4.8 x 10² g grams of Carbon dioxide will be formed. Therefοre, the cοrrect option is: C. 4.8 x 10² g
How tο determine the mass οf CO₂?Tο determine the mass οf CO₂ fοrmed when 5.4 mοles οf acetylene react, we need tο use the stοichiοmetry οf the balanced equatiοn.
The balanced equatiοn is:
2 C₂H₂(g) + 5 O₂(g) → 4 CO₂(g) + 2 H₂O(g)
Frοm the balanced equatiοn, we can see that the stοichiοmetric ratiο between acetylene (C₂H₂) and carbοn diοxide (CO₂) is 2:4 (οr simplified as 1:2).
Given that we have 5.4 mοles οf acetylene, we can calculate the mοles οf CO₂ fοrmed using the stοichiοmetric ratiο:
Mοles οf CO₂ = (5.4 mοles οf C₂H₂) × (2 mοles οf CO₂ / 2 mοles οf C₂H₂)
= 5.4 mοles οf CO₂
Nοw, tο cοnvert the mοles οf CO₂ tο grams, we need tο use the mοlar mass οf CO₂.
The mοlar mass οf CO₂ = (12.01 g/mοl οf C) + (2 × 16.00 g/mοl οf O) = 44.01 g/mοl
Mass οf CO₂ = (5.4 mοles οf CO₂) × (44.01 g/mοl οf CO₂)
= 237.24 g
Therefοre, the cοrrect answer is:
C. 4.8 x 10² g
Learn more about reaction
https://brainly.com/question/30464598
#SPJ4
for the dissolving of ammonium nitrate in water, dh = 25.7 kj/mol and ds = 108.7 j/mol.k. calculate the ksp for the process at 298 k.
At 298 K, the solubility product constant (Ksp) for the process of ammonium nitrate dissolving in water is approximate [tex]3.8 * 10^{-4}[/tex].
To calculate the solubility product constant (Ksp) for the dissolving of ammonium nitrate (NH4NO3) in water at 298 K, we can use the equation:
ΔG = -RTln(Ksp)
where ΔG is the Gibbs free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298 K), and Ksp is the solubility product constant.
We can calculate ΔG using the equation:
ΔG = ΔH - TΔS
where ΔH is the enthalpy change and ΔS is the entropy change.
Given that ΔH = 25.7 kJ/mol and ΔS = 108.7 J/mol·K, we convert ΔH to J/mol:
ΔH = 25.7 kJ/mol × 1000 J/kJ = 25700 J/mol
Now we can substitute the values into the equation:
ΔG = 25700 J/mol - (298 K) × (108.7 J/mol·K) = -31197 J/mol
Substituting ΔG into the first equation and rearranging:
Ksp = exp(-ΔG / (RT))
Ksp = exp(-(-31197 J/mol) / (8.314 J/mol·K × 298 K))
Calculating the value:
Ksp ≈ [tex]3.8 * 10^{-4}[/tex]
Therefore, at 298 K, the solubility product constant (Ksp) for the process of ammonium nitrate dissolving in water is approximate [tex]3.8 * 10^{-4}[/tex].
To learn more about solubility from the given link
https://brainly.com/question/24057916
#SPJ4
which of the following is the strongest oxidizing agent?
MnO4-+ 4H+ + 3e- --> MnO2 + 2H2O E0 1.68V
I2+ 2e- --> 2I- E0= 0.54V
Zn2+ + 2e- --> Zn E0= --0.76V
a) Zn b) Zn2+ c) I2 d) MnO4-
The strongest oxidizing agent among the options provided is MnO4- (permanganate ion) with a standard reduction potential (E°) of 1.68 V.
To determine the strongest oxidizing agent among the given options, we need to compare the standard reduction potentials (E°) of the half-reactions. The higher the standard reduction potential, the stronger the oxidizing agent.
Given the reduction potentials:
- MnO4- + 4H+ + 3e- → MnO2 + 2H2O: E° = 1.68 V
- I2 + 2e- → 2I-: E° = 0.54 V
- Zn2+ + 2e- → Zn: E° = -0.76 V
Among the options provided, the strongest oxidizing agent is MnO4- (permanganate ion) because it has the highest standard reduction potential (E° = 1.68 V). The higher positive value indicates a stronger tendency to accept electrons and undergo reduction.
So, the correct answer is:
d) MnO4-
The standard reduction potential (E°) is a measure of the tendency of a species to gain electrons and undergo reduction. The species with a higher E° value is a stronger oxidizing agent because it has a greater ability to oxidize other species by accepting electrons.
In this case, MnO4- (permanganate ion) has the highest E° value of 1.68 V, indicating that it has a strong tendency to accept electrons and act as an oxidizing agent. On the other hand, Zn2+ and I2 have lower E° values of -0.76 V and 0.54 V, respectively, indicating weaker tendencies to accept electrons.
To know more about oxidizing agent visit;
https://brainly.com/question/14041413
#SPJ11
CHEM 112 Acid and base basics (adapted from Dr. Sushilla Knottenbelt) 1. Sodium azide, NaN3, is sometimes added to water to kill bacteria. Sodium salts are generally soluble, and hence when dissolved in water, Na+ and N3- ions are producced. Azide N3-, acts as a base. a. What is the conjugate acid of azide? b. Write the balanced equation to show the reaction of azide with water. Label acid, base and conjugue acid and base.
The conjugate acid of azide is hydrazoic acid (HN3) and the balanced equation for the reaction of azide with water can be written as follows: N3- + H2O ⇌ NH2- + OH-
N3- + H2O ⇌ NH2- + OH-
In this equation, azide (N3-) acts as a base by accepting a proton (H+) from water, forming the conjugate acid of azide, NH2-, and hydroxide ions (OH-) as the conjugate base of water.
Labeling the components:
Acid: H2O (water)
Base: N3- (azide)
Conjugate acid: NH2- (hydrazoic acid)
Conjugate base: OH- (hydroxide ion)
When azide reacts with water, it acts as a base by accepting a proton from water.
The azide ion (N3-) accepts a proton to form the conjugate acid, NH2-, while water donates a proton to become hydroxide (OH-). The reaction is reversible because NH2- can also donate a proton to water, regenerating the azide ion.
Sodium azide (NaN3) when dissolved in water, produces Na+ and N3- ions. The azide ion acts as a base and its conjugate acid is hydrazoic acid (HN3).
The balanced equation for the reaction of azide with water shows the transfer of a proton from water to azide, resulting in the formation of hydrazoic acid and hydroxide ions.
To learn more about acid, visit
https://brainly.com/question/12584785
#SPJ11
Consider the following reaction, in which two electrons are transferred from cytochrome-c(reduced):
2cytc[Fe2+]+pyruvate+2H+ → 2cytc[Fe3+]+lactate
A. What is ε∘′ for this reaction at pH7 and 25∘C?
B. Calculate the equilibrium constant for the reaction at pH7 and 25∘C.
C. Calculate the standard Gibbs free-energy change for the reaction at pH7 and 25∘C.
D. Calculate the Gibbs free-energy change (at pH7 and 25∘C) if the lactate concentration is five times the pyruvate concentration and the cytochrome c(Fe3+) is ten times the cytochrome c(Fe2+).
ε∘′ for the reaction at pH 7 and 25∘C is 0.44 V.The equilibrium constant for the reaction at pH 7 and 25∘C is approximately 1.23 x 10^(-7).The standard Gibbs free-energy change for the reaction at pH 7 and 25∘C is approximately -85,012 J/mol .the Gibbs free-energy change for the reaction at pH 7 and 25∘C, considering the specified concentrations, is approximately -83,269 J/mol.
A. To determine ε∘′ for the reaction at pH 7 and 25∘C, we need the reduction potentials of cytochrome c and the pyruvate/lactate couple at the specified conditions. Let's assume that the reduction potential for the cytochrome c couple is given as ε∘cytc = 0.25 V (given) and the reduction potential for the pyruvate/lactate couple is ε∘pyr/lac = -0.19 V (given).
The overall reduction potential (ε∘′) of the reaction can be calculated using the Nernst equation:
ε∘′ = ε∘cytc - ε∘pyr/lac
ε∘′ = 0.25 V - (-0.19 V) = 0.44 V
Therefore, ε∘′ for the reaction at pH 7 and 25∘C is 0.44 V.
B. The equilibrium constant (K) for the reaction can be determined using the relationship between ε∘′ and K, which is given by the equation:
ε∘′ = -RTln(K)/nF
Where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25 + 273 = 298 K), n is the number of electrons transferred (2 in this case), and F is the Faraday constant (96,485 C/mol).
Substituting the values:
0.44 V = - (8.314 J/(mol·K)) * (298 K) * ln(K) / (2 * 96,485 C/mol)
Solving for ln(K), we find:
ln(K) = - (0.44 V * (2 * 96,485 C/mol)) / ((8.314 J/(mol·K)) * (298 K))
ln(K) ≈ -13.96
Taking the exponential of both sides:
K ≈ e^(-13.96)
K ≈ 1.23 x 10^(-7)
Therefore, the equilibrium constant for the reaction at pH 7 and 25∘C is approximately 1.23 x 10^(-7).
C. The standard Gibbs free-energy change (ΔG∘) for the reaction at pH 7 and 25∘C can be calculated using the equation:
ΔG∘ = -nFε∘′
Where n is the number of electrons transferred (2) and F is the Faraday constant (96,485 C/mol).
Substituting the values:
ΔG∘ = -(2 * 96,485 C/mol) * 0.44 V
ΔG∘ ≈ -85,012 J/mol
Therefore, the standard Gibbs free-energy change for the reaction at pH 7 and 25∘C is approximately -85,012 J/mol.
D. To calculate the Gibbs free-energy change (ΔG) under the given conditions, we can use the equation:
ΔG = ΔG∘ + RTln(Q)
Where ΔG∘ is the standard Gibbs free-energy change (-85,012 J/mol), R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (25 + 273 = 298 K), and Q is the reaction quotient.
Assuming the reaction is at equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (K), which we calculated earlier as approximately 1.23 x 10^(-7).
ΔG = -85,012 J/mol + (8.314 J/(mol·K)) * (298 K) * ln(1.23 x 10^(-7))
ΔG ≈ -83,269 J/mol
Therefore, the Gibbs free-energy change for the reaction at pH 7 and 25∘C, considering the specified concentrations, is approximately -83,269 J/mol.
To know more about equilibrium visit:
https://brainly.com/question/984092
#SPJ11
What is the purpose of adding base in the aldol condensation reaction? Choose the best answer. a Protonate a-carbon to generate electrophile b To generate intermediate enol c Deprotonate a-carbon to generate electrophile d Protonate a-carbon to generate nucleophile e Deprotonate a-carbon to generate nucleophile f To neutralize acid
The purpose of adding base in the aldol condensation reaction is c) Deprotonate α-carbon to generate electrophile.
What is aldol condensation?
Aldol condensation is a reaction in organic chemistry that involves the condensation of two carbonyl compounds, typically an aldehyde and a ketone, to form a β-hydroxy carbonyl compound.
In the aldol condensation reaction, a base is added to deprotonate the α-carbon of the carbonyl compound, typically an aldehyde or a ketone. The deprotonation of the α-carbon generates an enolate ion, which is an excellent nucleophile. This deprotonation step is crucial in generating the reactive electrophile necessary for the aldol condensation reaction.
By deprotonating the α-carbon, the base increases the electron density on the carbon atom, making it more nucleophilic and prone to react with another carbonyl compound. This enables the formation of a new carbon-carbon bond, resulting in the formation of an aldol product.
Therefore, the purpose of adding a base in the aldol condensation reaction is to deprotonate the α-carbon and generate an electrophilic enolate ion, which can then react with another carbonyl compound to form the desired product.
To know more about aldol condensation, refer here:
https://brainly.com/question/31558115
#SPJ4
Which molecule exhibits the greatest London dispersion forces?
a. C2H2
b. CHA
c. CH3CH2CH2CH3
d. CH3CH3
e. CH3CH2CH3
The molecule that exhibits the greatest London dispersion forces is e) CH3CH2CH3.
The London dispersion force is an intermolecular force that arises from the temporary dipole moment induced by instantaneous fluctuations in electron distribution within a molecule. The strength of the London dispersion force increases with the increasing size of the molecule and its polarizability.
In this case, the molecule that exhibits the greatest London dispersion forces is CH3CH2CH3, a branched alkane. As the molecule gets larger, the number of electrons increases, which results in greater dispersion forces between molecules due to the greater polarization. Additionally, the branching of the molecule decreases its surface area, which also increases the polarizability of each molecule. Therefore, CH3CH2CH3 has greater London dispersion forces than the other molecules, which are either smaller, less polarizable, or both.
The magnitude of London dispersion forces depends on the size and polarizability of the molecule. In this case, the molecule with the greatest London dispersion forces is CH3CH2CH3 due to its larger size and higher polarizability than the other molecules provided.
To know more about London dispersion forces, visit:
https://brainly.com/question/30763886
#SPJ11
Draw a structural formula of an alkene or alkenes (if more than one) that undergo acid-catalyzed hydration and without rearrangement give 2-butanol as the MAJOR product. O Your do not have to consider stereochemistry. O You do not have to explicitly draw H atoms.
O If more than one structure fits the description, draw them all. O Draw one structure per sketcher. Add additional sketchers using the dropdown menu in the bottom right corner. O Separate structures with + signs from the dropdown menu.
The alkene structure that undergoes acid-catalyzed hydration without rearrangement to give 2-butanol as the major product is:
CH₃CH=CHCH₃
Which alkene structure gives 2-butanol upon acid-catalyzed hydration?Acid-catalyzed hydration is a reaction in which an alkene reacts with water in the presence of an acid catalyst to form an alcohol. In this case, we want to obtain 2-butanol as the major product without any rearrangement. The given alkene, CH₃CH=CHCH₃, meets this criterion.
When this alkene undergoes acid-catalyzed hydration, the double bond is broken, and water adds to the carbon atoms, resulting in the formation of 2-butanol. No rearrangement of the carbon skeleton occurs during the reaction, ensuring that 2-butanol is the major product. Acid-catalyzed hydration and alkene reactions for further understanding.
Learn more about:Acid-catalyzed hydration
brainly.com/question/30891657
#SPJ11
For the following reaction, if NH3 is used up at a rate of 0.30Mmin, what is the rate of formation of N2?
2NH3→N2+3H2
Answer should be in two significant figures.
the - ve sign will be ignored since it's at the product side and product increase
0.15Mmin
you can support by rating brainly it's very much appreciated ✅
The rate of formation of N2 (to two significant figures) is 0.15 M/min. Therefore, the rate of formation of N2 is half of the rate of consumption of NH3.
According to the balanced chemical equation, the molar ratio of NH3 to N2 is 2:1. To find the rate of formation of N2, we will use the stoichiometric coefficients from the balanced chemical equation: 2NH3 → N2 + 3H2. Since the ratio of NH3 to N2 is 2:1, the rate of formation of N2 is half the rate of consumption of NH3. Given the rate of NH3 consumption is 0.30 M/min, we can set up a ratio: (Rate of N2 formation) / (Rate of NH3 consumption) = 1 / 2. Solving for the rate of N2 formation, we get: Rate of N2 formation = (1/2) * 0.30 M/min = 0.15 M/min.
By solving the given reaction the rate of formation of N2 is 0.15 M/min in two significant figures.
To know more about rate of formation visit:
brainly.com/question/27549044
#SPJ11
John Dalton's original atomic theory contain the following key ideas. Which part(s) of these ideas were was incorrect?
A. elements are made of tiny indivisible particles called atoms
B. Atoms are unchanged in chemical reaction
C. Atoms can join together in whole number ratios to form compounds
D. The atoms of each element are unique
John Dalton's original atomic theory proposed that elements are made up of tiny indivisible particles called atoms, and that these atoms are unchanged in chemical reactions. It also stated that atoms can combine in whole number ratios to form compounds, and that the atoms of each element are unique.
However, with the advancements in science, we now know that one part of Dalton's theory was incorrect. Atoms are not indivisible, as they can be split into smaller subatomic particles, such as protons, neutrons, and electrons. Furthermore, atoms can also be changed in chemical reactions, as they can lose or gain electrons, or undergo nuclear reactions.
Therefore, while John Dalton's theory provided a solid foundation for the understanding of atoms and elements, it was incomplete and required further refinement through scientific exploration and discovery.
To learn more about atomic visit;
https://brainly.com/question/1566330
#SPJ11
what formula represents the compound formed from calcium and dihydrogen phosphate?
The compound formed from calcium and dihydrogen phosphate is known as calcium dihydrogen phosphate or calcium phosphate monobasic.
Its chemical formula is Ca(H2PO4)2, which indicates that it contains one calcium ion (Ca2+) and two dihydrogen phosphate ions (H2PO4-). This compound is commonly used as a food additive, fertilizer, and as a source of phosphoric acid. It is also found naturally in some minerals and rocks. Calcium dihydrogen phosphate is an important compound in biological systems as it plays a role in bone and tooth formation, cell signaling, and energy metabolism.
The formula representing the compound formed from calcium and dihydrogen phosphate is Ca(H2PO4)2. This compound is called calcium dihydrogen phosphate. It is an ionic compound, where calcium is the cation (Ca2+) and dihydrogen phosphate is the anion (H2PO4-). In this compound, one calcium ion is bonded to two dihydrogen phosphate ions, maintaining charge neutrality. Calcium dihydrogen phosphate is often used in fertilizers and as a food additive.
To learn more about calcium visit;
https://brainly.com/question/30954368
#SPJ11
Triphenylmethane can be prepared by reaction of benzene and chloroform in the presence of AlCl3. Propose a mechanism for the reaction:
The reaction of benzene and chloroform in the presence of AlCl₃ to produce triphenylmethane is known as the Friedel-Crafts alkylation reaction.
Here is a proposed mechanism for this reaction
Step 1: Formation of the Lewis acid complex
AlCl₃ acts as a Lewis acid and accepts an electron pair from chloroform (CHCl₃) to form a complex between AlCl₃ and CHCl₃.
AlCl₃ + CHCl₃ ⟶ AlCl₄⁻ + CH₂Cl₂
Step 2: Activation of benzene
The Lewis acid complex formed in Step 1 reacts with benzene (C₆H₆), leading to the activation of benzene towards electrophilic attack.
AlCl₄⁻ + C₆H₆ ⟶ AlCl₃ + C₆H₆⁺
Step 3: Electrophilic attack
The electrophilic benzene (C₆H₆⁺) reacts with another molecule of benzene to form a cyclohexadienyl cation intermediate.
C₆H₆+ + C₆H₆ ⟶ C₆H₆C₆H₆⁺
Step 4: Rearrangement
The cyclohexadienyl cation undergoes rearrangement to form the more stable carbocation intermediate.
C₆H₆C₆H₆⁺ ⟶ C₆H₅C₆H₇⁺
Step 5: Loss of proton and regeneration of the catalyst
The carbocation loses a proton to regenerate the aromaticity of the benzene ring, and AlCl₃ acts as a catalyst by accepting the proton.
C₆H₅C₆H₇⁺ ⟶ C₆H₅C₆H₆ + H⁺ (proton)
H⁺ + AlCl₃ ⟶ AlCl₄⁻
Overall reaction
Benzene + Chloroform ⟶ Triphenylmethane
It's important to note that this is a simplified mechanism and may not capture all the intricacies of the actual reaction. Additionally, there may be variations in the reaction mechanism depending on reaction conditions and other factors.
To know more about Friedel-Crafts alkylation here
https://brainly.com/question/32219561
#SPJ4
63 C u has an atomic mass of 62.9298 amu 62.9298 a m u and an abundance of 69.09% . The other isotope, 65Cu 65 C u , has an abundance of 30.91% . The average atomic mass between these two isotopes is 63.546 amu 63.546 a m u . Calculate the actual atomic mass of 65Cu 65 C u .
The actual atomic mass of 65Cu is 64.9278 amu , based on the given information about the abundances and the average atomic mass.
To calculate the actual atomic mass of 65Cu, we can use the information provided. Let's denote the atomic mass of 63Cu as x. According to the given data, the abundance of 63Cu is 69.09%, and the abundance of 65Cu is 30.91%.
The average atomic mass between these two isotopes is given as 63.546 amu. This average atomic mass can be expressed as the weighted average of the atomic masses of each isotope, taking into account their respective abundances:
(0.6909 * x) + (0.3091 * 65) = 63.546
Simplifying the equation, we get:
0.6909x + 20.0595 = 63.546
0.6909x = 63.546 - 20.0595
0.6909x = 43.4865
Dividing both sides of the equation by 0.6909, we find:
x = 43.4865 / 0.6909
x ≈ 62.9298 amu
Therefore, the atomic mass of 63Cu is approximately 62.9298 amu.
Since the atomic mass of 65Cu is the actual atomic mass that we need to calculate, we subtract the atomic mass of 63Cu from the average atomic mass:
65Cu atomic mass = Average atomic mass - 63Cu atomic mass
65Cu atomic mass = 63.546 amu - 62.9298 amu
65Cu atomic mass ≈ 0.6162 amu
Thus, the actual atomic mass of 65Cu is approximately 64.9278 amu.
The actual atomic mass of 65Cu is approximately 64.9278 amu, based on the given information about the abundances and the average atomic mass.
To know more about atomic mass ,visit:
https://brainly.com/question/3187640
#SPJ11
How many grams of Ca (OH)2 are needed to make a 0.600 M solution with a volume of 200.0 mL? Your answer should have three significant figures. Provide your answer below:
In order to create a 0.600 M solution with a 200.0 mL volume, 8.89 grammes of Ca(OH)₂ are required.
To calculate the amount of Ca(OH)₂ needed, we'll use the formula:
Amount (in moles) = Concentration (in M) × Volume (in L)
First, convert the volume from milliliters to liters:
200.0 mL = 200.0 mL × (1 L / 1000 mL) = 0.200 L
Next, substitute the given values into the formula:
Amount (in moles) = 0.600 M × 0.200 L = 0.120 moles
Now, calculate the molar mass of Ca(OH)₂:
Ca: 40.08 g/mol
O: 16.00 g/mol (2 atoms)
H: 1.01 g/mol (2 atoms)
Molar mass of Ca(OH)₂ = 40.08 g/mol + 16.00 g/mol × 2 + 1.01 g/mol × 2 = 74.09 g/mol
Finally, calculate the mass of Ca(OH)₂ needed:
Mass = Amount (in moles) × Molar mass = 0.120 moles × 74.09 g/mol = 8.89 grams
To know more about molar mass
https://brainly.com/question/12127540
#SPJ4
Which of the following gases has the highest average speed at 400K?
a NO₂
b SO₂
c SF₂
d OF₂
NO₂ has the lowest molar mass. Therefore, it will have the highest average speed at 400K among the listed gases. So, the correct answer is option (a) NO₂.
The average speed of a gas molecule is directly proportional to its temperature. As all the given gases have the same temperature of 400K, the average speed will be determined by their molar mass. The lighter the gas molecule, the higher the average speed. Out of the given options, OF₂ has the lowest molar mass (32 g/mol) while the other three have higher molar masses. Therefore, OF₂ will have the highest average speed at 400K. Hence, option d is the correct answer.
At 400K, the gas with the highest average speed will be the one with the lowest molar mass, as lighter molecules move faster at the same temperature. The molar masses of the given gases are:
a) NO₂ - 46 g/mol
b) SO₂ - 64 g/mol
c) SF₂ - 70 g/mol
d) OF₂ - 66 g/mol
Comparing these values, NO₂ has the lowest molar mass. Therefore, it will have the highest average speed at 400K among the listed gases. So, the correct answer is option (a) NO₂.
To learn more about temperature visit;
https://brainly.com/question/7510619
#SPJ11
When considering the equilibrium between copper(I) carbonate and its aqueous ions, what could be added to solution to promote the precipitation of copper(I) carbonate?
copper(I) sulfide
calcium carbonate
potassium carbonate
copper(II) sulfide
To promote the precipitation of copper(I) carbonate when considering its equilibrium with its aqueous ions, you should add potassium carbonate to the solution.
When potassium carbonate (K2CO3) is added to the solution, it will dissociate into potassium ions (K+) and carbonate ions (CO3^2-). The presence of additional carbonate ions will shift the equilibrium of the reaction between copper(I) carbonate and its aqueous ions, causing more copper(I) carbonate to precipitate out of the solution.
The reaction is as follows:
Cu2CO3(s) ⇌ 2Cu+(aq) + CO3^2-(aq)
By adding potassium carbonate (K2CO3), you increase the concentration of CO3^2- ions, which then shifts the equilibrium to the left, favoring the formation of the solid copper(I) carbonate.
To know more about copper(I) carbonate visit:
https://brainly.com/question/25251015
#SPJ11
Name the elements that would have the following ground-state electron configurations: A. 1s^2 2s^2 2p^5 B. 1s^2 2s^2 2p^6 3s^2 C. 1s^2 2s^2 2p^6 3s^2 3p^3 D. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 E. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d
Fluorine (F) Neon (Ne) Phosphorus (P) Potassium (K) Calcium (Ca)
The electron configuration 1s^2 2s^2 2p^5 corresponds to the element fluorine (F) with atomic number 9.
The electron configuration 1s^2 2s^2 2p^6 3s^2 corresponds to the element neon (Ne) with atomic number 10.
The electron configuration 1s^2 2s^2 2p^6 3s^2 3p^3 corresponds to the element phosphorus (P) with atomic number 15.
The electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 corresponds to the element potassium (K) with atomic number 19.
The electron configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d corresponds to the element calcium (Ca) with atomic number 20.
The given electron configurations correspond to the following elements: A) Fluorine (F), B) Neon (Ne), C) Phosphorus (P), D) Potassium (K), and E) Calcium (Ca).
To learn more about atomic number, visit
brainly.com/question/11353462
#SPJ11
Which of the following compounds will be more soluble in acidic solution than in pure water?
a. Cul
b. Ca(CIO4)2
c. PbBr2
d. FeS
e. None of the above will be more soluble in acidic solution.
None of the above will be more soluble in acidic solution than in pure water. The correct answer is e.
When considering the solubility of compounds, the solubility behavior in acidic solution is not solely determined by the nature of the compound itself. Instead, it depends on the specific interactions between the compound and the acid present in the solution.
(a) Cul (copper(I) iodide) is generally insoluble in both pure water and acidic solutions. It does not exhibit increased solubility in acidic conditions.
(b) Ca(CIO₄)₂ (calcium perchlorate) is a soluble salt that dissolves readily in both pure water and acidic solutions. The solubility of calcium perchlorate is not significantly affected by the presence of acids.
(c) PbBr₂ (lead(II) bromide) is sparingly soluble in pure water and generally exhibits similar solubility behavior in both pure water and acidic solutions.
(d) FeS (iron(II) sulfide) is insoluble in both pure water and acidic solutions and does not show enhanced solubility in acidic conditions.
In summary, the solubility behavior of the compounds listed is not significantly altered in acidic solutions compared to pure water. Therefore, none of the compounds listed will be more soluble in acidic solution.
Therefore, the correct option is E, None of the above will be more soluble in acidic solution.
To know more about solubility in acidic solution, refer here:
https://brainly.com/question/15463895#
#SPJ11
Part A Calculate the solubility of Au(OH)3 in water (Ksp=5.5×10−46) (pH=7).
Part B Calculate the solubility of Au(OH)3 in a solution maintained at a nitric acid concentration of 1.0 M .
Part A: The solubility of Au(OH)₃ in water is approximately 5.5 × 10⁻²⁵ M.
Part B: The solubility of Au(OH)₃ in a solution maintained at a nitric acid concentration of 1.0 M is approximately 5.5×10⁻⁴⁶ mol/L.
Part A:
To calculate the solubility of Au(OH)₃ in water, we need to determine the concentration of Au³⁺ ions in the solution. Since Au(OH)³ dissociates into Au³⁺ and OH⁻ ions, we can set up an equilibrium expression using the solubility product constant (Ksp):
Au(OH)³ ⇌ Au³⁺ + 3OH⁻
The Ksp expression for this reaction is:
Ksp = [Au³⁺][OH⁻]³
We are given that the Ksp value for Au(OH)₃ is 5.5 × 10⁻⁴⁶. Since water has a neutral pH of 7, we can assume that the concentration of OH⁻ ions is equal to the concentration of H⁺ ions, which is 10⁻⁷ M.
Let's assume the solubility of Au(OH)₃ is "s" M. Then, the concentration of Au³⁺ ions is also "s" M.
Using the equilibrium expression, we can write:
Ksp = [Au³⁺][OH⁻]³
5.5 × 10⁻⁴⁶ = (s)(10⁻⁷)³
Simplifying the equation:
5.5 × 10⁻⁴⁶ = s × 10⁻²¹
Dividing both sides by 10⁻²¹:
s = (5.5 × 10⁻⁴⁶)/(10⁻²¹)
s = 5.5 × 10⁻⁴⁶⁺²¹
s = 5.5 × 10⁻²⁵
The solubility of Au(OH)₃ in water can be estimated to be around 5.5 × 10⁻²⁵ M.
Part B:
When calculating the solubility of Au(OH)3 in a solution with a nitric acid concentration of 1.0 M, it is important to take into account the influence of nitric acid on the equilibrium of the solubility process. Nitric acid is a strong acid that dissociates completely in water to produce H⁺ ions.
The balanced equation for the dissociation of nitric acid (HNO₃) is:
HNO₃ (aq) → H⁺ (aq) + NO₃⁻(aq)
The H+ ions from nitric acid will react with the OH⁻ ions produced from the dissociation of Au(OH)₃. This reaction can shift the equilibrium of the solubility of Au(OH)₃.
Let's assume that x mol/L of Au(OH)₃ dissolves in the presence of nitric acid. Since Au(OH)₃ dissociates into 1 Au³⁺ ion and 3 OH⁻ ions, the concentrations of these ions can be expressed as follows:
[Au³⁺] = x mol/L
[OH⁻] = 3x mol/L
The concentration of H⁺ ions introduced by the nitric acid is 1.0 M. Therefore, the concentration of OH⁻ ions can be expressed as (1.0 - 3x) M.
Now, we can write the solubility product expression for Au(OH)3 including the effect of nitric acid:
Ksp = [Au3³⁺][OH⁻]³
Ksp = (x)(1.0 - 3x)³
We can solve this equation to find the value of x, which represents the solubility of Au(OH)3 in the presence of 1.0 M nitric acid. However, solving this equation analytically can be complex due to the cubic term. To simplify the calculation, we can make an assumption that 3x << 1.0, which means the concentration of OH- ions is significantly smaller than 1.0 M.
Using this approximation, we can neglect the contribution of 3x to 1.0, and the equation becomes:
Ksp ≈ (x)(1.0)³
Ksp ≈ x
Since Ksp = 5.5×10⁻⁴⁶, we can equate it to x and solve for x:
x ≈ 5.5×10⁻⁴⁶
Therefore, the solubility of Au(OH)₃ in a solution maintained at a nitric acid concentration of 1.0 M is approximately 5.5×10⁻⁴⁶ mol/L.
Learn more about solubility at: https://brainly.com/question/23946616
#SPJ11
ist the possible effects of inhaling excessive amounts of pinacolone (3,3-dimethylbutan-2-one).
The possible effects of inhaling excessive amounts of pinacolone (3,3-dimethylbutan-2-one) can include respiratory irritation and central nervous system (CNS) depression.
1. Respiratory Irritation: Inhaling excessive amounts of pinacolone vapor or aerosol may cause irritation to the respiratory system. This can result in symptoms such as coughing, shortness of breath, chest discomfort, and throat irritation. Prolonged or repeated exposure to high concentrations of pinacolone can potentially lead to more severe respiratory effects.
2. Central Nervous System (CNS) Depression: Pinacolone is a solvent and can have CNS depressant effects when inhaled in excessive amounts. It can affect the normal functioning of the brain and nervous system, leading to symptoms such as dizziness, drowsiness, confusion, and in severe cases, loss of consciousness.
It is important to note that the specific effects of inhaling excessive amounts of pinacolone can vary depending on the concentration, duration of exposure, individual sensitivity, and other factors. It is always recommended to handle chemicals in a well-ventilated area and follow proper safety precautions to minimize the risk of inhalation exposure. If someone experiences significant exposure or develops symptoms after inhaling pinacolone, it is crucial to seek medical attention promptly.
To know more about nervous system , refer here :
https://brainly.com/question/8695732#
#SPJ11
Which of the following statements are FALSE regarding the Lewis structures of XeFO^3- and XeFO^3+ ?
Question options:
a. The octet rule is violated in XeFO^3+.
b. There is a formal charge on O in XeFO^3+.
c. There is a formal charge on O in XeFO^3-.
d. The Xe atom in XeFO^3+ has a single non-bonding electron pair.
e. The xenon atom in XeFO^3- has a single non-bonding electron pair.
Based on the provided statements, the FALSE statements regarding the Lewis structures of XeFO^3- and XeFO^3+ are:
a. The octet rule is violated in XeFO^3+.
c. There is a formal charge on O in XeFO^3-.
Regarding statement a, the octet rule refers to the tendency of atoms to gain, lose, or share electrons to achieve a stable electron configuration with eight valence electrons. In XeFO^3+, the xenon atom (Xe) is surrounded by three fluorine atoms (F) and one oxygen atom (O), forming a trigonal pyramidal structure. Each fluorine atom forms a single bond with the xenon atom, and the oxygen atom forms a double bond with the xenon atom. The xenon atom in XeFO^3+ has a total of eight valence electrons, satisfying the octet rule.
Regarding statement c, in XeFO^3-, the xenon atom (Xe) is surrounded by three fluorine atoms (F) and one oxygen atom (O), forming a trigonal pyramidal structure. Each fluorine atom forms a single bond with the xenon atom, and the oxygen atom forms a double bond with the xenon atom. The oxygen atom in XeFO^3- has a formal charge of -1, as it has six valence electrons (from the lone pair and the shared electrons in the double bond) and seven assigned electrons (from the bonds). Therefore, there is a formal charge on O in XeFO^3-, making statement c FALSE.
The remaining statements, b, d, and e, are not mentioned as FALSE, so they are either TRUE or not provided in the question.
To learn more about octet rule click here;
brainly.com/question/1202688
#SPJ11
For the equilibrium 2IBr(g)⇌I2(g)+Br2(g) Kp=8.5×10−3 at 150 ∘C.
a) If 2.4×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of IBr after equilibrium is reached? Express your answer to two significant figures and include the appropriate units.
b) If 2.4×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of I2 after equilibrium is reached? Express your answer to two significant figures and include the appropriate units
c) If 2.4×10−2 atm of IBr is placed in a 2.0-L container, what is the partial pressure of Br2 after equilibrium is reached? Express your answer to two significant figures and include the appropriate units.
To solve this problem, we'll use the provided equilibrium expression and the given initial pressure of IBr to calculate the partial pressures of each species at equilibrium. Let's proceed with the calculations:
a) Partial pressure of IBr after equilibrium is reached:
Let's assume that the equilibrium partial pressure of IBr is x atm. Using the equilibrium expression, we have:
Kp = [I2][Br2] / [IBr]^2
Substituting the given values into the expression, we have:
8.5×10^(-3) = (x)(x) / (2.4×10^(-2))^2
Simplifying and solving for x, we get:
x^2 = 8.5×10^(-3) * (2.4×10^(-2))^2
x^2 = 4.464×10^(-6)
x ≈ 6.68×10^(-3) atm
Therefore, the partial pressure of IBr after equilibrium is reached is approximately 6.68×10^(-3) atm.
b) Partial pressure of I2 after equilibrium is reached:
Since the stoichiometric coefficient for IBr in the balanced equation is 2, the partial pressure of I2 will also be x atm at equilibrium.
Therefore, the partial pressure of I2 after equilibrium is reached is also approximately 6.68×10^(-3) atm.
c) Partial pressure of Br2 after equilibrium is reached:
Using the stoichiometric coefficient of 1 for Br2, we know that the partial pressure of Br2 will also be x atm at equilibrium.
Therefore, the partial pressure of Br2 after equilibrium is reached is approximately 6.68×10^(-3) atm.
In summary:
a) The partial pressure of IBr after equilibrium is reached is approximately 6.68×10^(-3) atm.
b) The partial pressure of I2 after equilibrium is reached is also approximately 6.68×10^(-3) atm.
c) The partial pressure of Br2 after equilibrium is reached is approximately 6.68×10^(-3) atm.
To know more about stoichiometric , refer here :
https://brainly.com/question/32088573#
#SPJ11
Based on the guidance in Section 2 of experiment EMS, determine the experimental Rydberg constant, in m', for an emission line from the hydrogen spectrum that has a wavelength of 94 nm. a. 1.09422 x 10 m-1 b. 9.13889 x 10-8 m-1 c. -1.09422 x 10-2 m-1 d. 1.09422 x 1016 m m-1 e. 1.09422 x 10-2 m-1 f. -1.09422 x 10 m-1
Based on the guidance in Section 2 of experiment EMS, determine the experimental Rydberg constant, in m', for an emission line from the hydrogen spectrum that has a wavelength of 94 nm id: 1.06383 x 10^7 m^(-1). The correct option is (a).
The experimental Rydberg constant, denoted as R', can be determined using the formula:
R' = 1 / λ
where λ is the wavelength of the emission line from the hydrogen spectrum.
In this case, the wavelength given is 94 nm. To find the experimental Rydberg constant, we can substitute this value into the formula:
R' = 1 / 94 nm
To simplify the units, we convert nanometers to meters:
R' = 1 / (94 x 10^(-9) m)
Calculating this expression gives us:
R' ≈ 1.06383 x 10^7 m^(-1)
It's important to note that the answer options provided in the question do not match the calculated value. The closest option is a. 1.09422 x 10^(-1) m^(-1), but it is missing the correct exponent. The correct answer should have a positive exponent of 10 to match the calculated value
To know more about "Rydberg constant" refer here:
https://brainly.com/question/29673026#
#SPJ11
decide whether the lewis structure proposed for each molecule is reasonable or not. nh2
The Lewis structure proposed for the molecule NH₂ is not reasonable. The reason is that nitrogen (N) has five valence electrons, and hydrogen (H) has one valence electron each. In total, there are four valence electrons in the NH₂ molecule. However, in the proposed Lewis structure, there is no lone pair of electrons on nitrogen, which means it has only three valence electrons surrounding it.
To achieve stability, nitrogen needs to have a complete octet (eight valence electrons) by sharing or gaining electrons. One possible Lewis structure for NH₂ is as follows:
H
|
N
|
H
In this structure, the nitrogen atom shares one electron with each of the two hydrogen atoms, forming two covalent bonds. Additionally, the nitrogen atom has one lone pair of electrons, which provides a total of eight valence electrons around nitrogen, satisfying the octet rule.
Therefore, The Lewis structure proposed for the molecule NH₂ is not reasonable.
To know more about Lewis structure, refer here:
https://brainly.com/question/32265997#
#SPJ11
Write a balanced equation for the combustion of gaseous methane?
Another potential future fuel is methanol (CH3OH). Write a balanced equation for the combustion of gaseous methanol? Express your answer as a chemical formula. Another potential future fuel is methanol (CH3OH). Use bond energies to calculate the enthalpy of combustion of methanol in kJ/mol. Express your answer in kiloJoules to three significant figures. Use bond energies to calculate AH rxn for this reaction: N2(g) + 3H2(g) rightarrow 2NH3(g). Express your answer in kiloJoules to two significant figures.
The balanced equation for the combustion of gaseous methane is: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g).
What is the balanced equation for the combustion of methane?The combustion of gaseous methane (CH4) involves the reaction of methane with oxygen gas (O2) to produce carbon dioxide (CO2) and water vapor (H2O). To balance the equation, we ensure that the same number of atoms of each element is present on both sides of the equation.
In the balanced equation, we have one carbon atom on both sides, four hydrogen atoms on both sides, and four oxygen atoms on both sides, which indicates that the equation is balanced.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
The combustion of methanol (CH3OH) follows a similar pattern. The balanced equation for the combustion of gaseous methanol is:
2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)
To calculate the enthalpy of combustion of methanol, we can use bond energies. The enthalpy change (∆H) of a reaction can be determined by calculating the difference between the energy required to break the bonds of the reactants and the energy released when new bonds are formed in the products.
To calculate the enthalpy change (∆H) for the reaction, we need the bond energies of the relevant bonds involved. By summing up the bond energies of the bonds broken minus the bond energies of the bonds formed, we can determine the enthalpy of combustion of methanol.
Learn more about Combustion
brainly.com/question/31123826
#SPJ11
calculate the ratio of the concentrations of h2o2 (3.0% and 2.25%)in these two experiments. show your work (include units). hint: this is just the ratio of one concentration over another.
The ratio of the concentrations of H^2O^2 (3.0% and 2.25%) in the experiments are 1.097.
What is a concentration?
Concentration refers to the amount of solute present in a given amount of solvent or solution. It represents the ratio of the quantity of solute to the quantity of solvent or solution.
Experiment 1:
H2O2 concentration: 3.0% (w/v)
Reaction rate: 0.0622 mL/s
Experiment 2:
H2O2 concentration: 2.25% (w/v)
Reaction rate: 0.0566 mL/s
Since the reaction rates are given in mL/s, we need to convert them to concentrations in order to compare them.
For Experiment 1:
Reaction rate = 0.0622 mL/s
Concentration = (Reaction rate / 1000) / Time (s)
Concentration = (0.0622 mL/s / 1000) / 1s
Concentration = 0.0000622 g/mL
For Experiment 2:
Reaction rate = 0.0566 mL/s
Concentration = (Reaction rate / 1000) / Time (s)
Concentration = (0.0566 mL/s / 1000) / 1s
Concentration = 0.0000566 g/mL
Ratio of concentrations = Concentration Experiment 1 / Concentration Experiment 2
Ratio of concentrations = (0.0000622 g/mL) / (0.0000566 g/mL)
Ratio of concentrations ≈ 1.097
Therefore, the ratio of the concentrations of H2O2 in Experiment 1 (3.0%) to Experiment 2 (2.25%) is approximately 1.097.
To learn more about Concentration,
https://brainly.com/question/4184101
#SPJ4
Complete question:
Calculate the ratio of the concentrations of H^2O^2 (3.0% and 2.25%) in these two experiments. Show your work (include units).
Experiment 1 with 3.0% H^2O^2: Reaction rate = .0622 mL/s
Experiment 2 with 2.25% H^2O^2: Reaction rate = .0566 mL/s
Estimate the value of the equilibrium constant at 610 KK for each of the following reactions. ΔG∘fΔGf∘ for BrCl(g)BrCl(g) is −−1.0 kJ/molkJ/mol. The standard molar entropy, S∘S∘, for BrCl(g)BrCl(g) is 240.0 J/mol⋅KJ/mol⋅K.
Part A
2NO2(g)⇌N2O4(g).2NO2(g)⇌N2O4(g). ΔH∘fΔHf∘ for N2O4(g)N2O4(g) is 9.16 kJ/molkJ/mol.
Express your answer using three significant figures.
K= ???
Part B
Br2(g)+Cl2(g)⇌2BrCl(g).Br2(g)+Cl2(g)⇌2BrCl(g). ΔH∘fΔHf∘ for BrCl(g)BrCl(g) is 14.6 kJ/molkJ/mol.
Express your answer using three significant figures.
K= ???
Part A: K = Unable to estimate due to missing information.
Part B: K ≈ 1.221 (approximately 1.221)
Part A: Missing information, unable to estimate.Part B: K ≈ 1.221 (approximately 1.221)To calculate the equilibrium constant (K) at 610 K for each of the reactions, we can use the relationship between K, ΔG°, and temperature (T):
ΔG° = -RT ln(K)
Where:
ΔG° is the standard Gibbs free energy change
R is the gas constant (8.314 J/mol·K)
T is the temperature in Kelvin (610 K)
ln represents the natural logarithm
Part A: 2NO2(g) ⇌ N2O4(g)
We need the ΔG°f value for N2O4(g). However, the given information only provides ΔH°f for N2O4(g), not ΔG°f. We can use the relationship between ΔG°f and ΔH°f:
ΔG°f = ΔH°f - TΔS°
Where:
ΔH°f is the standard enthalpy of formation
ΔS° is the standard molar entropy
Since we don't have ΔS° for N2O4(g), we cannot calculate ΔG°f or subsequently the equilibrium constant (K). Therefore, we cannot provide an estimate for K in this case.
Part B: Br2(g) + Cl2(g) ⇌ 2BrCl(g)
Given:
ΔH°f for BrCl(g) = 14.6 kJ/mol
ΔG°f for BrCl(g) = -1.0 kJ/mol
S° for BrCl(g) = 240.0 J/mol·K
To calculate K, we'll use the equation:
ΔG° = -RT ln(K)
First, convert the given values to J/mol:
ΔH°f = 14.6 kJ/mol = 14,600 J/mol
ΔG°f = -1.0 kJ/mol = -1,000 J/mol
Substituting the values into the equation, we get:
-1,000 J/mol = -(8.314 J/mol·K)(610 K) ln(K)
Now solve for K:
ln(K) = (-1,000 J/mol) / [-(8.314 J/mol·K)(610 K)]
ln(K) = 0.200493
K = e^(0.200493)
Using a calculator, we find:
K ≈ 1.221
Therefore, the estimated equilibrium constant for the reaction Br2(g) + Cl2(g) ⇌ 2BrCl(g) at 610 K is approximately 1.221.
Learn more about Equilibrium
brainly.com/question/30694482
#SPJ11
.PART B:
If 7.50 mL of 0.125 M HCl are added to 100 mL of the original buffer . . .
What is the concentration of NH3 in the buffer *after* the addition of HCl?
[Choose], 0.140 M, 0.148 M, 9.28, 0 M, 0.00938 M, 0.131 M, 4.80, 7.00, 0.159 M, 9.21, 8.11
What is the concentration of NH4Cl in the buffer *after* the addition of HCl?
[Choose], 0.140 M, 0.148 M, 9.28, 0 M, 0.00938 M, 0.131 M, 4.80, 7.00, 0.159 M, 9.21, 8.11
What is the pH of the buffer solution *after* the addition of HCl?
[Choose], 0.140 M, 0.148 M, 9.28, 0 M, 0.00938 M, 0.131 M, 4.80, 7.00, 0.159 M, 9.21, 8.11
After addition of HCl in the equation, pH = 9.21 and After addition of NaOH in the solution, pH = 9.31 .
Given the buffer contains 50 mL of 0.3 M NH₃
therefore moles of NH₃ = 50 mL × 0.3 mol / 1000 mL
= 0.0150 moles.
similarly moles of NH₄Cl = 0.015 moles.
The addition of 7.5 mL of 0.125 M HCl = 7.5 mL × 0.125 mol / 1000 mL
= 9.375 × 10⁻⁴ moles of HCl.
The added HCl will react with NH₃ to form NH₄Cl
NH₃ + HCl 1 NH₄Cl
initial 0.015 0.0009375 0.015
change -0.0009375 - 0.0009375 + 0.0009375
final 0.0140625 0 0.0159375
The final volume of the solution = 100 mL + 7.5 mL = 107.5 mL
= 0.1075 L
Therefore the concentration of NH₃ after addition of HCl = mol / L
= 0.0140625 mol / 0.1075 L
= 0.131 M
similarly the conc of NH₄Cl after addition of HCl = mol / L
= 0.0159375 mol / 0.1075 L
= 0.148 M
By Henderson equation,
pOH = pKb + log [salt/base]
= 4.74 + log [0.148 / 0.131]
= 4.74 + 0.053
= 4.793
pH = 14 - pOH = 14 - 4.793
= 9.21
The addition of 7.5 mL of 0.125 M NaOH = 7.5 mL × 0.125 mol / 1000 mL
= 9.375 × 10⁻⁴ moles of NaOH.
The added NaOH will react with NH₄Cl to form NH₃
NH₄Cl + NaOH 1 NH₃ + NaCl + H₂O
initial 0.015 0.0009375 0.015
change -0.0009375 - 0.0009375 + 0.0009375
final 0.0140625 0 0.0159375
The final volume of the solution = 100 mL + 7.5 mL = 107.5 mL
= 0.1075 L
Therefore the concentration of NH₃ after addition of NaOH = mol / L
= 0.0159375 mol / 0.1075 L
= 0.148 M
similarly the concentration of NH₄Cl after addition of NaOH = mol / L
= 0.0140625 mol / 0.1075 L
= 0.131 M
By Henderson equation,
pOH = pKb + log [salt/base]
= 4.74 + log [0.131 / 0.148]
= 4.74 - 0.053
= 4.69
pH = 14 - pOH = 14 - 4.693
= 9.31
Thus
After addition of HCl in the equation :
[NH₃] = 0.131
[NH₄Cl] = 0.148
pH = 9.21
After addition of NaOH in the solution :
[NH₃] = 0.148
[NH₄Cl] = 0.131
pH = 9.31
Buffer solution :A buffer is a solution that can resist changing its pH when acidic or basic ingredients are added. It can neutralize small amounts of added acid or base, maintaining a relatively stable pH in the solution. This is significant for processes and additionally responses which require explicit and stable pH ranges.
Learn more about buffer solution :
brainly.com/question/13076037
#SPJ4
the crystal field splitting energy (δo) of co(cn)63– is 6.74 × 10–19 j. what is the wavelength of light absorbed by this complex ion?
The complex ion [Co(CN)₆]³⁻ absorbs light at a wavelength of approximately 9.86 × 10⁻⁷ meters.
To calculate the wavelength of light absorbed by the complex ion [Co(CN)₆]³⁻, we can use the relationship between energy (E) and wavelength (λ) given by the equation:
E = hc/λ
where h is Planck's constant (6.626 × 10⁻³⁴ J·s) and c is the speed of light (2.998 × 10⁸ m/s).
Given:
Crystal field splitting energy (Δ₀) = 6.74 × 10⁻¹⁹ J
We know that the absorbed energy corresponds to the energy difference (ΔE) between the ground state and an excited state. In this case, the absorbed light leads to the transition between the d-orbitals of the central cobalt ion in the complex.
ΔE = Δ₀
We can rearrange the equation to solve for the wavelength:
λ = hc/ΔE
Substituting the values:
λ = (6.626 × 10⁻³⁴ J·s × 2.998 × 10⁸ m/s) / (6.74 × 10⁻¹⁹ J)
λ ≈ 9.86 × 10⁻⁷ m
Therefore, the wavelength of light absorbed by the [Co(CN)₆]³⁻ complex ion is approximately 9.86 × 10⁻⁷ meters.
Learn more about wavelength here:
https://brainly.com/question/30532991
#SPJ4
In order to correctly report the accuracy of a number, scientists will use significant
figures. It can be confusing to determine which digits are significant and which are
not. Which one of these numbers is the one that must be questioned?
A:The zeros
B:numbers between 1-9
C:numbers which include a decimal point
D:any digits written in scientific notation.
The zeros is the one that must be questioned (option A)
What is a significant figure?Significant figures pertain to the digits within a number that possess precise accuracy. It is important to note that not all zeros in a number carry significance.
For instance, when considering the number 100, the zeros merely serve as placeholders, rendering only two significant figures. Conversely, in the case of 100.0, the zeros following the decimal point are considered significant, thus granting a total of three significant figures.
Learn about significant figure here https://brainly.com/question/24630099
#SPJ1
Which of the following pairs of solutions will form a buffer? Select all that apply 0.100 M HCl and 0.100 M HNO3 O 0.100 M NH4Cl and 0.100 M NH3 O 0.100 M HCl and 0.100 M Naci O 0.100 MHCN and 0.100 M LICN
The pair of solutions that will form a buffer is 0.100 M NH4Cl and 0.100 M NH3.
To determine whether a pair of solutions will form a buffer, we need to check if there is a weak acid and its conjugate base, or a weak base and its conjugate acid, present in the solution.
1. 0.100 M HCl and 0.100 M HNO3: Both HCl and HNO3 are strong acids and fully dissociate in water. Therefore, this pair of solutions will not form a buffer.
2. 0.100 M NH4Cl and 0.100 M NH3: NH4Cl is a salt of a weak base (NH3) and a strong acid (HCl). NH3 can act as a weak base and NH4+ as its conjugate acid. Thus, this pair of solutions will form a buffer.
3. 0.100 M HCl and 0.100 M NaCl: Both HCl and NaCl are strong acids and fully dissociate in water. Hence, this pair of solutions will not form a buffer.
4. 0.100 M HCN and 0.100 M LICN: HCN is a weak acid, but LICN is not its conjugate base. Therefore, this pair of solutions will not form a buffer.
The pair of solutions that will form a buffer is 0.100 M NH4Cl and 0.100 M NH3, as it consists of a weak acid (NH4+) and its conjugate base (NH3).
To know more about dissociate visit :
https://brainly.com/question/31106631
#SPJ11
