Answer:
[Na₂S₂O₃] = 0.83 M
[Na₂S₂O₃] = 0.86 m
Mole fraction = 0.015
Explanation:
Na₂S₂O₃ 12 % by mass. This data means, that 12 g of solute are contained in 100 g of solution.
Let's find out the volume of solution, with density to determine molarity.
Solution density = Solution mass / Solution volume
1.1003 g/cm³ = 100 g / Solution volume
100 g / 1.1003 g/cm³ = Solution volume → 90.88 mL (1cm³ = 1mL)
Now, that we have volume, we can calculate molarity
Molarity is mol/L
90.88 mL = 0.09088 L
12 g / 158.12 g/mol = 0.0759 moles
0.0759 moles / 0.09088 L = 0.83 M
Total mass of solution = 100 g
12 g + Solvent mass = 100 g
Solvent mass = 100 g - 12 g → 88 g
Molality = moles of solute /1kgof solvent
88 g = 0.088 kg
0.0759 moles / 0.088 kg = 0.86 m
As solvent mass is 88 g, let's determine solvent's moles for mole fraction
88 g / 18 g/mol = 4.89 moles
Mole fraction = moles of solute / moles of solutes + moles of solvent
Mole fraction = 0.0759 mol / 0.0759 mol + 4.89 moles = 0.015
Three atoms of iron (Fe) are to be combined with four molecules of oxygen (O2). How many total iron atoms and oxygen molecules will be combined during the formation of Fe3O4
Final answer:
In the formation of Fe3O4, three atoms of iron (Fe) combine with four molecules of oxygen (O2), involving a total of three iron atoms and eight oxygen atoms to maintain the stoichiometric ratio as per the compound's chemical formula.
Explanation:
The question relates to a chemical reaction where three atoms of iron (Fe) combine with four molecules of oxygen (O2) to form the compound Fe3O4. This compound, known as magnetite, is an example of an iron oxide where the ratio of iron to oxygen atoms is maintained as per the stoichiometry of the compound's chemical formula. In this case, three atoms of iron will combine with oxygen to utilize four oxygen molecules (which equates to eight oxygen atoms), resulting in the formula Fe3O4. This shows the direct application of stoichiometry in understanding chemical formulas and the ratios in which elements combine to form compounds.
The solid XY decomposes into gaseous X and Y: XY(s) m X(g) + Y(g) Kp = 4.1 (at 0 °C) If the reaction is carried out in a 22.4 L container, which initial amounts of X and Y will result in the formation of solid XY?a) 5 mol X; 0.5 mol Yb) 2.0 mol X; 2.0 mol Yc) 1 mol X; 1 mol Yd) none of the above
Answer:
None of the conditions will favor either the forward reaction or backward reaction , hence the answer is D
Explanation:
The principle of chemical Equilibrium is applied here, where the concentration of the reactants or the forward reaction is same as the concentration of the products or the backward reaction. The equilibrium constants is also involved here, K can be in terms of pressure (Kp) or concentration (Kc) hence equilibrium constant is the ration of the concentration of the products to the concentration of the reactants raised to the power of the coefficient of the reactants and products.Partial pressure , total pressure and the mole fraction relationship is also appliedThe step by step explanation is as shown in the attachment below.It's challenging to determine the precise initial amounts of gaseous elements X and Y in the reaction, XY(s) → X(g) + Y(g), with the given Kp and the conditions provided. Based on the data provided, the accurate answer would be 'none of the above'.
Explanation:This question involves equilibrium in chemical reactions, specifically the concept of Kp, the equilibrium constant under constant pressure circumstances. In its simplest form, it represents a ratio of the concentrations of products and reactants, each raised to the power of their respective stoichiometric coefficients in the balanced chemical equation. In this context, if the reaction XY(s) → X(g) + Y(g) has a Kp = 4.1, it means that when equilibrium is reached, the ratio [X][Y]/[XY] is 4.1.
Given that one mole of gas at 0°C and 1 atmosphere of pressure occupies a volume of 22.4L (standard molar volume), we can establish the initial conditions of X and Y using the stoichiometric ratio. However, without additional specifics on the initial conditions of the solid XY such as its volume or weight, it's challenging to determine precise initial amounts of X and Y. Therefore, the accurate answer from the given choice would be (d) none of the above.
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CFC-11 (trichlorofluoromethane, CCl3F) has been used for many years as the working fluid in refrigerators. Given its heat of vaporization is 26.88 kJ/mol and its entropy of vaporization is 90.51 J/(mol ⋅ K), what is the boiling point of CFC-11?
Answer:
297.0 K
Explanation:
Given data
Enthalpy of vaporization (ΔHvap): 26.88 kJ/mol = 26.88 × 10³ J/molEntropy of vaporization (ΔSvap): 90.51 J/(mol ⋅ K)Boiling point (Tb): ?We can find the boiling point of CFC-11 using the following expression.
ΔSvap = ΔHvap/Tb
Tb = ΔHvap/ΔSvap
Tb = (26.88 × 10³ J/mol)/(90.51 J/(mol ⋅ K))
Tb = 297.0 K
A scientist adds a solution of HCl (aq) to a buffer comprised of CH3COOH (aq) and CH3COONa (aq). Which net ionic reaction would initially occur?
Explanation:
Net ionic between HCl and the buffer equation
CH3COOH --> acetic acid, CH3COO- acetate , H+ proton from HCl
there is neutralization of the acetate, and an increase in acid
the ionic equation
CH3COO-(aq) + H+(aq) + Cl-(aq) --> CH3COOH(aq) + Cl-(aq)
Net ionic:
CH3COO-(aq) + H+(aq) --> CH3COOH(aq).
Final answer:
H+ ions from added HCl react with the acetate ions in a buffer to form acetic acid, preventing a significant pH change.
Explanation:
When a strong acid like HCl is added to a buffer made of acetic acid (HC₂H₃O₂) and sodium acetate (CH₃COONa), the hydrogen ions (H+) from HCl will react with the acetate ions (C₂H₃O₂-) from sodium acetate in the buffer. The net ionic reaction for this process is:
H*(aq) + C₂H₃O₂-(aq) → HC₂H₃O₂(aq)
The reaction shows that H+ ions are intercepted by the buffer's acetate anions to form acetic acid (HC₂H₃O₂), preventing a significant change in pH and maintaining the buffer's purpose.
Choose the aqueous solution below with the highest freezing point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. Choose the aqueous solution below with the highest freezing point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable.
a. 0.200 m Ba(NO3)2
b. 0.200 m Mg(ClO4)2
c. 0.200 m HOCH2CH2OH
d. 0.200 m Na3PO3
Answer: c. 0.200 m HOCH2CH2OH
Explanation:
The collective properties of solvents expressed that the freezing point of water is lowered when solute particles are dissolved in it.
When more particles are dissolved, the more the freezing point is deepened and lowered.
Here, all the molarities are equal, the deciding factor in measuring the lowering will be which substance produces the fewest particles in solution; then we relate it to lowering of the freezing point.
Ba(NO3)2 ---> Ba2+ + 2NO3(1-)
1 mol releases 3 mol of ions.
Mg(ClO4)2 ---> Mg2+ + 2ClO4(1-)
1 mol releases 3 mol of ions.
HOCH2CH2OH is covalent and doesn't ionize.
1 mol in water is just 1 mol of molecules.
Na3PO3 ---> 3Na+ + PO3(1-)
1 mol releases 4 mol ions.
The substance that produces the fewest particles in solution is the 0.200 m HOCH2CH2OH
The solution with the highest freezing point will be the one that forms the fewest particles when dissolved. In this case, that would be the ethylene glycol solution (HOCH2CH2OH), because it does not ionize in solution and therefore has a van't Hoff factor of 1.
Explanation:In order to understand which solution has the highest freezing point, think about a concept called freezing point depression. The freezing point depression is due to the number of solute particles in a given volume of solvent; more particles will result in a lower freezing point. The key to the number of particles is the van't Hoff factor (i), which shows how many particles a solute breaks up into when dissolved.
Selected solution (a) Ba(NO3)2 is a salt which breaks up into three ions in solution: Ba2+ and 2NO3-, so i=3. Solution (b) Mg(ClO4)2 also breaks up into three ions: Mg2+ and 2ClO4-, so i=3. Solution (c) HOCH2CH2OH, ethylene glycol, is a nonelectrolyte and doesn't ionize in solution, so i=1. Solution (d) Na3PO3 breaks up into four ions: 3Na+ and PO33-, so i=4.
Therefore, the solution with the highest freezing point will be the one that forms the fewest particles when dissolved, which is 0.200 m HOCH2CH2OH.
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Salts are always ________. Select one: a. single covalent compounds b. hydrogen bonded c. ionic compounds d. double covalent compounds
Answer:
C
Explanation:
The chemical formula of salt is NaCl (sodium chloride).
Na is a metal and Cl is a non-metal.
This means that NaCl is an ionic compound.
Covalent compound is a compound made up of non-metal atoms only.
Salts are always ionic compounds. Therefore, option C is correct.
An ionic compound is a type of chemical compound that is composed of ions held together by electrostatic forces of attraction.
Ionic compounds play vital in various applications, such as in the formation of salts, the electrolysis of compounds, and the construction of solid-state electronic devices.
Salts are a specific type of ionic compound. They are compounds formed when an acid reacts with a base through a chemical reaction called neutralization.
Salts consist of cations (positively charged ions) and anions (negatively charged ions) held together by ionic bonds.
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What is the final pH of a solution when 0.1 moles of acetic acid is added to water to a final volume of 1L?
Answer:
The pH of solution is 2.88 .
Explanation:
The reaction is :
[tex]CH_3COOH-->CH_3COO^-+H^+[/tex]
We know, [tex]K_a[/tex] for this reaction is = [tex]1.76\times 10^{-5}.[/tex]
Also, since volume of water is 1 L.
Therefore, molarity of solution is equal to number of moles.
Also, [tex]K_a=\dfrac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Let, amount of [tex]CH_3COO^- and\ H^+[/tex] produce is x.
So,
[tex]K_a=\dfrac{[x][x]}{[0.1]}\\1.76\times 10^{-5}=\dfrac{[x][x]}{[0.1]}[/tex]
[tex]x=0.0013\ mol.[/tex]
We know, [tex]pH=-log{x}=-log(0.0013)=2.88[/tex]
Therefore, pH of solution is 2.88 .
Hence, this is the required solution.
Final answer:
The final pH of a 1L solution containing 0.1 moles of acetic acid is determined by using the hydrogen ion concentration resulting from its dissociation. The pH is calculated to be approximately 2.879 following the calculation procedures outlined for weak acids.
Explanation:
Calculating pH of an Acetic Acid Solution
The final pH of a solution can be determined by calculating the hydrogen ion concentration ([H+]) resulting from the dissociation of acetic acid in water. Given that we have 0.1 moles of acetic acid added to water to make 1L of solution, the molarity of acetic acid is 0.100 M. Acetic acid (CH₃COOH) is a weak acid, and its dissociation in water is represented by the following equation:
CH₃COOH <--> CH₃COO⁻ + H⁺
We use the acid dissociation constant (Ka) to solve for [H+]. For acetic acid, Ka = 1.8 x 10⁻⁵ M. The calculation involves setting up an ICE table and using the quadratic formula, as outlined in the subject's textbook or reference. The calculation typically results in a hydrogen ion concentration of 1.32 x 10⁻³ M.
The pH is found by taking the negative logarithm of the hydrogen ion concentration:
pH = -log(1.32 x 10⁻³) = 2.879
Therefore, the pH of a 0.100 M solution of acetic acid is approximately 2.879.
Why does ΔG for a reaction scale with reaction quantity but E does not? For example, ΔG0 rxn for the combustion of 1 mol of hydrogen is 1 × –237 kJ∕mol = –237 kJ, CHAPTER EXERCISES 73 while ΔG0 rxn for the combustion of 2 mol of hydrogen is 2 × –237 kJ∕mol = –474 kJ. In both
Answer:
This is as a result of their property type
ΔG is extensive and E is Intensive. The explanation is as given below
Explanation:
Basically both ΔG and the cell potential or the electromotive force (E.M.F) has some disparity especially in their spontaneity, for spontaneous reaction ΔG = -ve while E = +ve and vice versa. But the most important disparity is their state function i.e one is intensive and the other is extensive property.
ΔG is an example of an extensive property, they are properties whose value is dependent on the volume or the size of the system. other examples are mass, volume etc.
E on the other hand is an intensive property, they are properties whose value is not dependent on the size of the system. As such, this differences explains why ΔG for a reaction scale with a reaction quantity and E does not.
The change in Gibbs free energy (ΔG) for a reaction scales with reaction quantity because it is an extensive property, while the standard electrode potential (E) does not scale with reaction quantity because it is an intensive property.
Explanation:The change in Gibbs free energy (ΔG) for a reaction scales with reaction quantity because it is an extensive property, meaning it depends on the amount of substance involved in the reaction. On the other hand, the standard electrode potential (E) does not scale with reaction quantity because it is an intensive property, meaning it is independent of the amount of substance.
For example, in the combustion of hydrogen, the standard Gibbs free energy change (ΔG°) for the reaction is -237 kJ/mol for 1 mole of hydrogen and -474 kJ/mol for 2 moles of hydrogen. Since ΔG is an extensive property, it doubles when the quantity of hydrogen doubles. However, the standard electrode potential (E°) remains the same regardless of the quantity of hydrogen.
In summary, the scaling of ΔG with reaction quantity and the independence of E from reaction quantity are due to the differences in their nature as extensive and intensive properties, respectively.
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Draw the sulfur‑containing product of the oxidation reaction between two 2‑methyl‑1‑propanethiol molecules. Include all hydrogen atoms.
Final answer:
The sulfur-containing product of the oxidation of two 2-methyl-1-propanethiol molecules is a disulfide, where the sulfur atoms form a bond between the two original thiol molecules, resulting in a dimer.
Explanation:
The student has asked for the sulfur-containing product of the oxidation reaction between two 2-methyl-1-propanethiol molecules. In the presence of mild oxidizing agents such as oxygen, sulfur atoms can be oxidized from the sulfhydryl (S-H) groups present in 2-methyl-1-propanethiol. When two such sulfur atoms, each with an unpaired electron, come into contact, they can form a sulfur-sulfur bond (disulfide bond), resulting in the dimerization of the two thiol molecules.
The chemical formula of 2-methyl-1-propanethiol is CH3CH2CH(SH)C(CH3)2. Upon oxidation, the sulfur atoms from two of these molecules will bond together to form a disulfide. The resulting molecule will be CH3CH2CH(S-S)CH2C(CH3)2.
A block of metal has a width of 3.2 cmcm, a length of 17.1 cmcm, and height of 5.0 cmcm . Its mass is 1.2 kgkg . Calculate the density of the metal. Express your answer to two significant figures and include the appropriate units.
Answer:
≅ 4.39 gm/cm^3
Explanation:
Given data:
width of block =3.2 cm
length = 17.1 cm
height = 5.0 cm
Mass = 1.2 kg
Therefore volume= width×length×height = 3.2×17.1×5= 273.6 cm^3
now, density = mass/volume
calculating density in gm/cm^3
= 1200/273.6 = 4.38596
≅ 4.39 gm/cm^3
N2O5(g) → NO3(g) + NO2(g)
This reaction is 1st order with half-life = 18.0 seconds. How long will it take for 95% of a sample of N2O5(g) to react?
A. 42.2 seconds
B. 132 seconds
C. 109 seconds
D. 77.8 seconds
E. 158 seconds
Answer:
It will take 77.8 seconds for 95% of sample to react
Explanation:
As the given reaction obeys 1st order therefore-
[tex][N_{2}O_{5}]=[N_{2}O_{5}]_{0}\times (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]
Where [tex][N_{2}O_{5}][/tex] is the concentration of [tex]N_{2}O_{5}[/tex] after "t" time, [tex][N_{2}O_{5}]_{0}[/tex] is the initial concentration of [tex]N_{2}O_{5}[/tex] and [tex]t_{\frac{1}{2}}[/tex] is half life
Here, [tex]\frac{[N_{2}O_{5}]}{[N_{2}O_{5}]_{0}}=\frac{100-95}{100}=0.05[/tex] and [tex]t_{\frac{1}{2}}=18.0 seconds[/tex]
So, [tex]0.05=(\frac{1}{2})^{\frac{t}{18.0seconds}}[/tex]
or, [tex]t=77.8 seconds[/tex]
So, it will take 77.8 seconds for 95% of sample to react
Azobisisobutyronitrile (AIBN) is commonly used as a radical initiator. Use correct arrow formalism to show this process.
Answer: AIBN is good radical initiator
Explanation:
A radical in organic chemistry refers to any specie having a single unshared electron. An initiator is any specie capable of producing radicals thus starting up a chain reaction. Azobisisobutyronitrile is a good initiator basically owing to its structure. It forms radicals by breaking up to release nitrogen gas as shown in the image attached. The two radicals formed both contain the -CN group which stabilizes the radical.
A sample of gas occupies a volume of 10.5 L when the pressure is 14.3 psi. Use Boyle’s Law to calculate the pressure (psi) when the volume is reduced to 8.55L. Show the calculation.
Answer: 17.6psi
Explanation:
V1 = 10.5 L
P1 = 14.3 psi
V2 = 8.55L
P2 =?
P1V1 = P2V2
14.3 x 10.5 = P2 x 8.55
P2 = (14.3 x 10.5) / 8.55
P2 = 17.6psi
Answer:
The final pressure is 17.56 psi
Explanation:
Step 1: Data given
Initial volume = 10.5 L
Initial pressure = 14.3 psi
Final volume = 8.55 L
Step 2: Calculate the new pressure
P1V1 = P2V2
⇒ P1 = The initial pressure = 14.3 psi
⇒ V1 = The initial volume = 10.5L
⇒ P2 = The final pressure = TO BE DETERMINED
⇒ V2 = The final volume = 8.55 L
14.3 * 10.5 = P2*8.55
P2 = 17.56 psi
The final pressure is 17.56 psi
Identify the functional group(s) that appear in the molecule cortisone (a major hormone synthesized by the human adrenal cortex).
Answer:
Ketone, alcohol and unsaturation
Explanation:
The functional groups are the groups that identify the organic function of the molecule. For example, the hydroxyl (-OH) represents alcohol, the carbonyl (represented in the figure below) can represent a ketone or an aldehyde, and so each function has its representation.
The cortisone molecule is represented below, and each group is marked in blue. They are:
a, c, and d = carbonyl of ketone;
b = unsaturation (double bond) of alkene;
e, and f = hydroxyl of alcohol.
For the following formula, C7H10NBr, calculate the IHD and select all the types of unsaturation that might be present in the molecule based on the IHD.
The molecule C7H10NBr has an Index of Hydrogen Deficiency (IHD) of 4, indicating 4 degrees of unsaturation. These could be a combination of double bonds, triple bonds, or rings.
Explanation:The Index of Hydrogen Deficiency (IHD) or Degree of Unsaturation for a molecule can be calculated using the formula IHD = 1/2(2C + 2 + N - X - H), where C is the number of Carbons, N the number of Nitrogens, X the number of Halogens, and H the total Hydrogen count. For the molecule C7H10NBr, applying the formula would yield:
IHD = 1/2(2*7 + 2 + 1 - 1 - 10) = 4
This means there are 4 degrees of unsaturation present in the molecule. The types of unsaturation could be double bonds, triple bonds, or rings. In this case, with a value of 4, it could be 4 double bonds, or 2 double bonds and 2 rings, or 2 rings and 1 triple bond, or 1 ring and 3 double bonds, or 2 triple bonds, and many more combinations.
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The Index of Hydrogen Deficiency for the compound C₇H₁₀N Br is 4, which indicates that the molecule can have a combination of 4 double bonds or 4 rings or a mix of both.
Explanation:The Index of Hydrogen Deficiency (IHD) is a count of how many molecules of H2 are missing from an alkane having the same number of carbon atoms. It's used to identify the number of rings and/or double bonds in organic compounds. For the compoundC₇H₁₀N Br, consider that Br adds 1 to the hydrogen count, and N subtracts 1. Thus, the IHD calculation would be: IHD= (2*C + 2 + N - X - H)/2 = [2*7 + 2 +1 -0 -10]/2 = 4
Since it has an IHD of 4, it indicates that the molecule can have a combination of 4 double bonds or 4 rings or a mix of both.
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For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because ________.
Answer:
Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.
Explanation:
In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.
[tex]\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)[/tex].
When [tex]\rm H_2[/tex] is added to [tex]\rm H_2C\text{=}CH_2[/tex] (ethene) under heat and with the presence of a catalyst, [tex]\rm H_3C\text{-}CH3[/tex] (ethane) would be produced.
Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, [tex]\Delta S < 0[/tex].
The equation for the change in Gibbs Free Energy for a particular reaction is:
[tex]\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})[/tex].
For a particular reaction, the more negative [tex]\Delta G[/tex] is, the more spontaneous ("favorable") the reaction would be.
Since typically [tex]\Delta S < 0[/tex] for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.
[tex]T[/tex] (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of
How many grams of magnesium bromide, MgBr2, must be dissolved to prepare 400. mL of a 0.183 M aqueous solution of the salt?
Answer: 13.47g of MgBr2
Explanation:
MM of MgBr2 = 24 + (2 x 80) = 24 + 160 = 184g/mol
Mass conc. Of MgBr2 = 0.183 x184 = 33.672g
33.672g of MgBr2 dissolves in 1000mL
Therefore Xg of MgBr2 will dissolve in 400mL
Xg of MgBr2 = ( 33.672 x 400)/1000
Xg of MgBr2 = 13.47g
Using molecular orbital theory, the bond order of the C-C bond in the C2 molecule is _____. g
Answer:
The bond order for C2 molecule is 2.
Explanation:
Bond order can be defined as the half of the difference between the number of electrons in the bonding orbital and the number of electrons in the antibonding orbitals. It can be represented mathematically by; .
Bond order,n= [number of electrons in the bonding molecular orbitals(BMO) - the number or electrons in the anti-bonding molecular orbitals(AMO) ] / 2.
The electronic configuration of the C2 molecule is given below;
C2 = (1sσ)^2 (1s^*σ)^2 (2sσ)^2 (2s^*σ)^2 (2pπ)^4.
The ones with the (*) are known as the Anti-bonding molecular orbitals while the ones without (*) are known as the bonding molecular orbitals. Hence, we have 8 Electrons from the bonding molecular orbitals and 4 Electrons from the anti-bonding molecular orbitals.
So, from the formula given above, the bond order of C2 molecule is;
===> 8-4/2= 4/2.
===> 2.
The bond order of the C-C bond in the C₂ molecule is 1. This is calculated using molecular orbital theory by identifying bonding and antibonding electrons and applying the bond order formula.
The bond order of the C-C bond in the C₂ molecule can be calculated using molecular orbital theory. First, we identify the number of bonding and antibonding electrons in the molecule. The molecular orbital configuration for C₂ is (σ2s)² (σ*2s)² (π2p)⁴. This gives us 4 bonding electrons (from π2p) and 2 antibonding electrons (from σ*2s).
To calculate the bond order, we use the formula:
Bond order = (Number of bonding electrons - Number of antibonding electrons) / 2
Substituting the numbers, we get Bond order = (4 - 2) / 2 = 1.
Therefore, the bond order of the C-C bond in the C₂ molecule is 1.
Consider the five balanced chemical reactions listed below, all using O2 as a reactant. Normally, O2 is an excess reagent for reactions because there is a large amount of it available in the air. For this question, suppose that you have5.73 moles of O2 and1.70 moles of any of the other reactants in every equation. Select those equations below in which O2 would be the limiting reactant.
A.C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)B.C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)C.H2S(g) + 2 O2(g) → SO3(g) + H2O(g)D.4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)E.None because in all the reactions O2 is in excess
Answer: option E. None because in all the reactions O2 is in excess
Explanation:
In the above reactions, only Reaction A and Reaction B have oxygen (O2) as the limiting reactant. For the rest of the reactions, the other compound will be used up before all the O2 is spent.
Explanation:In every reaction, you have 5.73 moles of O2 and 1.70 moles of the other reactants. You can determine which of the two reactants are limiting (which will be consumed first) by comparing the number of moles of that reactant divided by its stoichiometric coefficient (the numbers in front of the chemicals in the reaction) to the same calculation for the other reactant.
For reaction A, C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g), the ratios are 5.73 moles O2/6 and 1.70 moles C6H12O6/1. So O2 is limiting. For reaction B, C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g), the ratios are 5.73 moles O2/5 and 1.70 moles C3H8/1. So O2 is limiting. For reaction C, H2S(g) + 2 O2(g) → SO3(g) + H2O(g), the ratios are 5.73 moles O2/2 and 1.70 moles H2S/1. In this case, H2S is limiting, not O2. For reaction D, 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g), the ratios are 5.73 moles O2/5 and 1.70 moles NH3/4. In this case, NH3 is limiting, not O2. Learn more about Limiting Reactant here:
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A reaction has an equilibrium constant of 7.9×10³ at 298 K. At 713 K, the equilibrium constant is 0.77.
Do you predict that the enthalpy of the reaction is positive or negative? Why?
a. Positive, because endothermic reactions shift toward products at higher temperatures
b. Negative, because exothermic reactions shift toward reactants at higher temperatures
c. Positive, because endothermic reactions shift toward reactants at higher temperatures
d. Negative, because exothermic reactions shift toward products at higher temperatures
Answer:
option b is correct
Explanation:
Taking the Van't Hoff equation
d ln Keq / dT = ΔH/(RT²)
then
Keq increases with increasing temperature (d ln Keq / dT>0) when ΔH>0 and decreases with increasing temperature (d ln Keq / dT<0) when ΔH<0
if Keq decreases (reactions shift toward reactants) from 7.9×10³ to 0.77 when temperature increases from 298 K to 713 K (d ln Keq / dT<0) → ΔH<0 ( exothermic reaction)
therefore option b is correct
The enthalpy of the reaction is predicted to be negative, indicating it is an exothermic reaction, because the equilibrium constant decreases with an increase in temperature, a characteristic of exothermic reactions. Therefore, the correct answer is b.
The question asks if the enthalpy of the reaction is positive or negative based on changes in the equilibrium constant at different temperatures. The equilibrium constant decreases from 7.9×10³ at 298 K to 0.77 at 713 K. According to the principle that for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature, and conversely, for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, we can deduce the nature of the reaction. As the equilibrium constant decreases with an increase in temperature, the reaction must be exothermic. Therefore, the correct answer is b. Negative, because exothermic reactions shift toward reactants at higher temperatures.
If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount of salt?
Answer: The amount of water required to prepare given amount of salt is 398.4 mL
Explanation:
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:
[tex]0.16M=\frac{16\times 1000}{251\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{16\times 1000}{251\times 0.16}=398.4mL[/tex]
Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL
Raoult's law accounts for the fact that the vapor pressure of a solvent will decrease as the mole fraction of the solvent is decreased. In considering the mole fraction, it is important to consider the total moles of dissolved particles. Remember: a particle can be a dissolved molecule or ion. Which aqueous solutions would have the lowest vapor pressure
Raoult's law accounts for the fact that the vapor pressure of a solvent will decrease as the mole fraction of the solvent is decreased. In considering the mole fraction, it is important to consider the total moles of dissolved particles. Remember: a particle can be a dissolved molecule or ion. Which aqueous solutions would have the lowest vapor pressure.
0.1 M [tex]NH_4NO_3(aq)[/tex] , 0.1 M [tex]NaF(aq)[/tex], 0.1 M [tex]LiNO_3(aq)[/tex], 0.1 M [tex]Na_3PO_4(aq)[/tex] and 0.1 M [tex]HC_2H_3O_2(aq)[/tex]
Answer: 0.1 M [tex]Na_3PO_4[/tex]
Explanation:
As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
The formula for relative lowering of vapor pressure will be,
[tex]\frac{p^o-p_s}{p^o}=i\times x_2[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex]= relative lowering in vapor pressure
i = Van'T Hoff factor
[tex]x_2[/tex] = mole fraction of solute
1. For 0.1 M [tex]NH_4NO_3[/tex]
[tex]NH_4NO_3\rightarrow NH_4^{+}+NO_3^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]
2. For 0.1 M [tex]NaF[/tex]
[tex]NaF\rightarrow Na^{+}+F^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]
3. For 0.1 M [tex]LiNO_3[/tex]
[tex]LiNO_3\rightarrow Li^{+}+NO_3^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions. and concentration of ions will be [tex]1\times 0.1+1\times 0.1=0.2[/tex]
4. For 0.1 M [tex]Na_3PO_4[/tex]
[tex]Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}[/tex]
, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]3\times 0.1+1\times 0.1=0.4[/tex]
5. For 0.1 M [tex]HC_2H_3O_2(aq)[/tex]
[tex]HC_2H_3O_2(aq)\rightarrow CH_3COO^{-}+H^{+}[/tex] [/tex]
, i= 2 as it is a electrolyte and dissociate to give two ions, concentration of ions will be [tex]2\times 0.1=0.2[/tex]
Thus as concentration of solute is highest for 0.1 M [tex]Na_3PO_4[/tex] , the vapor pressure will be lowest.
Mechanism for enzyme catalyzed reactions to explain the kinetics of enzyme-substrate reactions, michaelis final rate from -ra to ________
Answer:
ra to r+a
Explanation: In an enzyme catalyzed reaction, enzyme binds with the active side(3D) of the substrate substrate. It provides an alternative path for the reaction to take place with lower activation energy. in the reaction the kinetic energy of the molecules increases so reaction takes place at a higher rate. when the reaction is completed enzyme leaves the active side.
½ H₂(g) + ½ I₂(s) --> HI(g) ΔH= 26 kJ/mol
½ H₂(g) + ½ I₂(g) --> HI(g) ΔH= -5.0 kJ/mol
Based on the information above, what is the enthalpy change for the sublimation of iodine, represented by I₂(s) --> I₂(g)?
a) 15 kJ/mol
b) 21 kJ/mol
c) 31 kJ/mol
d) 42 kJ/mol
e) 62 kJ/mol
The enthalpy change for the sublimation of iodine, represented by I₂(s) → I₂(g), is 31 kJ/mol.
Explanation:The enthalpy change for the sublimation of iodine, represented by I₂(s) → I₂(g), can be determined by comparing the enthalpy changes of the two given reactions. Since the equation ½ H₂(g) + ½ I₂(g) → HI(g) has a lower enthalpy change (-5.0 kJ/mol) compared to the equation ½ H₂(g) + ½ I₂(s) → HI(g) (26 kJ/mol), it means that the phase change from solid iodine to gaseous iodine requires an additional amount of energy. Thus, the enthalpy change for the sublimation of iodine is the difference between the two reaction enthalpies, which is 31 kJ/mol.
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Gas is $2.50/gal at Station A and 85 cents/L at Station B. A) Convert the cost of gas at Station B to $/gal. B) At which station would you buy gas?
Answer:
85 cents/L is equal to 3.2176$/gallon.
gas wiuld be heaper to buy from the station A.
Explanation:
As 85 cents/L is more than $2.5/gallon therefore buying gas from station A would be cheaper than the other one.
Chlorofluorocarbons (CFCs) are no longer used as refrigerants because they destroy the ozone layer.Trichlorofluoromethane (CCl3F) boils at 296.95 K, and its molar heat of vaporization is 24.8 kJ/mol.
Explanation:
It is known that entropy is the measure of degree of randomness present in a substance due to the movement of its molecules.
More randomly the atoms are moving from one place to another more will be the entropy of system.
In order to calculate the entropy, we divide the amount of heat transferred by the temperature at which heat transfer occurs.
As the given temperature is 296.95 K and heat of vaporization is 24.8 kJ/mol. Therefore, calculate the molar entropy of vaporization as follows.
Molar entropy of vaporization = [tex]\frac{24.8 kJ/mol}{296.95 K}[/tex]
= 0.08351 kJ/mol K
Thus, we can conclude that molar entropy of vaporization of [tex]CC_{3}F[/tex] is 0.08351 kJ/mol K.
CFCs, including Trichlorofluoromethane, have been banned due to their destructive effect on the ozone layer. UV light causes them to release chlorine atoms, which react with and deplete ozone. The harmful side effect of this is increased solar radiation reaching Earth.
Explanation:Chlorofluorocarbons, or CFCs, like Trichlorofluoromethane, were previously widely used in a variety of industries, such as refrigeration and aerosols. However, the stable nature of these compounds, combined with their ability to break down and release chlorine atoms under ultraviolet light, led to significant depletion of the ozone layer in the stratosphere. This process happens because as CFCs are broken down by UV light, they produce chlorine atoms. These chlorine atoms then react with ozone molecules, leading to a net decrease in stratospheric ozone. This ozone depletion had severe implications, such as increased risks from solar radiation, which prompted the worldwide effort to phase out the use of CFCs under the Montreal Protocol.
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What are the coefficients for the following reaction when it is properly balanced? ?potassium iodide +?lead (II) acetate lead (II) iodide +-potassium acetate 3,2,2,1 1,1,2,2 none of these
Answer: none of the above
Explanation:
It should be 2,1,1,2 to give a balanced chemical reaction
To balance the chemical equation between potassium iodide and lead (II) acetate, producing lead (II) iodide and potassium acetate, the coefficients must be 2 (for potassium iodide), 1 (for lead (II) acetate), 1 (for lead (II) iodide), and 2 (for potassium acetate).
Explanation:The question is asking for the correct coefficients that balance the chemical equation between potassium iodide and lead (II) acetate to produce lead (II) iodide and potassium acetate. The balanced chemical equation is:
2KI + Pb(C2H3O2)2 → PbI2 + 2KC2H3O2
The resultant balanced equation shows that two molecules of potassium iodide react with one molecule of lead (II) acetate to yield one molecule of lead (II) iodide and two molecules of potassium acetate. Therefore, the coefficients for the balanced equation are 2, 1, 1, 2 respectively.
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We consider a spherical air bubble rising through a water tank. The pressurepinside the bubble isequal to its surrounding pressure. The volumeVof the bubble is connected to the pressure throughthe isentropic relationWhy does the volume of a water bubble increase as it moves higher to the liquid surface?
The answer is a change in internal energy causes work to be done and heat to flow into the system.
Explanation:
Boyle's law says, PV=RT
Here P represents the pressure, V represents the volume and T represents the temperature. R is a constant. The volume of an ideal gas is inversely proportional to its pressure if the temperature is constant.When a bubble is present in deep water it has water pressure and atmospheric pressure. Then the Volume increases when water pressure raises which is proportional to the depth reduces.But we should not finalize the volume of the bubble will be four-time as great as at the top than the bottom. if the bottom of the lake is at four atmospheres, the temperature will not be equal to the top. If the bubble travels from the bottom to the top or vice-versa, it's going to lose or gain heat in a way that must be quite hard to measure.Carbon dioxide and an unknown gas start to effuse from a container and the carbon dioxide takes 4.69 times as long to escape as the unknown gas. What is the identity of the unknown gas?
Answer: the gas is lighter than CO2 and it is hydrogen
Explanation:Please see attachment for explanation
Polonium-209, an alpha emitter, has a half-life of 102 years.How many alpha particles are emitted in 1.0 s from a 1.0 ngsample of 209Po?
The number of alpha particles emitted from a 1.0ng sample of Polonium-209 in 1.0s depends on the specifics of the isotope's half-life and the initial amount of the isotope. The emissions of these alpha particles are part of the isotope's decay process.
Explanation:The half-life concept is key in understanding the decay of radioactive isotopes like polonium-209. Given that polonium-209 is an alpha emitter, it releases alpha particles during its decay process. These alpha particles are essentially helium-4 nuclei (two protons and two neutrons).
Understanding the number of alpha particles emitted in a certain time requires the understanding of the isotope's half-life and atomic decay. However, without knowing the specific numbers of atoms present in the given 1.0 ng sample of 209Po, it is difficult to provide a precise number of alpha particles emitted in 1.0 s. You would need to make use of the radioactive decay formula N = N0e^-λt where N0 represents the initial amount of the isotope, N represents the remaining amount of the isotope, λ represents the decay constant, and t represents time.
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