Answer:
[tex]PO_4^{-3} < NO_3^- < Na^+[/tex]
Explanation:
Solutions.
a) 100 ml 1 M of Na3PO4
b) 100 ml 1 M of AgNO3
c) Mixture: 200 ml 0.5 M of Na3PO4 and 0.5 M of AgNO3
Reaction:
[tex] 3 Ag^+ + PO_4^{-3} \longrightarrow Ag_3PO_4[/tex]
So if silver ion is consumed almost completely:
[tex][Ag^+]=0 M[/tex]
[tex][PO_4^{-3}]=0.5 M-0.5 M \frac{1 mol PO4}{3 mol Ag}=0.33 M[/tex]
[tex][Na^+]=0.5 M * \frac{3 mol Na}{mol Na_3PO_4}=1.5 M[/tex]
[tex][NO_3^-]=0.5 M[/tex]
In increasing order of concentration:
[tex]PO_4^{-3} < NO_3^- < Na^+[/tex]
The correct order of ions remaining in solution after mixing Na3PO4 with AgNO3, by increasing concentration, is phosphate (PO43-), nitrate (NO3-), and sodium (Na+), which is option (A).
Explanation:When 100 ml of 1.0 M Na3PO4 is mixed with 100 ml of 1.0 M AgNO3, a reaction occurs where silver phosphate (Ag3PO4), a yellow precipitate, forms due to its low solubility and Ag+ ions become almost absent in the solution. The remaining ions in the solution will be Na+, NO3-, and a very small amount of PO43-. Since all of the Ag+ reacts to form the precipitate, we can assume that there are three times as many Na+ ions compared to PO43- ions originally, because each formula unit of Na3PO4 produces three sodium ions. Therefore, this leaves Na+ with the highest concentration in solution. NO3- concentration remains unchanged because it is a spectator ion and does not participate in the reaction.
The correct order of ions in solution by increasing concentration is PO43- < NO3- < Na+, which corresponds to option (A).
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George is writing an essay about the role of observation and inference in the development of the atomic theory. He wants to explain why it was more difficult to observe the presence of neutrons in atoms. Which two statements should he include in his essay?
George should include two main points: First, the hypothesis of the existence of neutrons arose from the need to explain the remaining mass in an atom's nucleus not accounted for by protons. Second, the detection of neutrons was especially difficult due to their lack of charge, which required more advanced techniques to observe.
Explanation:In his essay, George should include the following statements:
The nucleus of an atom contains most of its mass, and although the number of protons was determined, they only accounted for half, or less, of that mass. This led to the hypothesis of the existence of neutral particles in the nucleus — known as neutrons.Detecting neutrons was more difficult because they were uncharged particles. It wasn't until 1932 that physicist James Chadwick found evidence of neutrons, uncharged, subatomic particles with a mass approximately the same as that of protons.Understanding the nature of neutrons was a significant step in the development of atomic theory, as it helped explain isotopes and variants of a particular chemical element that differ in neutron number.
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Classify the following as an endothermic or exothermic reaction:
a. making popcorn in a microwave oven.
b. a burning match.
c. boiling water.
d. burning rocket fuel.
e. the reaction inside a heat pack
Answer:
a. making popcorn in a microwave oven. Endothermic
b. a burning match. Exothermic
c. boiling water. Endothermic
d. burning rocket fuel. Exothermic
e. the reaction inside a heat pack Exothermic
Explanation:
In order to answer, we need to review the definitions of exothermic and endothermic reactions.
Exothermic reactions give out heat. They cause increase in the energy of the system.
Endothermic reactions absorb heat. They cause decrease in the energy of system.
By this definition,
a. making popcorn in a microwave oven. Endothermic as heat energy is provided to the corn which causes it to pop.
b. a burning match. Exothermic as heat energy is given out by a burning match.
c. boiling water. Endothermic as heat energy is provided to the water which causes it to boil.
d. burning rocket fuel. Exothermic as heat energy is given out by burning fuel.
e. the reaction inside a heat pack. Exothermic as reaction which takes place inside heat pack gives out heat. This heat provides comfort to painful joints and muscles.
Energy is absorbed in an endothermic reaction while energy is released in an exothermic reaction.
An exothermic process is a process in which energy is released. This implies that heat is evolved in the process. In an endothermic process, heat is absorbed in the process. We shall now classify the following process as endothermic or exothermic accordingly;
making popcorn in a microwave oven - Endothermic a burning match - Exothermicboiling water - Endothermicburning rocket fuel - Exothermic the reaction inside a heat pack - ExothermicLearn more: https://brainly.com/question/4612545
Why does ionization energy decrease moving down a group in the periodic table?
Answer:
Electron shielding
Explanation:
Ionization energy decreases moving down a group in the periodic table because of a phenomen known as Electron shielding, in which valence electrons do not interact with the positively charged nucleus as strongly as inner electrons do, because these inner electrons shield the valence electrons. This means it's easier for these valence electrons to leave the atom the more inner electrons are between them and the nucleus, this translates into a decreased ionization energy value.
Ionization energy decreases moving down a group due to the increased distance of the valence electrons from the nucleus, the greater shielding effect of inner electrons, and the higher principal quantum number of the valence electrons, which reduces the effective nuclear charge.
Ionization energy refers to the amount of energy required to remove an electron from an atom in the gaseous state. A key periodic trend observed is that ionization energy decreases as we move down a group in the periodic table. There are several reasons for this trend:
As atomic size increases down a group, the outermost, or valence, electrons are further from the nucleus, experiencing a weaker attraction due to increased distance.The shielding effect comes into play; inner shell electrons block the pull of the positively charged nucleus on the outer electrons, making them easier to remove.The increase in principal quantum number (n) down a group means that valence electrons are in higher energy levels, further away from the nucleus, thereby decreasing effective nuclear charge (Zeff) on these electrons.For example, within Group 1 of the periodic table, which contains the alkali metals, we notice a significant drop in ionization energy from lithium to cesium as each succeeding element has more filled inner electron shells that shield the outer electron. Furthermore, the increase in atomic number down a group does not proportionately increase the nuclear charge's effect on the valence electrons due to the reasons mentioned above.
In summary, the larger atomic radius and greater shielding by inner electrons result in a reduced attraction between the valence electrons and the nucleus, leading to a decrease in ionization energy down a group.
Calculate the grams of so2 gas present at stp in a 5.9 l container.
Answer: The mass of sulfur dioxide gas at STP for given amount is 16.8 g
Explanation:
At STP conditions:
22.4 L of volume is occupied by 1 mole of a gas.
So, 5.9 L of volume will be occupied by = [tex]\frac{1mol}{22.4L}\times 5.9L=0.263mol[/tex]
Now, to calculate the mass of a substance, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of sulfur dioxide gas = 0.263 mol
Molar mass of sulfur dioxide gas = 64 g/mol
Putting values in above equation, we get:
[tex]0.263mol=\frac{\text{Mass of sulfur dioxide gas}}{64g/mol}\\\\\text{Mass of sulfur dioxide gas}=(0.263mol\times 64g/mol)=16.8g[/tex]
Hence, the mass of sulfur dioxide gas at STP for given amount is 16.8 g
Look up the density of n-butyl chloride (1-chlorobutane). Assume that this alkyl halide was prepared instead of the bromide. Decide whether the alkyl chloride would appear as the upper or the lower phase at each stage of the separation procedure: after the reflux, after the addition of water, and after the addition of sodium bicarbonate.
Answer:
See explanation below
Explanation:
First, you need to know the density of each compound in order to know this.
The density of 1-chlorobutane is 0.88 g/mL,
The density of water is 1 g/mL
The density of sodium bicarbonate is 2.2 g/cm3.
therefore, the one that has a greater density will always go at the lower phase.
In this case, after the reflux, it will stay in the lower phase, basically because you don't have another solvent with a greater density than the butane.
After adding water, it will be in the upper phase, water has a greater density.
After adding bicarbonate, it will be in the upper phase too.
What type of compound involves the transfer of electrons?
Answer: Electrovalent or Ionic Compounds
Explanation:
Electrovalent Compounds Form bonds that are characterised by transfer of electrons from metallic atoms to non-metal licenses atoms during a chemical reaction.
The metallic atom after donating their valence electrons, become positively charged, while the non-metal license atoms becomes negatively charged after acquiring extra electrons.
A typical example of electrovalent compounds can be found between the association of Group 1(Alkali Metals) elements and the Group 7(Halogen Family) elements.
Answer: The type of compound involves the transfer of electrons is called the ionic compounds.
Explanation: ionic compounds are compounds in which one atom or molecule completely transfers an electron to another.
Ions that have gained an electron are negatively charged and they are called anions while ions that have lost an electron are positively charged and they are called cations.
What sublevels are filling across the transition elements
The transition elements fill d sublevels, coming after the s sublevel of the same principal energy level. Lanthanides begin filling the 4f sublevel after the 6s, positioned two principal energy levels behind. Many transition element compounds display bright colors from d electron transitions.
Explanation:The sublevels that are filling across the transition elements are primarily the d sublevels. The electron configurations of these elements have their outermost s sublevel either completely filled or missing one electron. However, the defining characteristic of transition elements is the filling of the inner d sublevel, which typically occurs after the s sublevel of the same principal energy level has been filled. When discussing the f-block elements, specifically the lanthanides, these elements begin filling their 4f sublevels after the 6s sublevel. This is due to the behaviour of electron filling, where the f sublevels are two principal energy levels behind the current one being filled.
Many transition element compounds are known for their brightly colored appearances as a result of inner-level d electron transitions. Unlike the transition elements, the lanthanides are not grouped together in the periodic table and instead are inserted in a separate row, reflecting their unique electron configurations and properties.
In Experiment #6 you learned that Zn will dissolve in acidic aqueous solutions as Zn2+, but Cu will not dissolve because Cu is less active than H2. A stronger oxidizing agent than H+ is required to oxidize Cu. Aqueous solutions of ________________ are strong oxidizing agents that are capable of oxidizing a number of less active metals, including Cu.
Answer:family fdhdfgfdsh
Nitric acid aqueous solutions are strong oxidizing agents capable of oxidizing a variety of less active metals, including Cu.
What is Nitric acid?The inorganic compound nitric acid has the formula HNO3. It is a mineral acid that is extremely corrosive. The compound is colorless, but older samples have a yellow cast due to decomposition into nitrogen oxides. The majority of commercially available nitric acid has a water concentration of 68%. Nitric acid is used to make ammonium nitrate, a key component of fertilizers. It is also used to make explosives such as nitroglycerin and trinitrotoluene (TNT), as well as to oxidize metals. Nitric acid, with the chemical formula HNO3, is a strong acid. It is also known as niter's spirit and aqua fortis. It is colorless in its pure form, but as it ages, it takes on a yellow cast.
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A voltaic cell is constructed with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell. Measurement shows that the silver electrode is positive.
(a) Write balanced half-reactions and the overall spontaneous reaction. (Type your answer using the format [NH4]+ for NH4+. Use the lowest possible coefficients.)
reduction:
__Ag+(aq) + __e- -> ___(s)
oxidation:
__(s) ->__Pb2+(aq) +__e-
overall reaction:
__Ag+(aq) +__(s) ->__(s) + __ Pb2+(aq)
(b) The cation flow is towards which electrode?
lead OR silver
What process occurs at the lead electrode?
oxidation OR reduction
In which direction do the electrons flow?
from the lead electrode to the silver electrode OR from the silver electrode to the lead electrode
Answer:
a) Reduction:
Ag⁺(aq) + e⁻ → Ag(s)
Oxidation:
Pb(s) → Pb⁺²(aq) + 2e⁻
Overall reaction:
2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺
b) Silver; oxidation; from the lead electrode to the silver electrode.
Explanation:
a) Ag⁺ had lost 1 electron, so need to gain 1 electron to become Ag(s). Pb needs to lose 2 electrons to become Pb⁺².
Reduction:
Ag⁺(aq) + e⁻ → Ag(s)
Oxidation:
Pb(s) → Pb⁺²(aq) + 2e⁻
Overall reaction:
2Ag⁺(aq) + Pb(s) → 2Ag(s) + Pb²⁺ (it will need 2Ag⁺ to gaind the 2 electrons released by Pb)
b) The cation formed in the redox reaction is Pb²⁺, so, to equilibrate the charges, it will flow towards the silver (Ag) electrode.
The lead (Pb) is being oxidized, so oxidation is happening at it.
The electrons flow from the oxidation (anode) to the reduction (cathode), so they flow from the lead electrode to the silver electrode.
In the voltaic cell, the Ag+ is reduced to Ag in the silver half-cell, while Pb is oxidized to Pb2+ in the lead half-cell. The cations flow towards the silver electrode and the electrons flow from the lead to the silver electrode. Hence, the overall reaction is 2Ag+(aq) + Pb(s) -> 2Ag(s) + Pb2+(aq).
Explanation:In a voltaic cell with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell, the silver half-cell acts as the cathode or reduction half-cell which gains electrons, while the lead half-cell acts as the anode or oxidation half-cell and loses electrons. Therefore, the balanced half-reactions and overall spontaneous reactions are:
(a) Balanced Half-Reactions and Overall Reaction:
Reduction: Ag+(aq) + 1e- -> Ag(s)
Oxidation: Pb(s) -> Pb2+(aq) + 2e-
Overall Reaction: 2Ag+(aq) + Pb(s) -> 2Ag(s) + Pb2+(aq)
(b) The Cation Flow and Electrons Flow:
The cation flow is towards the silver electrode and the electron flow is from the lead electrode to the silver electrode. In the voltaic cell, the process that occurs at the lead electrode is oxidation.
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Water has the following thermodynamic values: ΔH°fus of H2O = 6.02 kJ/mol ΔH°vap of H2O = 40.7 kJ/mol heat capacity of solid H2O = 2.09 J/g°C heat capacity of liquid H2O = 4.18 J/g°C heat capacity of gaseous H2O = 1.97 J/g°C
How much energy (in kJ) is required to raise the temperature of 25.0 g of H2O from -129°C to 262°C?
Enter your answer in units of kJ to three significant figures.
Answer:
Qtotal = 90.004 kJ
Explanation:
To start resolving the problem we need to first convert the kJ/mol units from the thermodynamic values to J/g, so that we can work with the units of the heat capacity values. We know that the molar mass of water is 18.015 g/mol, so with this we do the respective conversion:
ΔH°fus of H2O = (6.02 kJ/mol) (1 mol/18.015g) (1000J/kJ) = 334.165 J/g
ΔH°vap of H2O = (40.7 kJ/mol) (1mol/18.015g) (1000J/kJ) = 2259.228 J/g
Now we need to find out the heat energy required to rise the temperature (specific heat capacity) and the energy required for each change of phase (specific latent heat), and add everything up. For this we will require the specific heat capacity and latent heat equations:
Q = mCΔT ; where m = mass, C = Hear capacity, ΔT = change of temperature
Q = mL ; where m = mass, L = specific latent heat
First change of phase (solid to liquid - fusion)
Q1 = (25g) (2.09 J/g°C) (0°C - (-129°C) = 6740.25 J
Q2 = (25g) (334.165 J/g) = 8354.125 J
Second change of phase (liquid to gas - vaporization)
Q3 = (25g) (4.18 J/g°C) (100°C - 0°C = 10450 J
Q4 = (25g) (2259.228 J/g) = 56480.7 J
Rise of temperature of the gaseous water
Q5 = (25g) (1.97 J/g°C) (262°C - 100°C = 7978.5 J
Finally we add everything up:
Qtotal = Q1 + Q2 + Q3 + Q4 + Q5 = 6740.25 J + 8354.125 J + 10450 J + 56480.7 J + 7978.5 J = 90003.575 J = 90.004 kJ
Change of state has to do with the movement from one state of matter to another.
Change of stateChange of state has to do with the movement from one state of matter to another. We need to first change the kJ/mol units to J/g,knowing that the molar mass of water is 18.015 g/mol hence:
ΔH°fus of H2O = (6.02 kJ/mol) (1 mol/18.015g) (1000J/kJ) = 334.165 J/g
ΔH°vap of H2O = (40.7 kJ/mol) (1mol/18.015g) (1000J/kJ) = 2259.228 J/g
For each phase change;
Q = mCΔT
where
m = mass, C = Hear capacity, ΔT = change of temperatureAnd
Q = mL
where
m = mass, L = specific latent heatFirst, solid to liquid change
Q1 = (25g) (2.09 J/g°C) (0°C - (-129°C) = 6740.25 J
Q2 = (25g) (334.165 J/g) = 8354.125 J
Second liquid to gas change
Q3 = (25g) (4.18 J/g°C) (100°C - 0°C) = 10450 J
Q4 = (25g) (2259.228 J/g) = 56480.7 J
Then,
Q5 = (25g) (1.97 J/g°C) (262°C - 100°C) = 7978.5 J
Finally:
Qtotal = Q1 + Q2 + Q3 + Q4 + Q5 = 6740.25 J + 8354.125 J + 10450 J + 56480.7 J + 7978.5 J = 90003.575 J = 90.004 kJ
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In cellular chemical pathways, the product(s) of any particular reaction are often quickly consumed by the next reaction in the pathway. This would tend to keep the product concentration ___ and drive the reaction ____.
Answer:
Tend to keep the product concetration low and therefore drive the reaction righward
Explanation:
The fact the products of a reaction are quickly consumed by the next one would tend to keep the product concetration low and therefore drive the reaction righward (to the products).
This happens because the system will not achive equilibrium between the reactants and the product, and will keep producing it util the system achives equilibrium or the reactants dry out.
It turns out that the van der Waals constant b equals four times the total volume actually occupied by the molecules of a mole of gas.
Using this figure, calculate the fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C. Assume b=0.0322 L/mol
The fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C is 92.0%. The van der Waals constant b is 0.0322 L/mol.
The van der Waals constant b is defined as four times the total volume actually occupied by the molecules of a mole of gas 1. To calculate the fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C, we can use the following formula:
V_real = V_ideal - nb
where V_real is the real volume of the gas, V_ideal is the ideal volume of the gas, n is the number of moles of the gas, and b is the van der Waals constant. At 0 C, the ideal volume of one mole of any gas is 22.4 L 2. Therefore, the ideal volume of Ar atoms is 22.4 L/mol.
To calculate the real volume of Ar atoms, we need to know the number of moles of Ar atoms present in the container. We can use the ideal gas law to calculate the number of moles of Ar atoms:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Substituting the given values, we get:
n = PV/RT = (230 atm * V)/(0.0821 L atm/mol K * 273 K) = 9.03 V
Substituting the values of n and b into the formula for V_real, we get:
V_real = V_ideal - nb = 22.4 L/mol - 0.0322 L/mol * 4 * 9.03 mol = 20.6 L
Therefore, the fraction of the volume in a container actually occupied by Ar atoms at 230 atm pressure and 0 C is:
(V_real/V) * 100% = (20.6 L/V) * 100% = 92.0%
Based on the diagram below, which would have the most inertia?
The sphere on the left, because it has more mass
The sphere on the right because it has less mass
The sphere on the left because it is a bigger diameter
The sphere on the right because it is a smaller diameter
Answer:
The sphere on the left has the most inertia because it has more mass.
Explanation:
Inertia is a property of matter of a substance.
According to Newton's first law of motion, a body continues to stay in the state of rest or constant velocity unless acted upon a external force.
The amount of inertia that an object possess is proportional to the mass of the object.
The sphere on the left is of 300 kg and that on the right is of 30 kg.
Clearly, the sphere on the left has more mass.
Therefore, the sphere on the left has the most inertia.
Answer:
it would be the sphere on the left that has more mass
Explanation:
because the more weight that is applied to a force with being a small object appling force upon it it will be harder to move
A large sport utility vehicle has a mass of 2700 kg.Calculate the mass of CO2 emitted into the atmosphere upon accelerating the SUV from 0.0 mph to 67.0 mph. Assume that the required energy comes from the combustion of octane with 30\% efficiency. (Hint: Use KE=1/2mv^2 to calculate the kinetic energy required for the acceleration.)
Answer:
[tex]m_{CO2}=260.7 g CO2[/tex]
Explanation:
First of all we need to calculate the energy required:
[tex]KE= 0.5*m*v^2[/tex]
where:
[tex]m=2700kg[/tex]
[tex]v=67 mph=29.95 m/s[/tex]
[tex]KE= 0.5*2700kg*(29.95)^2[/tex]
[tex]KE= 1210953 J=1210.953 kJ[/tex]
Octane's combustion enthalpy: [tex]\Delta H_{comb}=- 5450 kJ/mol[/tex]
The reaction:
[tex]C_8H_{18} + 25/2 O_2 longrightarrow 8 CO_2 +9 H_O[/tex]
Mass of CO2:
[tex]m_{CO2}=\frac{1210.953 kJ}{5450mol}*\frac{1}{0.3}*\frac{8 mol CO2}{1 mol}*\frac{44 g CO2}{mol CO2}[/tex]
[tex]m_{CO2}=260.7 g CO2[/tex]
Electrons do not usually flow through the electron-transport chain to O2, unless ADP is simultaneously phosphorylated to _____.
Answer:
ATP
Explanation:
The main purpose for cellular respiration, is to finally obtain ATP (Adenosine Triphosphate), this process occurres through the electron transport chain: this is the final step of the aerobic respiration, and takes place when energy from NADH and FADH₂ (both products from Krebs Cycle) is transferred to ATP
This process occurres within the inner membrane of the mitochondria: while protons (H⁺) pass through the ATP synthase (this protein acts as a “tunnel” where H⁺ go through), which uses the difference of protons (H⁺) concentration between the matrix (between 2 mitochondrial membranes) and the inner matrix of mitochondria.
The ATP synthase also acts as an enzyme, creating ATP using ADP + Pi (inorganic phosphorus)
The electrons used to help with this process, finally attach to O₂ (oxygen) to form H₂O
Answer: The correct answer is : ATP
Explanation: The phosphorylation reaction is a type of metabolic reaction that results in the formation of (ATP) or (GTP) by the direct transfer and donation of a phosphoryl (PO3) group to (ADP) or (GDP) from a phosphorylated reactive intermediate. The breath is an ATP generating process in which an inorganic compound serves as the ultimate e-acceptor. O2 is delivered by blood flow.
What is the density of a block of marble that occupies 310 cm3 and has a mass of 853 g?1. Diamond has a density of 3.26 g/cm3 .2.What is the mass of a diamond that has a volume of 0.350 cm3 ?3. What is the volume of a sample of liquid mercury that has a mass of 76.2 g, given that the density of mercury is 13.6 g/mL?4. What is the density of a sample of ore that has a mass of 74.0 g and occupies 20.3 cm3 ?
Answer:53gm
Explanation:
To calculate the density of a substance, divide its mass by its volume. For the mass of a diamond or volume of mercury, multiply or divide, respectively, the given quantity by the substance's density. Densities are significant as they indicate how much matter is contained within a space.
The density of a substance is defined as its mass per unit volume. The formula for density (d) is d = mass (m) / volume (v), where the mass is measured in grams (g) and the volume in cubic centimeters (cm3) for solids and liquids, or in milliliters (mL) as 1 mL equals 1 cm3.
To find the density of a block of marble, we use the formula with the given values: d = 853 g / 310 cm3.
To find the mass of a diamond with a known density, multiply the volume by the density: mass = 0.350 cm3
x 3.26 g/cm3.
The volume of liquid mercury given its mass and density can be calculated by rearranging the formula: volume = 76.2 g / 13.6 g/mL.
Lastly, to find the density of a sample of ore, apply the formula: d = 74.0 g / 20.3 cm3.
Remember that densities can vary greatly among different materials and are particularly high for substances such as gold and mercury.
A certain substance X condenses at a temperature of 123.3 degree C . But if a 650. g sample of X is prepared with 24.6 g of urea ((NH2)2 CO) dissolved in it, the sample is found to have a condensation point of 124.3 degree C instead. Calculate the molal boiling point elevation constant Kb of X.Round your answer to 2 significant digits.
Answer:
The molal boiling point elevation constant is 1.59 ≈ 1.6 [tex]Kkgmol^{-1}[/tex]
Explanation:
To solve this question , we will make use of the equation ,
Δ[tex]T_{b} = i*K_{b} *m[/tex]
where ,
Δ[tex]T_{b}[/tex] is the change in boiling point of the substance [tex]X[/tex] ( °[tex]C[/tex] or [tex]K[/tex])[tex]i[/tex] is the Vant Hoff Factor which = 1 in this case ( no unit )[tex]K_{b}[/tex] is the mola boiling point elevation constant of X ( [tex]Kkgmol^{-1}[/tex])[tex]m[/tex] is the molality of the solution which has [tex](NH_{2})_{2} CO[/tex] as the solute and [tex]X[/tex] as the solution ([tex]molkg^{-1}[/tex])Δ[tex]T_{b}[/tex] = [tex]124.3 -123.3 = 1[/tex] °[tex]C[/tex] or [tex]K[/tex];[tex]i[/tex]=1;[tex]m[/tex]= [tex]\frac{moles of solute}{weight of solvent(kg)}[/tex][tex]molkg^{-1}[/tex]∴ [tex]m = \frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]
as the weight of [tex](NH_{2})_{2} CO[/tex] is [tex]60g[/tex] and thus number of moles = [tex]\frac{24.6}{60}[/tex]and the weight of solvent in [tex]kg[/tex] is [tex]\frac{650}{1000}[/tex]4. [tex]K_{b}[/tex] ⇒ ?
∴
[tex]1=1*K_{b} *\frac{\frac{24.6}{60} }{\frac{650}{1000} }[/tex]
⇒ [tex]K_{b}[/tex] = [tex]1.59[/tex] ≈ 1.6 [tex]Kkgmol^{-1}[/tex]
The molal boiling point elevation constant (Kb) of substance X is 4.1.
Explanation:The molal boiling point elevation constant (Kb) can be calculated using the formula: ΔT = Kb × m
Where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant, and m is the molality of the solution.
In this case, the change in boiling point (ΔT) is 1 degree C (124.3 - 123.3), the molality (m) can be calculated by dividing the molal mass of urea by the mass of the solvent water, which gives a value of 0.0246 kg urea / 0.100 kg water = 0.246 mol/kg, and the formula becomes: 1 = Kb × 0.246
Now, rearrange the equation to solve for Kb: Kb = 1 / 0.246 = 4.07
Rounding to 2 significant digits, the molal boiling point elevation constant Kb of substance X is 4.1.
A sample of hydrogen gas was collected over water at 36°C and 855 mm Hg. The volume of the container was 6.50 L. Calculate the mass of H2(g) collected. (Vapor pressure of water = 44.6 torr at 36°C)
Answer:
The amount of hydrogen gas collected will be 0.5468 g
Explanation:
We are given:
Vapor pressure of water = 44.6 torr = 44.6 mm Hg
Total vapor pressure = 855 mm Hg
Vapor pressure of hydrogen gas = Total vapor pressure - Vapor pressure of water = (855 - 44.6) mmHg = 810.4 mmHg
To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 810.4 mmHg
V = Volume of the gas = 6.50 L
T = Temperature of the gas = [tex]36^oC=[36+273]K=309K[/tex]
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = ?
Putting values in above equation, we get:
[tex]810.4mmHg\times 6.50L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 309K\\\\n=\frac{810.4\times 6.50}{62.3637\times 309}=0.2734\ mol[/tex]
To calculate the mass from given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of hydrogen gas = 0.2734 moles
Molar mass of hydrogen gas = 2 g/mol
Putting values in above equation, we get:
[tex]0.2734mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.2734mol\times 2g/mol)=0.5468g[/tex]
Hence, the amount of hydrogen gas collected will be 0.5468 g
Part C What is a reaction mechanism?
What is a reaction mechanism?
A reaction mechanism is an overall reaction that occurs with necessary values such as activation energy, temperature which explain the experimentally determined rate law.
A reaction mechanism is a rate-determining step of the reaction that describe how fast an overall reaction occurs and explain the limiting factors of reactions rate.
A reaction mechanism is a series of elementary reactions that describe how an overall reaction occurs and explain the experimentally determined rate law.
Answer:
A reaction mechanism is a series of elementary reactions that describe how an overall reaction occurs and explain the experimentally determined rate law.
Explanation:
A reaction machanism is a succession of steps to go from the reactants to the products of a reaction.
Each of this steps is a different elementary reaction and produces an intermediary product. This intermediaries not always can be seen in real life due to the high rate of reaction of following steps.
This intermediary reactions where created to explain the experimental determined rate law of some reactions, that doesn't fit in a one-step reaction.
In conclusion we can say that: A reaction mechanism is a series of elementary reactions that describe how an overall reaction occurs and explain the experimentally determined rate law
Final answer:
A reaction mechanism explains the step-by-step sequence in which reactants are converted into products, detailing elementary reactions that make up the entire process, including the rate-determining step that dictates the overall rate of the reaction.
Explanation:
A reaction mechanism is the detailed process by which a chemical reaction occurs, broken down into a series of elementary steps. Each of these steps involves a certain number of reactant species, as detailed by the molecularity (unimolecular, bimolecular, or termolecular) of the reaction. The overall rate of the chemical reaction is determined by the rate-determining step, which is the slowest step within the reaction mechanism. The rate laws for the elementary reactions must align with the experimentally determined rate law to confirm the proposed reaction mechanism as plausible.
Catalysts are substances that alter the reaction mechanism, offering an alternative pathway with a lower activation energy, which in turn affects the reaction rate without impacting the chemical equilibrium of the reaction.
A balanced chemical equation alone does not convey the complexity of the reaction mechanism. For example, the decomposition of ozone is a multi-step process, not evident from the simple overall equation. The mechanism provides a detailed description of these steps, much like a roadmap of a journey illustrating every turn and stop, rather than just showing the start and end points.
Which procedure will most increase the energy output of a nuclear reactor? exposing the core to the outside world removing fissile material from the core decreasing the amount of steam produced raising the control rods out of the reactor
Raising the control rods out of the reactor
The reactor regulates the number of neutrons that are involved in the chain reaction. This is accomplished by the reactor absorbing some of the neutrons, produced in the splitting of the atoms, in its walls.
Explanation:
If the rods are taken out of the reactor, the rods would heat up very fast and most probably an explosion would occur. This is because most of the neutrons produced in the splitting of the radioactive atoms in the rod would go ahead and bombard other atoms in the rods hence spiking up the chain reaction rate. This would release a lot of energy at a go.
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Answer:
Raising the control rods out of the reactor
Explanation:
Raising the rods will allow the chain reaction to flow more freely, therefore increasing the energy output of the nuclear reactor
A sample of gas has a volume of 1.9L and a temperature of 21 degrees celsius. Heat is applied to the sample, leading to an increase in temperature to 27 degrees celsius. The sample is held at constant pressure. Solve for the final volume.
Answer:
1.94 L
Explanation:
21°C = 21 +273 = 294 K
27°C = 27 + 273 = 300 K
T1/V1 = T2/V2
294 K/1.9 L = 300 K/x L
x = (1.9*300)/294 ≈ 1.94 L
A flask with a mass of 321.9 g is filled with 15.7 mL of carbon tetrachloride. The mass of the flask and carbon tetrachloride is found to be 523.6 g. From this information, calculate the density of carbon tetrachloride. According to this problem, the density of CCl4 is Answer g/mL.
Answer:
Density of carbon tetrachloride = 12.8 g/mL
Explanation:
Given :
[tex]m_{flask}=321.9\ g[/tex]
[tex]m_{flask}+m_{CCl_4}=523.6\ g[/tex]
Mass of carbon tetrachloride: -
[tex]m_{flask}+m_{CCl_4}=523.6\ g[/tex]
[tex]m_{CCl_4}=523.6-m_{flask}\ g=523.6-321.9\ g=201.7\ g[/tex]
Mass of carbon tetrachloride = 201.7 g
Given, Volume = 15.7 mL
Considering the expression for density as:
[tex]Density=\frac {Mass}{Volume}[/tex]
So,
[tex]Density=\frac {201.7\ g}{15.7\ mL}[/tex]
Density of carbon tetrachloride = 12.8 g/mL
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.2-L bulb, then filled it with the gas at 2.00 atm and 27.0 ∘C and weighed it again. The difference in mass was 5.1 g.
Identify the gas. Express your answer as a chemical formula.
Answer:
N2
Explanation:
We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get
Given:
P = 2atm
1atm = 101,325pa
2atm = 202,650pa
T = 27 degrees Celsius = 27 + 273.15 = 300.15K
V = 2.2L
R = molar gas constant = 8314.46 L.Pa/molK
PV = nRT
Rearranging n = PV/RT
Substituting these values will yield:
n = (202,650 * 2.2)/(8314.46* 300.15)
n = 0.18 moles
To get the molar mass, we simply divide the mass by the number of moles.
5.1/0.18 = 28.5g/mol
This is the closest to the molar mass of diatomic nitrogen N2.
Hence, the gas is nitrogen gas
The diatomic gas could be identified using the ideal gas law and the given conditions. The calculated molar mass matched with the molar mass of Oxygen, so the diatomic gas is likely Oxygen (O2).
Explanation:By examining the given conditions and the difference in mass, we can identify the gas X2 using the ideal gas law. The ideal gas law is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Here, the pressure P = 2.00 atm, the volume V = 2.2 L, R = 0.0821 L.atm/mol.K and T = 27°C = 300.15 K. Inserting these values gives us the number of moles of gas. The molar mass of the gas can be calculated by dividing the mass of the gas by the number of moles. Using the molar mass and comparing it to the periodic table, the diatomic gas appears to be Oxygen (O2).
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If 50.00 mL of 1.05 M sodium hydroxide is added to 25.00 mL of 1.88 M hydrochloric acid, with both solutions originally at 24.66°C, what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of 1.02 g/mL and a specific heat of 3.98 Jg⁻¹°C⁻¹.)
Final answer:
To find the final solution temperature, we need to calculate the heat exchanged by the reaction and the heat exchanged by the solution and set them equal to each other. By plugging in the given values and solving the equation, we find that the final solution temperature will be 24.77 °C.
Explanation:
To find the final solution temperature, we can use the principle that the heat given off by the reaction is equal to that taken in by the solution. We need to calculate the heat exchanged by the reaction and the heat exchanged by the solution and set them equal to each other.
First, we calculate the heat exchanged by the reaction using the equation:
q_reaction = C_reaction * ΔT_reaction
where C_reaction is the heat capacity of the reaction solution and ΔT_reaction is the change in temperature of the reaction.
Next, we calculate the heat exchanged by the solution using the equation:
q_solution = m_solution * C_solution * ΔT_solution
where m_solution is the mass of the solution, C_solution is the specific heat of the solution, and ΔT_solution is the change in temperature of the solution.
Now we can set the two heat exchanges equal to each other and solve for the final solution temperature:
q_reaction = q_solution
C_reaction * ΔT_reaction = m_solution * C_solution * ΔT_solution
Plugging in the given values:
C_reaction = C_solution = 3.98 Jg⁻¹°C⁻¹
m_solution = (50.00 mL of NaOH * 1.02 g/mL) + (25.00 mL of HCl * 1.02 g/mL) = 76.50 g
ΔT_reaction = (28.9 °C - 24.66 °C) = 4.24 °C
ΔT_solution = ?
Now we can solve for ΔT_solution:
3.98 Jg⁻¹°C⁻¹ * 4.24 °C = 76.50 g * 3.98 Jg⁻¹°C⁻¹ * ΔT_solution
ΔT_solution = (3.98 Jg⁻¹°C⁻¹ * 4.24 °C) / (76.50 g * 3.98 Jg⁻¹°C⁻¹) = 0.1107 °C
Finally, we calculate the final solution temperature:
Final Temperature = 24.66 °C + 0.1107 °C = 24.77 °C
The final temperature of the solution after the reaction is approximately 33.51°C.
To find the final temperature of the solution after the neutralization reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl), we can follow these steps:
The reaction between NaOH and HCl can be written as:
[tex]\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \][/tex]
Calculate the moles of NaOH and HCl:
[tex]\[ \text{Moles of NaOH} = 1.05 \, \text{M} \times 0.05000 \, \text{L} = 0.0525 \, \text{moles} \] \[ \text{Moles of HCl} = 1.88 \, \text{M} \times 0.02500 \, \text{L} = 0.0470 \, \text{moles} \][/tex]
Since HCl is the limiting reagent (0.0470 moles compared to 0.0525 moles of NaOH), the reaction will produce 0.0470 moles of water.
2.Calculate the heat released during the reaction:
The enthalpy change for the neutralization of strong acid and base (like HCl and NaOH) is typically [tex]\(-57.3 \, \text{kJ/mol}\).[/tex]
The total heat released q can be calculated as:
[tex]\[ q = \text{moles of HCl} \times \Delta H_{\text{neutralization}} \] \[ q = 0.0470 \, \text{moles} \times -57.3 \, \text{kJ/mol} = -2.6931 \, \text{kJ} = -2693.1 \, \text{J} \][/tex]
(The negative sign indicates that the heat is released, but we will use the magnitude for temperature calculation.)
3. Determine the total mass of the solution:
The total volume of the solution is:
[tex]\[ \text{Volume} = 50.00 \, \text{mL} + 25.00 \, \text{mL} = 75.00 \, \text{mL} \][/tex]
Given the density of the solution is 1.02 g/mL, the total mass (\(m\)) is:
[tex]\[ m = 75.00 \, \text{mL} \times 1.02 \, \text{g/mL} = 76.50 \, \text{g} \][/tex]
4.Calculate the temperature change:
[tex]\[ \Delta T = \frac{q}{mc} \] \[ \Delta T = \frac{2693.1 \, \text{J}}{76.50 \, \text{g} \times 3.98 \, {J/gC}} = \frac{2693.1}{304.47} \approx 8.85 \°C} \][/tex]
5.Calculate the final temperature:
The initial temperature of both solutions is 24.66°C. Thus, the final temperature [tex](\(T_f\))[/tex] is:
[tex]\[ T_f = 24.66 \, \°C} + 8.85 \, \°C} = 33.51 \, \°C} \][/tex]
So, the final temperature of the solution after the reaction is approximately 33.51°C.
A 20.0–milliliter sample of 0.200–molar K2CO3 solution is added to 30.0 milliliters of 0.400–molar Ba(NO3)2 solution. Barium carbonate precipitates. The concentration of barium ion, Ba2+, in solution after reaction is_________.
Answer:
[Ba^2+] = 0.160 M
Explanation:
First, let's calculate the moles of each reactant with the following expression:
n = M * V
moles of K2CO3 = 0.02 x 0.200 = 0.004 moles
moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles
Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.
Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3
As you can see, 0.04 moles of K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of
0.012 - 0.004 = 0.008 moles of Ba(NO3)2
These moles are in total volume of 50 mL (30 + 20 = 50)
So finally, the concentration of Ba in solution will be:
[Ba] = 0.008 / 0.050 = 0.160 M
A 670.-g piece of copper tubing is heated to 95.3°C and placed in an insulated vessel containing 52.5 g of water at 36.5°C. Assuming no loss of water and heat capacity of 10.0 J/K for the vessel, what is the final temperature (c of copper = 0.387 J/g · K)?
Answer:
Final temperature will be T = 67.68°C
Explanation:
The heat evolved by the copper tubing will be absrobed by both water and the vessel used.
The heat evolved by the copper tubing will be:
Heat = [tex]Q1=massXspecificheatX(changeintemperature)[/tex]
Mass = 670 g
Specific heat = 0.387 J/g · K
Change in temperature = Initial - Final
[tex]Q1=670X0.387X(ChangeinTemperature)[/tex]
The heat absorbed by water will be
[tex]Q2=massXspecificheatXchangeintemperature[/tex]
mass = 52.5
Specific heat = 4.184 J/g · K
the heat absorbed by vessel will be:
[tex]Q3=heatcapacityXchange intemperature[/tex]
Heat capacity = 10J/K
Final temperature of all the three will be same (say T)
[tex]Q1=Q2+Q3[/tex]
[tex]670X0.387X(ChangeinTemperature)=massXspecificheatXchangeintemperature+heatcapacityXchange intemperature[/tex]
[tex]670X0.387X(95.3-T)=(52.5X4.184X(T-36.5))+(10X(T-36.5)[/tex]
[tex]259.29(95.3-T)=219.66(T-36.5)+10(T-36.5)[/tex]
[tex]24710.337-259.29T=219.66T-8017.59+10T-365[/tex]
[tex]33092.59=488.95T[/tex]
T = 67.68°C
All isotopes of an element with an atomic number greater than ___ are radioactive.
Answer:
83 .
Explanation:
Radioactive elements: It is defined as the atoms which contains unstable nucleus because of the constantly change and imbalance of energy in the nucleus. Radioactivity of an atom is showing by when the nucleus of an atom loose a neutron, it gives an energy and this process is called to be radioactivity.
Isotopes elements containing different number of neutrons and same number of protons. All isotopes are not considered as radioisotopes. All isotopes of an element with an atomic number greater than 83 are radioactive means, they are having unstable nucleus.
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when [ Zn2+] = 3.5 M and [Pb2+] = 2.0⋅10−4 M.
Pb2+ (aq) + Zn (s) → Zn2+ (aq) + Pb (s)
A) 0.84
B) 0.76
C) 0.50
D) 0.63
E) 0.39
Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
[tex]Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)[/tex]
The half-cell reactions are:
Oxidation half reaction (anode): [tex]Zn\rightarrow Zn^{2+}+2e^-[/tex]
Reduction half reaction (cathode): [tex]Pb^{2+}+2e^-\rightarrow Pb[/tex]
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
[tex]E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}[/tex]
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = [tex]25^oC=273+25=298K[/tex]
n = number of electrons in oxidation-reduction reaction = 2
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.63 V
[tex]E_{cell}[/tex] = cell potential for the reaction = ?
[tex][Zn^{2+}][/tex] = 3.5 M
[tex][Pb^{2+}][/tex] = [tex]2.0\times 10^{-4}M[/tex]
Now put all the given values in the above equation, we get:
[tex]E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}[/tex]
[tex]E_{cell}=0.50V[/tex]
Therefore, the cell potential for this reaction is 0.50 V
Using relative enthalpy and entropy values, determine how the process is affected after each of the following temperature or pressure changes. Consider that a more effective reaction produces more product or more product in a shorter amount of time.
Reaction: SO₂ (g) + 2H₂S (g) ↔ 3S(s) + 2H₂O (g)Substance | ΔG kJ/mol | ΔH kJ/molH₂O(g) | -228.6 | -241.8H₂O(l) | -237.1 | -285.8SO₂(g) | -300.4 | -296.9SO₃(g) | -370.4 | -395.2H₂S(g) | -33.01 | -20.17S(s) | 0 | 0Categorize into: "More Effective" ~ "Less Effective" ~ "Equally Effective"a. Temp. decreases while maintaining container sizeb. Temp. increases while maintaining container sizec. Pressure decreases while maintaining container sized. Pressure increases while maintaining container size
Changes in pressure and temperature can affect a reaction's effectiveness. A drop in temperature or a rise in pressure makes the given reaction more effective, while a rise in temperature or a drop in pressure makes it less effective.
Explanation:In the given reaction, the relative enthalpy and the entropy can provide insights on how the temperature and pressure changes can affect the reaction.
a. Temperature decreases: When temperature decreases, reactions that are exothermic (ΔH < 0) become more effective because the system attempts to increase its temperature by favoring the exothermic reaction. Therefore, in this case, considering ΔHSO₂(g) and ΔHH₂S(g) are both negative, the reaction is 'More Effective' at lower temperatures.b. Temperature increases: The increase in temperature generally favors the endothermic processes (ΔH > 0) and any process with positive ΔS (entropy). But in this case, the reaction is exothermic and creating more order (ΔS is negative because gases are forming a solid), so an increase in temperature would make the reaction 'Less Effective.'c. Pressure decreases: When pressure decreases, it favors the side of the equation with more gaseous moles. Here, there are 3 gas moles on the reactant side and 2 on the product side. Thus, reducing pressure would favor the reactant side, making the reaction 'Less Effective.'d. Pressure increases: Increasing pressure favors the side with fewer moles of gas, which is the product side in this case, so the reaction becomes 'More Effective.'Learn more about Chemical reaction effectiveness here:https://brainly.com/question/34566703
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More effective reaction conditions for the specified equilibrium include decreasing temperature and increasing pressure, leading to more product formation. Increasing temperature and decreasing pressure are less effective. Effectiveness is based on shifting equilibrium towards the product side.
The reaction in question is SO₂ (g) + 2H₂S (g) ↔ 3S(s) + 2H₂O (g). To determine the effectiveness of the reaction under different temperature and pressure changes, one must consider how each change affects the equilibrium. Here’s the analysis:
Temp. decreases while maintaining container size: As the reaction is exothermic (ΔH for H₂S and SO₂ are negative), decreasing the temperature shifts the equilibrium to the right, producing more products. Thus, this condition is More Effective.Temp. increases while maintaining container size: With an exothermic reaction, increasing the temperature shifts the equilibrium to the left, producing fewer products. Thus, this condition is Less Effective.Pressure decreases while maintaining container size: Since the number of gas molecules decreases from reactants to products (3 moles to 2 moles), decreasing pressure shifts the equilibrium to the left. Therefore, this condition is Less Effective.Pressure increases while maintaining container size: Increasing pressure favors the side with fewer gas molecules. Therefore, the equilibrium shifts to the right, making this condition More Effective.Given the partial equation ? MnO4−+ ? SO32− → ? Mn2++ ? SO42− what must the coefficients be so that the electrons are balanced? Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4, where 1 indicates the absence of a coefficient).
Answer:
The coefficients should be: 2, 5, 2, 5
Explanation:
Given redox reaction: MnO₄⁻ + SO₃²⁻ → Mn²⁺+ SO₄²⁻
To balance the given redox reaction in acidic medium, the oxidation and the reduction half-reactions should be balanced first.
Reduction half-reaction: MnO₄⁻ → Mn²⁺
Oxidation state of Mn in MnO₄⁻ is +7 and the oxidation state of Mn in Mn²⁺ is +2. Therefore, Mn accepts 5e⁻ to get reduced from +7 to +2 oxidation state.
⇒ MnO₄⁻ + 5e⁻ → Mn²⁺
Now the total charge on reactant side is (-6) and the total charge on product side is +2. Therefore, to balance the total charge, 8H⁺ must be added to the reactant side.
⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺
To balance the number of hydrogen and oxygen atoms, 4H₂O must be added to the product side.
⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O .....equation 1
Oxidation half-reaction: SO₃²⁻ → SO₄²⁻
Oxidation state of S in SO₃²⁻ is +4 and the oxidation state of S in SO₄²⁻ is +6. Therefore, S loses 2e⁻ to get oxidized from +4 to +6 oxidation state.
⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻
Now the total charge on reactant side is (-2) and the total charge on product side is (-4). Therefore, to balance the total charge, 2H⁺ must be added to the product side.
⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻ + 2H⁺
To balance the number of hydrogen and oxygen atoms, 1 H₂O must be added to the reactant side.
⇒ SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ .....equation 2
Now, to cancel the electrons transferred, equation (1) is multiplied by 2 and equation (2) is multiplied by 5.
Balanced Reduction half-reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O ] × 2
⇒ 2MnO₄⁻ + 10e⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O .....equation 3
Balanced Oxidation half-reaction:
SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ ] × 5
⇒ 5SO₃²⁻ + 5H₂O → 5SO₄²⁻ + 10e⁻ + 10H⁺ .....equation 4
Now adding equation 3 and 4, to obtain the overall balanced redox reaction:
2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O
Therefore, the coefficients should be: 2, 5, 2, 5