{2.0 \times 10^{-39} }{27} } = 9.28 \times 10^{-11} mol/L[/tex]

Answer:

S = 9.28 E-11 M

Explanation:

          S                 S            3S

∴ Ksp Fe(OH)3 = 2.0 E-39

⇒ Ksp = [Fe3+]*[OH-]³ = (S)*(3S)³ = 27(S)∧4

⇒ 27(S)∧4 = 2.0 E-39

⇒ (S)∧4 = 7.407 E-41

⇒ S = (7.407 E-41)∧(1/4)

⇒ S = 9.28 E-11 M

Answer:

increasing fishing for them is thought to be causing problems for some of their natural predators, especially the auks which take them in deeper water. They are also tied as flies to catch fish.

Explanation:

Answer:

The following data point should be kept when making your calibration.

1.21E-05 M

1.60E-05 M

8.09E-06 M

4.04 E-06 M

Explanation:

From the graph it is obvious that the reading for the concentration 2.02E-06 M does not fall on the straight line. So this point should not be taken to construct the calibration plot.

Find attached of the graph.    

Answer:

Check the explanation

Explanation:

Acidipic Acid (which is an essential dicarboxylic acid for manufacturing purposes with about 2. 5 billion kilograms produced per year. It is mainly used for the production of nylon and its related materials.)

Going by the question, since the [tex]H_{2} CrO_{4}[/tex] is comparatively mild oxidizing agent than the [tex]CrO_{3}[/tex], it only oxidizes the carbon group

Kindly check the attached image below for the full explanation to the question above.

Answer:

the ans is option b(help navigate at night)

Answer:

1. 0.02 M

2. 0.01 M

3. 4×10⁻⁶

Explanation:

We know that V₁S₁ = V₂S₂

1.

Concentration of HCl = 0.05 M

end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

2.

2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)

[Ca²⁺] = 0.02 ÷ 2 = 0.01 M

3.

[tex]K_{sp}[/tex] = [Ca²⁺(aq)] [OH⁻(aq)]²

Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)

[tex]K_{sp}[/tex] = [0.01 × (0.02)²] = 4×10⁻⁶

4.

If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.

5.

[tex]K_{sp}[/tex] value describes the solubility of a particular ionic compound. The higher the [tex]K_{sp}[/tex] value, the higher the Solubility will be.

6.

This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility

Final answer:

False statements include: a false notion that ∆Ssys must always be positive for a spontaneous reaction, and a misunderstanding about ∆Euniv increasing in spontaneous reactions. True comprehension of spontaneity involves ∆H and ∆S, affecting Gibbs free energy (∆G), where a negative ∆G indicates spontaneity.

Explanation:

Among the given statements regarding spontaneous chemical reactions, we need to identify those that are false. Here’s the analysis:

a) It is false that if a chemical reaction is spontaneous then the ∆Ssys must always be positive. Spontaneity can also depend on the temperature and ∆Hsys (enthalpy change) of the system, not just ∆Ssys (entropy change).

b) It is true that if a chemical reaction with a negative ∆Ssys is spontaneous, then ∆Ssurr must be positive. This balances the entropy change leading to a positive ∆Suniv (entropy change of the universe), which is a requirement for spontaneity.

c) The statement that if a chemical reaction is spontaneous then the ∆Euniv must be increasing is false. The energy of the universe remains constant; what matters for spontaneity is the entropy change of the universe (∆Suniv), not energy change.

d) The idea that we don't need to worry about spontaneous reactions because they are rare is false. Many important chemical and physical processes are spontaneous, including combustion, corrosion, and the melting of ice at room temperature.

The correct understanding of spontaneity in reactions requires knowledge of both enthalpy (∆H) and entropy (∆S) changes and their effects on the Gibbs free energy (∆G). Generally, for a process to be spontaneous, ∆G must be negative.

The cell potential (Eocell) of the given reaction is 4.07V and the Gibbs free energy(∆Gorxn) is -785.42 KJ/mol suggesting the reaction occurs spontaneously and the cell in consideration is a voltaic cell.

To calculate the cell potential (Eocell), we use the formula Eocell = Eocathode - Eoanode. Here, for the reaction 2 Na+(aq) + 2 Cl−(aq)⟶2 Na(s) + Cl2(g), Na+ is being reduced to Na and Cl- is being oxidized to Cl2, therefore, Cl2 is the anode and Na+ is the cathode. So, the Eocell should be 1.36 V - (-2.71 V) = 4.07 V.

To calculate the Gibbs free energy (∆Gorxn), we use the formula ∆Gorxn = -n F Eocell, where n stands for the number of moles of electrons transferred in the redox reaction (which is 2) and F is the Faraday's constant (96500 C/mol). Therefore, ∆Gorxn = - (2 mol * 96500 C/mol * 4.07 V) = -785420 J/mol or -785.42 KJ/mol.

Since the E0cell value is positive, the reaction will occur spontaneously and represents a voltaic (or galvanic) cell which is a type of electrochemical cell where spontaneous redox reactions are used to convert chemical energy into electrical energy.

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Answer:

C= 3.01 x 10-23 molecules

Explanation:

n = N/NA

0.5 = N/6.02×10^23

N= 3.01×10^23

Answer:

1.34 KJ

Explanation:

To calculate the reaction free energy for the chemical reaction, find the solution below.

Answer:

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

Explanation:

Step 1: Data given

Initial temperature = 10.0 °C

Final temperature = 25.0 °C

Energy required = 30000 J

Mass of the object = 40.0 grams

Step 2: Calculate the specific heat capacity of the object

Q = m* c * ΔT

⇒With Q = the heat required = 30000 J

⇒with m = the mass of the object = 40.0 grams

⇒with c = the specific heat capacity of the object = TO BE DETERMINED

⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C

30000 J = 40.0 g * c * 15.0 °C

c = 30000 J / (40.0 g * 15.0 °C)

c = 50 J/g°C

The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)

Answer:

c. ΔH° is positive and ΔS° is positive.

Explanation:

Hello,

In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:

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

Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.

Best regards.

Final answer:

The reaction C(s) + CO2(g) → 2CO(g) being only spontaneous at high temperatures above 1100K indicates that the reaction has a positive ΔH° and a positive ΔS°, corresponding to option c.

Explanation:

The reaction C(s) + CO2(g) → 2CO(g) is spontaneous only at temperatures above 1100 K. This information allows us to infer the signs of ΔH° (enthalpy change) and ΔS° (entropy change). According to Gibbs free energy equation, ΔG = ΔH - TΔS, a reaction is spontaneous when ΔG is negative. Given that the reaction is only spontaneous at high temperatures, this suggests that ΔH° is positive (endothermic reaction) and ΔS° is positive (increase in disorder). At lower temperatures, the TΔS term is not sufficient to overcome the positive ΔH°, making ΔG positive and the reaction non-spontaneous. Hence, the correct conclusion is that the reaction is spontaneous at high temperatures because of a positive ΔH° and a positive ΔS°, indicating option c.

Answer:

6 orbitals

Explanation:

Each orbital holds 2 electrons so for 5 of 7 orbitals to have unpaired electrons, five orbitals will have one electron and one orbital will have two electrons making 7 total electrons and 6 total orbitals.

The molecule ClF5 needs five degenerate orbitals to house seven electrons, with five of them unpaired. This set of orbitals becomes the five sp³d hybrid orbitals, similar to phosphorus pentachloride, created from the hybridization of atomic orbitals. Similar hybridization happens in boron where the fifth electron occupies one of three 2p degenerate orbitals.

To accommodate seven electrons with five of them unpaired in the molecule ClF5, we need five degenerate orbitals. These are the 3s orbital, three 3p orbitals, and one 3d orbital to form the set of five sp³d hybrid orbitals, modeled after phosphorus pentachloride. They come from the hybridization of atomic orbitals, where valence electrons occupy the newly formed hybrid orbitals. In this case, with ClF5, the hybrid orbital overlaps with a fluorine orbital during the formation of the Cl-F bonds. The process is similar to how boron's fifth electron occupies one of the three degenerate 2p orbitals.

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Answer:

2.06 × 10⁻¹⁰

Explanation:

Let's consider the solution of a generic compound AB₂.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

We can relate the molar solubility (S) with the solubility product constant (Kps) using an ICE chart.

      AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

I                      0              0

C                    +S            +2S

E                      S              2S

The solubility product constant is:

Kps = [A²⁺] × [B⁻]² = S × (2S)² = 4 × S³ = 4 × (3.72 × 10⁻⁴)³ = 2.06 × 10⁻¹⁰

Answer:

It can be concluded that the third step of the reaction is very fast, in this way, it does not contribute to the rate law

Explanation:

Please, observe the solution in the attached Word document.

The rate law is defined as the molar concentration of reactants raised to the power of their stoichiometric coefficients. The law states the dependency of chemical reactions on reactants.

In the given decomposition of nitramide in water, it can be stated that the third step of the reaction is fast, and therefore, does not contribute to the rate law.

The rate-determining step of a reaction is a slow step.

The decomposition reaction can be written as:

O₂NNH₂ [tex]\rightarrow[/tex] O₂NNH⁻ + H⁺ (fast equilibrium)

The value of K is:

[tex]\text{K}_{\text {eq}}&=\dfrac{K_1}{K_{-1}}&=\dfrac{\text[{O}_2\text{NNH}^-][\text{H}^+]}{\text[{O}_2\text{NNH}_{-2}]}[/tex]

[tex]\begin{aligned}\text{Rate}=\text{K}_2\frac{\text{K}_1[\text{O}_2\text{NNH}_2]}{\text{K}_{-1}[\text{H}_2]}\end{aligned}[/tex]

The second step of the reaction:

O₂NNH₂ [tex]\rightarrow[/tex] O₂N + OH⁻

The rate can be given as:

Rate = K₂ [O₂NNH⁻]

Substituting equation 1, we get:

[tex]\begin{aligned} \text[{O}_2\text{NNH}^-] &= \dfrac{\text{K}_1[\text{O}_2\text{NNH}_2]}{\text{K}_{-1}[\text{H}^+]}\end{aligned}[/tex]

Now, writing the equation in the rate constant, we get:

[tex]\text {K}&=\dfrac{K_2K_1}{K_{-1}}[/tex]

Thus, it can be concluded that the third step is very fast and does not contribute to the rate law.

To know more about rate law, refer to the following link:

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Answer:

[tex]n_{C_6H_{10}}=0.03molC_6H_{10}[/tex]

Explanation:

Hello,

In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:

[tex]n_{C_6H_{10}}=3mLC_6H_{12}O*\frac{0.9624gC_6H_{12}O}{1mLC_6H_{12}O}*\frac{1molC_6H_{12}O}{100.158gC_6H_{12}O}*\frac{1molC_6H_{10}}{1molC_6H_{12}O} \\n_{C_6H_{10}}=0.03molC_6H_{10}[/tex]

Besides, the mass could be computed as well by using the molar mass of cyclohexene:

[tex]m_{C_6H_{10}}=0.03molC_6H_{10}*\frac{82.143gC_6H_{10}}{1molC_6H_{10}} \\\\m_{C_6H_{10}}=2.4gC_6H_{10}[/tex]

Even thought, the volume could be also computed by using its density:

[tex]V_{C_6H_{10}}=2.4gC_6H_{10}*\frac{1mLC_6H_{10}}{0.811gC_6H_{10}} \\V_{C_6H_{10}}=3mLC_6H_{10}[/tex]

Best regards.

Answer:

The theoretical yield of moles cyclohexene is 0.0288 moles

Explanation:

Step 1: Data given

Volume of cyclohexanol = 3 mL

Density cyclohexanol = 0.9624 g/mL

Molar mass of cyclohexanol = 100.158 g/mol

Sulfuric acid is in excess

Density of cyclohexene = 0.811 g/mL

Molar mass of cyclohexene = 82.143 g/mol

Step 2: The balanced equation

C6H12O + H2SO4 → C6H10 + H3O + HSO4

Step 3: Calculate mass cyclohexanol

Mass cyclohexanol = density cyclohexane * volume

Mass cyclohexanol = 0.9624 g/mL * 3mL

Mass cyclohexanol = 2.8872 grams

Step 4: Calculate moles cyclohexanol

Moles cyclohexanol = mass cyclohexanol / molar mass

Moles cyclohexanol = 2.8872 grams / 100.158 g/mol

Moles cyclohexanol = 0.0288 moles

Step 5: Calculate moles cyclohexene

For 1 mol cyclohexanol we need 1 mol H2SO4 to produce 1 mol cyclohexene

For 0.0288 moles cyclohexanol we'll have 0.0288 moles cyclohexene

The theoretical yield of moles cyclohexene is 0.0288 moles

Answer:

Explanation:

The mass of polymer in the order recieved is:

The mass of polymer in the 10% solution is:

So to solve this problem, we need to add (390-50) 340 lb of polymer that comes from the 20% solution.

We calculate the mass of the 20% solution that would contain 340 lb of polymer:

So far, in total we have a solution that weighs (1700+500) 2200 lb and contains 390 lb of polymer.

Thus, in order to reach the required mass of solution (and percentage of polymer by weight), we add (3000-2200) 800 lb of pure solvent.

Answer

The transport of the substance into the cell is reduced or reversed in the absence of ATP. I think........

Explanation:

Answer:

1.356x10⁻²¹ grams.

Explanation:

mass of carbon containing 68 atoms = 12g Carbon/6.02x10²³atoms = 1.993x10⁻²³ g/atom  x  68 atoms = 1.356x10⁻²¹ grams.

Answer:

The final balanced equation is :

[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]

Explanation:

[tex]SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq) [/tex]

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

[tex]Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)[/tex]

Balance the charge by adding 2 electrons on product side:

[tex]Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-[/tex]....[1]

Reduction :

[tex]SO_4^{2-}(aq)\rightarrow H_2SO_3(aq) [/tex]

Balance O by adding water on required side:

[tex]SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]

Now, balance H by adding [tex]H^+[/tex] on the required side:

[tex]SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]

At last balance the charge by adding electrons on the side where positive charge is more:

[tex]SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l) [/tex]..[2]

Adding [1] and [2]:

[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]

The final balanced equation is :

[tex]SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)[/tex]

The coefficients of the reactants and products in the balanced equation for the reaction between sulfate ion and tin(II) ion in acidic solution are 1, 1, 10, 1, 1, 5.

The balanced equation for the reaction between sulfate ion and tin(II) ion in acidic solution can be written as:

SO₄²⁻(aq) + Sn₂⁺(aq) + 10H⁺(aq) → H₂SO₃(aq) + Sn₄+(aq) + 5H₂O(l)

The coefficients of the reactants and products in the balanced equation are 1, 1, 10, 1, 1, 5 respectively.

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Answer:

Check the explanation

Explanation:

functional group found in the major organic product = alpha -beta unsaturated ketone

Reaction used to form this functional group = Michael condensation reaction

Also other reactions are - Aldol condensation , Robinson annulation reaction.

Kindly check the attached image below to see the step by step solution to the question above.

The balanced thermochemical equation for the reaction where Hydrogen gas reacts with Oxygen gas to form Water with the release of 242 kJ energy is 2H2(g) + O2(g) → 2H2O(g) ΔH = -242 kJ. This indicates an exothermic reaction.

The balanced thermochemical equation of the reaction where Hydrogen gas (H2) reacts with Oxygen gas (O2) to form Water (H2O) and 242 kJ (kilojoules) energy is released per mole of H2, can be formulated as follows:

2H2(g) + O2(g) → 2H2O(g) ΔH = -242 kJ

The negative sign before the energy value indicates that it's an exothermic reaction, meaning that energy is released during the reaction.

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Answer:

Explanation:The final homogenous solution, after cooling it to 40°C, will contain 47 g of potassium sulfate disolved in 150 g of water, so you can calculate the amount disolved per 100 g of water in this way:

[47 g of solute / 150 g of water] * 100 g of g of water = 31.33 grams of solute in 100 g of water.

So, when you compare with the solutiblity, 15 g of solute / 100 g of water, you realize that the solution has more solute dissolved with means that it is supersaturated.

To make a saturated solution, 15 grams of potassium sulfate would dissolve in 100 g of water.

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