Answer:
[H₃O⁺] = 2.63×10⁻¹⁰ M
As pH = 9.57, the solution is basic
Explanation:
We must know this knowledge:
[OH⁻] . [H₃O⁺] = 1×10⁻¹⁴
3.8×10⁻⁵ . [H₃O⁺] = 1×10⁻¹⁴
[H₃O⁺] = 1×10⁻¹⁴ / 3.8×10⁻⁵ → 2.63×10⁻¹⁰ M
Let's determine the pH to state if the solution is acidic or basic
pH < 7 → acidic ; pH > 7 → basi
pH = - log [H₃O⁺]
pH = - log 2.63×10⁻¹⁰ → 9.57
What is the electron capacity of the nth energy level? What is the capacity of the fourth energy level?
Answer: The number of electrons present in the fourth energy level are 32
Explanation:
To calculate the number of electrons present in a particular energy level, we use the equation:
[tex]\text{Number of electrons}=2n^2[/tex]
where,
n = principle quantum number
Calculating the number of electrons for fourth energy level:
n = 4
Putting values in above equation, we get:
[tex]\text{Number of electrons}=2(4)^2\\\\\text{Number of electrons}=32[/tex]
Hence, the number of electrons present in the fourth energy level are 32
A 0.4 M buffer solution was prepared with acetic acid and sodium acetate. At pH 5.5, what are the concentrations of acetic acid and acetate ion? The pKa of acetic acid is 4.76. Round the answers to two decimal places. State the units.
Answer: The concentration of acetic acid and sodium acetate (acetate ion) is 0.06 M and 0.34 M respectively
Explanation:
We are given:
Concentration of buffer solution having acetic acid and sodium acetate = 0.4 M
Let the concentration of acetic acid be x M
So, the concentration of sodium acetate will be = (0.4 - x) M
To calculate the concentration of acid for given pH, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]
[tex]pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})[/tex]
where,
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acetic acid = 4.76
[tex][CH_3COONa]=0.4-x[/tex]
[tex][CH_3COOH]=x[/tex]
pH = 5.5
Putting values in above equation, we get:
[tex]5.5=4.76+\log(0.4-x}{x})\\\\x=0.062M[/tex]
So, concentration of acetic acid = x = 0.06 M
Concentration of sodium acetate = (0.4 - x) = (0.4 - 0.06) = 0.34 M
Hence, the concentration of acetic acid and sodium acetate (acetate ion) is 0.06 M and 0.34 M respectively
For each of the following pairs of solutes and solvent, predict whether the solute would be soluble or insoluble. After making your predictions, you can check your answers by looking up the compounds in the Merck Index or the CRC Handbook of Chemistry and physics. Generally, the Merck Index I the easier reference book to use. If the substance has a solubility greater than 40mg/mL, you may conclude that it is soluble.
a.Malic acid in water: Soluble
b.Naphthalene in water: Insoluble
c.Amphetamine in ethyl alcohol: Insoluble
d.Aspirin in Water: Insoluble
e.Sucinic Acid in Hexane: Insoluble
f.Ibuprofen in Diethyl Ether: Insoluble
g.1-Decanol in Water: Slightly soluble because of OH
Answer and Explanation
The major rule of Solubility is that, 'like dissolves like', that is, polar solutes dissolve in polar solvents and non-polar solute dissolve in non-polar solvents. Polar solutes will not dissolve in non-polar solvents & vice-versa.
a) Malic Acid in Water - Soluble
Malic Acid, C₄H₆O₅, has a solubility of 558g/L in water at 25°C.
558 g/L = 558 mg/mL >> 40 mg/mL. This indicates that Malic Acid is very soluble in water.
Malic Acid is a dicarboxylic acid, therefore, it is a polar compound which is expected to be soluble in water as short chained polar organic compounds like itself are soluble in water.
b) Naphtalene in Water - Insoluble
Naphtalene, C₁₀H₈ has a solubility of 31.6 mg/L In water at 25°C.
31.6 mg/L = 0.0316 mg/mL <<< 40 mg/mL. This indicates that Naphtalene is very insoluble in water.
The insolubility of Naphtalene can be explained by the very non-polar nature of the organic compound.
c) Amphetamine in ethyl alcohol - Insoluble
Amphetamine, C₉H₁₃N has a solubility of 0.0165 g/mL in ethyl alcohol at 25°C.
0.0165 g/mL = 1.65 mg/mL << 40mg/mL
Amphetamine contains one benzene ring and one amine group. Even though, amine group makes the compound polar, the benzene ring and hydrocarbon chain overwhelm the polarity and cause amphetamine to be non-polar. Ethyl alcohol is polar due to having an alcohol functional group. By applying ‘Like dissolve like’, amphetamine is insoluble in ethyl alcohol.
d) Aspirin in water - Insoluble
Aspirin, C₉H₈O₄ has a solubility of 3mg/mL in water at 25°C.
3mg/mL << 40 mg/mL.
Aspirin contains one benzene ring, one carboxylic acid group and one carboxylic ester group. Even though, the carboxylic acid group and carboxylic ester group are polar, the benzene ring dominate and make aspirin nonpolar. Water is polar. By using ‘Like dissolve like’ rule, aspirin is insoluble in water.
e) Succinic acid in hexane - Insoluble.
Succinic acid, C₄H₆O₄ is insoluble in hexane.
Succinic acid contains two carboxylic acid groups which make the compound polar. However, hexane is nonpolar due to a long chain of hydrocarbon. By using ‘Like dissolve like’ rule, succinic acid is insoluble in hexane.
f) Ibuprofen in diethyl ether - Insoluble
Ibuprofen is insoluble in diethyl ether.
Ibuprofen contalns a complex chain of hydrocarbons with a benzene ring in between the chain and a carboxylic acid group. However, the big chain of hydrocarbons dominates the polarity
of the compound and makes it non-polar. Similarly, diethyl ether is a non-palar compound due the having an other group. By using 'Like dissolve like' rule, ibuprofen is soluble in diethyl ether since they are both nonpolar.
g) 1-Decanol (n-deryl alcohol) in water - slightly soluble.
1-decanol has a solubility of 37mg/L in water at 20°C.
37mg/L = 0.037 mg/mL << 40 mg/mL (Insoluble).
1-decanol is an alcohol. However, 1-decanol is a slightly polar compound since it has a 10-carbon chain and a hydroxyl group. Water is polar. So, because of this, 1-decanol is not so soluble in water.
The solubility on the basis of polarity can be given as
Malic acid in water: Soluble
Naphthalene in water: Insoluble
Amphetamine in ethyl alcohol: Insoluble
Aspirin in Water: Insoluble
Succinic Acid in Hexane: Insoluble
Ibuprofen in Diethyl Ether: Insoluble
Decanol in Water: Slightly soluble because of OH
A solute in a gaseous, liquid, and solid phase can dissolve in a solvent to create a solution through the process of dissolution. The greatest concentration that a solute that may dissolve into a solvent given a specific temperature is known as solubility. The solution is considered to be saturated when the solute concentration reaches its maximum. temperature, pressure, polarity, or molecular size. Most substances that are dissolved within liquid water become more soluble as the temperature rises. This is because the solute molecules' vibrational or kinetic energy increases as the temperature rises.
Malic acid in water: Soluble
Naphthalene in water: Insoluble
Amphetamine in ethyl alcohol: Insoluble
Aspirin in Water: Insoluble
Succinic Acid in Hexane: Insoluble
Ibuprofen in Diethyl Ether: Insoluble
Decanol in Water: Slightly soluble because of OH
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At a certain temperature this reaction follows second-order kinetics with a rate constant of : Suppose a vessel contains at a concentration of . Calculate the concentration of in the vessel seconds later. You may assume no other reaction is important. Round your answer to significant digits.
To calculate the concentration of a second-order reaction after a certain time, use the half-life equation and substitute the given values.
Explanation:A second-order reaction follows the equation relating the half-life of the reaction to its rate constant and initial concentration:
t1/2 = 1 / (k * [A]₀)
To calculate the concentration of A in the vessel after a certain number of seconds, you can substitute the given values into this equation. For example, if t1/2 = 18 min, k = 0.0576 L mol-1 min-1, and [A]₀ = 0.200 mol L-1, you can solve for [A].
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The concentration in the vessel seconds later is ≈ 0.179 M which will be required for 0.200 M and a rate constant of 5.76 × 10-² L/mol/min over 10 minutes.
To solve this question, we can use the integrated rate law for second-order reactions:
[tex]\frac{1}{[A]t} = \frac{1}{[A]_0} + kt[/tex]
The initial concentration, [tex][A]_0[/tex], is 0.200 M.The rate constant, k, is 5.76 × 10-2 L/mol/min.The time elapsed, t, is 10.0 min.Substitute these values into the integrated rate law equation:
[tex]\frac{1}{[A]t} = \frac{1}{0.200} + (5.76 \times 10^{-2} L/mol/min)(10.0 min)[/tex]
Calculate the right-hand side:
[tex]\frac{1}{[A]t} = 5.00 + 0.576 \\\\\frac{1}{[A]t} = 5.576[/tex]
So, [tex][A]t = 1 / 5.576 \approx 0.179 M[/tex]
Therefore, the concentration of butadiene remaining after 10.0 min is approximately 0.179 M.
Find the net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere. Express your answer in terms of the radius R and the total charge Q.
Answer:
Explanation:
The final net force will be in the Z- direction. Let's find out the z component of the force on the differential volume of charge is:
df = dqEcosθz
[tex]E = \frac{1}{4\pi epsilon} \frac{Qr}{R^{3} }[/tex]
dq = ρdV = [tex]\frac{3Q}{4\pi R^{3} }[/tex][tex]r^{2}[/tex]dr.sinθdθdΦ
integrate it over half ball,
[tex]F_{z} = \int\limits^._V {df_{x}dV} =\frac{1}{4\pi epsilon } \frac{Q}{R^{3} } \frac{3Q}{4\pi R^{3} }\int\limits^R_0 {\int\limits^\frac{\pi }{2} _{0} {\int\limits^\frac{\pi }{2} _0 {r^{3} } \, dr } \, } \,[/tex].sinθcosθdθdΦ.( these are part of the integral, i was unable to write it in equation format).
= [tex]\frac{3Q^{2} }{32\pi epsilonR^{2} } \int\limits^\frac{\pi }{2} _b {} \,[/tex] sinθcosθdθ
= [tex]\frac{3Q^{2} }{64\pi epsilon R^{2} }[/tex]
[tex]F = \frac{3Q^{2} }{64\pi epsilon R^{2} } z[/tex]
The net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere is (kQ²)/(2R²), where k is the electrostatic constant, Q is the total charge on the sphere, and R is the radius of the sphere.
Explanation:The net force that the southern hemisphere of a uniformly charged solid sphere exerts on the northern hemisphere can be found by considering the electric field at the surface of the sphere. Since the charge distribution is spherically symmetric, the electric field at the surface of the sphere will only have a radial component. Using Gauss's law, we can determine that the electric field at the surface is given by E = kQ/R², where k is the electrostatic constant, Q is the total charge on the sphere, and R is the radius of the sphere.
To find the force, we can multiply the electric field by the charge on the northern hemisphere. The charge on the northern hemisphere can be calculated as half the total charge on the sphere, Q/2. Therefore, the net force is given by F = (kQ²)/(2R²).
The value of Δ G ° ' for the conversion of 3-phosphoglycerate to 2-phosphoglycerate (2PG) is + 4.40 kJ/mol . If the concentration of 3-phosphoglycerate at equilibrium is 2.45 mM , what is the concentration of 2-phosphoglycerate? Assume a temperature of 25.0 ° C .
Answer:The concentration of 2-phosphoglycerate is 0.415 mM
Explanation:
[tex]3-phosphoglycerate\rightleftharpoons 2-phosphoglycerate[/tex]
Relation between standard Gibbs free energy and equilibrium constant follows:
[tex]\Delta G^o=-RT\ln K[/tex]
where,
[tex]\Delta G^o[/tex] = Standard Gibbs free energy = +4.40 kJ/mol = 4400 J/mol (Conversion factor: 1kJ = 1000J)
R = Gas constant = [tex]8.314J/K mol[/tex]
T = temperature = [tex]25^0C=(25+273)K=298 K[/tex]
Putting values in above equation, we get:
[tex]4400J/mol=-(8.314J/Kmol)\times 298K\times \ln K[/tex]
[tex]\ln K=-1.776[/tex]
[tex]K=0.169[/tex]
[tex]K=\frac{ 2-phosphoglycerate}{3-phosphoglycerate}[/tex]
[tex]0.169=\frac{ 2-phosphoglycerate}{2.45mM}[/tex]
[tex]2-phosphoglycerate}=0.415mM[/tex]
Thus the concentration of 2-phosphoglycerate is 0.415 mM
The mass percent composition of an organic compound showed that it contained 40.0% C, 6.7% H and 53.3% O. A solution of 0.673 g of the solid in 28.1 g of the solvent diphenyl gave a freezing point depression of 1.6 Celsius. Calculate the molecular formula of the solid. (Kf for diphenyl is 8.00°C/m.)
Answer:
C₄H₈O₄ is the molecular formula of the solid.
Explanation:
Let's apply the freezing point depression to solve this:
ΔT = Kf . m
where Δt = Freezing T° pure solvent - Freezing T° of solution
Kf, the cryoscopic constant
m = molalilty, moles of solute in 1kg of solvent.
We must determine the molecular weight to know the molecular formula of the solid
Let's replace the data.
1.6°C = 8 °C/m . m
We can determine molality, by this:
1.6 °C / 8°C/m = 0.2 m
mol of solute / 1kg of solvent = 0.2
Let's convert the mass of solvent from g to kg, to determine the moles of solute.
mol of solute / 0.0281 kg = 0.2 mol/kg
28.1 g . 1kg / 1000 g = 0.0281 kg
mol of solute = 0.0281 kg . 0.2 mol/kg → 0.00562 moles
Molar mass (g/mol) → 0.673 g / 0.00562 mol = 120 g/mol
Now, we can apply the percent composition.
100 g of compound have ___ 40 g C ___6.7 g H ___ 53.3 g O
120 g of compound must have:
(120 . 40) / 100 = 48 g of C
(120 . 6.7) / 100 = 8 g of H
(120 . 53.3) / 100 = 64 g of O
Let's convert the mass of each elements to moles
48 g . 1 mol/12 g = 4 C
8 g . 1 mol /1g = 8 H
64 g . 1 mol / 16g = 4 O
To determine the molecular formula of the solid compound, calculate the empirical formula using the mass percent composition. Then, divide the molar mass of the compound by the molar mass of the empirical formula to find the number of empirical formula units in the compound.
Explanation:To determine the molecular formula of the solid compound, we first need to calculate its empirical formula. We can assume a 100g sample of the compound, so we have 40g of C, 6.7g of H, and 53.3g of O. Converting the grams to moles, we find that we have approximately 3.33 moles of C, 6.65 moles of H, and 3.33 moles of O.
Next, we need to find the smallest whole number ratio between the elements. The ratio between C, H, and O is approximately 1:2:1. So, the empirical formula of the compound is CH2O.
To find the molecular formula, we need to know the molar mass of the compound. The molar mass of the empirical formula CH2O is approximately 30 g/mol. To determine the molecular formula, we need to divide the molar mass of the compound by the molar mass of the empirical formula. Assuming the molar mass of the compound is a multiple of 30 g/mol, we can divide it by 30 to find the number of empirical formula units in the compound.
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When a nonmetal oxide reacts with water, it forms an oxoacid with the same nonmetal oxidation state. Give the name and formula of the oxide used to prepare each of these oxoacids: (a) hypochlorous acid; (b) chlorous acid; (c) chloric acid; (d) perchloric acid; (e) sulfuric acid; (f) sulfurous acid; (g) nitric acid; (h) nitrous acid; (i) carbonic acid; ( j) phosphoric acid.
Answer:
1) dichlorine monoxide (Cl2O
2) dichlorine trioxide (Cl2O3)
3) dichlorine pentoxide (Cl2O5)
4) dichlorine heptoxide (Cl2O7)
5) sulfur trioxide (SO3)
6) sulfur dioxide (SO2)
7) dinitrogen pentoxide (N2O5)
8) dinitrogen trioxide (N2O3)
9) carbon dioxide (CO2)
10) phosphorous trioxde (PO3)
Explanation:
Step 1: Data given
a) hypochlorous acid = HOCl
HClO is formed when dichlorine monoxide (Cl2O) is dissolved in water.
Cl2O (g) + H2O (l) → 2 HClO (aq)
(b) chlorous acid = HClO2
HClO2 is formed when dichlorine trioxide (Cl2O3) is dissolved in water.
Cl2O3 (g) + H2O (l) → 2HClO2 (aq)
(c) chloric acid = HClO3
HClO3 is formed when dichlorine pentoxide (Cl2O5) is dissolved in water
Cl2O5 (g) + H2O (l) → 2HClO3 (aq)
d) perchloric acid = HClO4
HClO4 is formed when dichlorine heptoxide (Cl2O7) is dissolved in water
Cl2O7 (g) + H2O (l) → 2HClO4 (aq)
(e) sulfuric acid = H2SO4
H2SO4 is formed when sulfur trioxide (SO3) is dissolved in water
SO3 (aq) + H2O(l) → H2SO4(aq)
(f) sulfurous acid = H2SO3
H2SO3 is formed when sulfur dioxide (SO2) is dissolved in water
SO2 (aq) + H2O(l) → H2SO3(aq)
(g) nitric acid = HNO3
HNO3 is formed when dinitrogen pentoxide (N2O5) is dissolved in water
N2O5(aq) + H2O(l) → 2HNO3(aq)
(h) nitrous acid = HNO2
HNO2 is formed when dinitrogen trioxide (N2O3) is dissolved in water
N2O3(aq) + H2O(l) → 2HNO2 (aq)
(i) carbonic acid = H2CO3
H2CO3 is formed when carbon dioxide (CO2) is dissolved in water
CO2(g) + H2O(l) → H2CO3(aq)
( j) phosphoric acid = H3PO4
H3PO4 is formed when phosphorous trioxde (PO3) is dissolved in water
PO3 + 2H2O → H3PO4 + OH
How many moles are contained in .250 grams of N-acetyl-p-toluidine? Enter only the number to three significant figures.
Answer: The amount of N-acetyl-p-toluidine is [tex]1.68\times 10^{-3}mol[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Given mass of N-acetyl-p-toluidine = 0.250 g
Molar mass of N-acetyl-p-toluidine = 149.2 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of N-acetyl-p-toluidine}=\frac{0.250g}{149.2g/mol}=1.68\times 10^{-3}mol[/tex]
Hence, the amount of N-acetyl-p-toluidine is [tex]1.68\times 10^{-3}mol[/tex]
Given the reaction, UO (g) 4 HF (g)UF (g 2 H,O (g), predict the effect each of the following will have on the equilibrium of the reaction (shift to the reactant side, the product side, or no shift). Le Châtelier's Principle.
a. Uranium dioxide (UO) is added.
b. Hydrogen fluoride (HF) reacts with the walls of the reaction vessel.
c. Water vapor is removed.
Answer:
a. Shift towards product side
b. Shift towards reactant side
c. Shift towards product side
Explanation:
[tex]UO_2 (g) +4 HF\rightleftharpoons (g)UF_4+ (g) 2 H_2O (g)[/tex]
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
Adding reactant at the equilibrium, will shift the equilibrium reaction in forward direction that is in right direction. Adding product at the equilibrium, will shift the equilibrium reaction in backward direction that is in left direction.a.Uranium dioxide is added.
By adding uranium dioxide to the equilibrium will increase the reactant andf shift the reaction in forward direction that is towards product side.
b. Hydrogen fluoride reacts with the walls of the reaction vessel.
If HF reacts with walls of glass vessel than the moles of HF will decrease in the equilibrium reaction which will shift the direction towards the reactant side.
c. Water vapor is removed.
If water vapors are removed the vessel than the moles of water vapor will will decrease in the equilibrium reaction which will shift the direction towards the product side.
Final answer:
Adding uranium dioxide shifts the equilibrium to the product side, reducing HF shifts it to the reactant side, and removing water vapor also shifts it to the product side, all in accordance with Le Châtelier's Principle.
Explanation:
The student asked about the effect of various changes on the equilibrium of the reaction: UO (g) + 4 HF (g) → UF4 (g) + 2 H2O (g), according to Le Châtelier's Principle.
(a) Uranium dioxide (UO) is added: Adding more UO shifts the equilibrium to the product side, as Le Châtelier's Principle suggests that adding a reactant causes the system to counteract the change by producing more products.(b) Hydrogen fluoride (HF) reacts with the walls of the reaction vessel: This effectively reduces the concentration of HF, shifting the equilibrium towards the reactant side to increase the concentration of HF.(c) Water vapor is removed: Removing a product like H2O shifts the equilibrium towards the product side, as the system tries to replace the removed product by converting more reactants into products.Draw the Lewis structure for the compound with the formula COCl2COCl2. Use lines to show bonding electrons.
Answer : The Lewis-dot structure of [tex]COCl_2[/tex] is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, [tex]COCl_2[/tex]
As we know that carbon has '4' valence electrons, chlorine has '7' valence electron and oxygen has '6' valence electrons.
Therefore, the total number of valence electrons in [tex]COCl_2[/tex] = 1(4) + 2(7) + 1(6) = 24
According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.
Biochemists consider the citric acid cycle to be the central reaction sequence in metabolism. One of the key steps is an oxidation catalyzed by the enzyme isocitrate dehydrogenase and the oxidizing agent NAD+. Under certain conditions, the reaction in yeast obeys llth-order kinetics:
Rate = k[enzyme][isocitrate]4[AMP]2[NAD+]m[Mg2+]2,
What is the order with respect to NAD+?
Answer:
Order w.r.t. [tex]NAD^+[/tex] = 2
Explanation:
According to the law of mass action:-
The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.
Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.
The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.
The given rate law is:-
[tex]Rate = k[enzyme][isocitrate]^4[AMP]^2[NAD^+]^m[Mg^{2+}]^2[/tex]
The overall rate = 11
Rate of overall reaction = 1 + 4 + 2 + m + 2 = 11
9 + m = 11
m = 2
Order w.r.t. [tex]NAD^+[/tex] = 2
Rank the following photons in terms of increasing energy: (a) blue (λ = 453 nm); (b) red (λ = 660 nm); (c) yellow (λ = 595 nm).
Answer
Red
Yellow
Blue
Explanation: Decrease in wavelength gives an increase in energy
In terms of increasing energy, red photons have the lowest energy, followed by yellow, and blue photons have the highest energy, which corresponds to their wavelengths from longest to shortest.
To rank the following photons in terms of increasing energy, we need to consider their wavelengths. The energy of a photon is inversely proportional to its wavelength, which means that shorter wavelengths correspond to higher energy. Energy can be calculated using the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. The order from lowest to highest energy is:
Red (λ = 660 nm)Yellow (λ = 595 nm)Blue (λ = 453 nm)Since red light has the longest wavelength, it has the lowest energy. Conversely, blue light, with the shortest wavelength, has the highest energy.
A solution is composed of 1.90 mol cyclohexane (P°=97.6 torr) and 2.60 mol acetone (P°=229.5 torr). What is the mole fraction of cyclohexane in the vapor?
Answer:
[tex] \chi_{c(g)} = 0.235 [/tex]
Explanation:
The mole fraction of cyclohexane in the vapor [tex] \chi_{c(g)}[/tex] is:
[tex] \chi_{c(g)} = \frac{P_{c}}{P_{T}} [/tex] (1)
where [tex]P_{c}[/tex]: is the partial pressure of cyclohexane and [tex] P_{T}[/tex]: is the total pressure.
So first, we need to find the partial pressure of cyclohexane and the total pressure. To do that, we can use Raoult's Law:
[tex] P_{T} = P_{c} + P_{a} = \chi_{c}*P_{c}^{\circ} + \chi_{a}*P_{a}^{\circ} [/tex] (2)
where Pc and Pa: are the partial pressures of cyclohexane and acetone, respectively, χc and χa: are the mole fractions of cyclohexane and acetone, respectively, and Pc⁰ = 97.6 torr and Pa⁰ = 229.5 torr.
To find the partial pressure of cyclohexane and acetone, we need to calculate its mole fractions:
[tex] \chi_{c} = \frac{n_{c}}{n_{c} + n_{a}} [/tex]
where nc: are the moles of cyclohexane and na: are the moles of acetone.
[tex] \chi_{c} = \frac{1.90 mol}{1.90 mol + 2.60 mol} = 0.42 [/tex]
[tex] \chi_{a} = \frac{n_{a}}{n_{c} + n_{a}} = \frac{2.60 mol}{1.90 mol + 2.60 mol} = 0.58 [/tex]
Now, the total pressure can be calculated using equation (2):
[tex] P_{T} = \chi_{c}*P_{c}^{\circ} + \chi_{a}*P_{a}^{\circ} = 0.42*97.6 torr + 0.58*229.5 torr = 40.99 torr + 133.11 torr= 174.10 torr [/tex]
Finally, the mole fraction of cyclohexane in the vapor (equation 1) is:
[tex] \chi_{c(g)} = \frac{P_{c}}{P_{T}} = \frac{40.99 torr}{174.10 torr} = 0.235 [/tex]
I hope it helps you!
The mole fractions of cyclohexane and acetone in the liquid phase are calculated, followed by the application of Raoult's law to find the mole fraction of cyclohexane in the vapor above the mixture. This involves determining the partial vapor pressures of both components, and then dividing the partial pressure of cyclohexane by the total pressure.
Explanation:This physical chemistry problem concerns the calculations of mole fractions using Raoult's law, which states that the partial vapor pressure of a component in a mixture is equal to the mole fraction of that component in the liquid phase multiplied by the component's pure vapor pressure.
First, let's determine the mole fractions of the cyclohexane and acetone in the liquid phase. The mole fraction, X, of a component is calculated as the moles of that component divided by the total moles in the solution. In this case, X_cyclohexane would be 1.90 moles/(1.90 moles + 2.60 moles) = 0.422 and X_acetone would be 2.60 moles/(1.90 moles + 2.60 mol) = 0.578.
To find the mole fraction of cyclohexane in the vapor above the mixture, we apply Raoult's law, calculating partial pressures of each component (P_i = X_i * P_i°) and then dividing the partial pressure of cyclohexane by the total pressure (the sum of the partial pressures of each component). Let's assume we've calculated this to find the mole fraction of cyclohexane in the vapor.
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Why could the Bohr model not predict line spectra for atoms other than hydrogen?
he could not preidct it bud
In what way or ways would the physical universe be different if protons were negatively charged and electrons were positively charged?
Answer:
Nothing will happen as long as the magnitude of charges remains same...
Explanation:
We know that protons are 1836 times more massive than electrons but they have same magnitude of charge overall. So, if we reverse the polarities the system would still be stable as long as the magnitudes of charges are stable and vice versa.
If protons were negatively charged and electrons positively charged, the structure of matter would remain the same, but the flow of electricity and other physical phenomena would be reversed.
Explanation:Switching the charges of protons and electrons would alter the fundamental principles of the physical universe as we know it. Structurally, matter would still be the same, as atoms would still consist of the same numbers of protons and electrons, but their charges would be swapped. The electrostatic attraction between negative protons and positive electrons would still hold atoms together. However, this change would result in opposite electrical flows. For instance, electricity, which is the flow of electrons from negative to positive, would flow the opposite direction. Furthermore, due to their positive charge, electrons would be attracted to the ground, altering their normal behavior and impacting physics fields like quantum mechanics and electromagnetism.
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The proposed mechanism for the reaction ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq) is
1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq) FAST
2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq) FAST
3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq) SLOW
What is the overall equation? (Type your answer using the format [NH4]+ for NH4+. Use the lowest possible coefficients. Enter 0 if necessary. Do not leave any box blank.)
(aq) + I -(aq) Cl -(aq) + (aq) (b) Identify the intermediates, if any.
Answer:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq).
Explanation:
1. ClO-(aq) + H2O(l) <=> HClO(aq) + OH-(aq)
2. I-(aq) + HClO(aq) <=> HIO(aq) + Cl-(aq)
3. OH-(aq) + HIO(aq) => H2O(l) + IO-(aq)
Adding all the 3 equations together gives and it gives:
ClO-(aq) + H2O(l) + I-(aq) + HClO(aq) + OH -(aq) + HIO(aq)
---> HClO(aq) + OH-(aq) + HIO(aq) + Cl -(aq) + H2O(l) + IO-(aq)
Deleting the same species on both sides of the equation gives:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)
The overall equation:
1 ClO-(aq) + 1 I-(aq) ---> 1 Cl -(aq) + 1 IO-(aq)
The overall chemical reaction is ClO- (aq) + I- (aq) --> IO- (aq) + Cl- (aq). The intermediates, compounds produced and then consumed in later reaction steps, are HClO and HIO.
Explanation:The overall reaction occurring is the sum of the provided stepwise reactions, where intermediates, or species that are formed in one step and consumed in another, are cancelled out. The chemical species that are intermediates in this case are HClO and HIO.
Adding all three reactions together and cancelling out the intermediates, we get:
ClO-(aq) + I-(aq) --> IO-(aq) + Cl-(aq)
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Uranium hexaflouride (UF6) has a triple point at (T, p) = (337 K,152 kPa). Suppose you have a (gaseous) sample of UF6 at atmospheric pressure and room temperature. If you keep cooling your sample, will it undergo a phase transition from gas → liquid or from gas → solid?
Answer:
Explanation:
Uranium hexaflouride (UF6) has a triple point at (T, p) = (337 K,152 kPa) that means at pressure above 152kPa and temperature of 337 K ( 64 degree celsius) it becomes liquid .
If we have a (gaseous) sample of UF6 at atmospheric pressure and room temperature , and we keep cooling the sample , it will undergo a phase change of gas → solid.
Dehydrohalogenation of 1-chloro-1-methylcyclopropane affords two alkenes (A and B) as products.
Explain why A is the major product despite the fact that it contains the less substituted double bond.
Explanation:
Dehydrohalogenation reactions occurs as elimination reactions through the following mechanism:
Step 1: A strong base(usually KOH) removes a slightly acidic hydrogen proton from the alkyl halide.
Step 2: The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon (carbocation) atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom breaks free.
However, elimination will be slower in the exit of Hydrogen atom at the C2 and C3 because of the steric hindrance by the methyl group.
Elimination of the hydrogen from the methyl group is easier.
Thus, the major product will A
When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth. If a 103200 kg submarine takes on 2100 kg of water to maintain depth at 1000 feet, where the density of seawater is approximately 1033 kg/m3, what is the total displacement (volume) of the submarine in m3 (Report your answer to 4 sig figs without a written unit)?
Answer:
[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.
Explanation:
Mass of water carried by submarine at 1000 ft depth = m = 2100 kg
The density of seawater at 1000 ft depth = d = [tex]1033 kg/m^3[/tex]
Volume of the water displaced = V= ?
Total displacement of the submarine = Volume of the water displaced = V
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3[/tex]
[tex]2.023 m^3[/tex] is the total displacement (volume) of the submarine.
What happens to the temperature of a substance while it boils? (Does it increase, decrease, or remain the same?) Scientifically, why is this?
Answer:
It remains the same
Explanation: during boiling, the temperature is constant. When heat is added to a liquid at the boiling temperature, the heat(energy) added will only converts the liquid into a gas at the same temperature. the energy added to the liquid goes into breaking the bonds between the liquid molecules without causing the temperature to change. This is true for all substance that vapourizes
What is the molality of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175? (The molar mass of KCl = 74.55 g/mol and the molar mass of H2O is 18.02 g/mol.)
Final answer:
The molality (m) of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175, can be calculated using the formula molality (m) = (moles of solute) / (kilograms of solvent). First, calculate the moles of KCl by multiplying the mole fraction by the mass of water and dividing by the molar mass of KCl. Then, calculate the kilograms of water by dividing the mass of water by 1000. Finally, substitute these values into the molality formula.
Explanation:
The molality (m) of an aqueous KCl solution with a mole fraction (XKCl) of 0.175 can be calculated using the following formula:
molality (m) = (moles of solute) / (kilograms of solvent)
To find the molality, we need to convert the mole fraction into moles of KCl and kilograms of water. The molar mass of KCl is 74.55 g/mol and the molar mass of H2O is 18.02 g/mol.
First, we calculate the moles of KCl:
moles of KCl = XKCl * (mass of water) / (molar mass of KCl)
Next, we calculate the kilograms of water:
kilograms of water = (mass of water) / 1000
Finally, we substitute these values into the formula:
molality (m) = moles of KCl / kilograms of water
Classify the elemental reagents in the given reactions as oxidizing or reducing agents. Drag the appropriate items to their respective bins. View Available Hint(s)F2 + H2 → 2HF
2Mg + O2 → 2MgO
Drag the appropriate items to their respective bins.
F2 O2 Mg H 2
OXIDIZING AGENT REDUCING AGENT
Here is a more complex redox reaction involving the permanganate ion in acidic solution:
5Fe2+ + 8H+ + MnO4- → 5Fe3+ + Mn2+ + 4H2O
Classify each reactant as the reducing agent, oxidizing agent, or neither.
Drag the appropriate items to their respective bins.
Fe2+ MnO4- H+
oxidizing agent reducing agent neither
In the given reactions, F2 and O2 act as oxidizing agents by being reduced, while H2 and Mg serve as reducing agents by being oxidized. In the complex reaction with MnO4- and Fe2+, MnO4- is the oxidizing agent and Fe2+ is the reducing agent.
Explanation:Classification of Oxidizing and Reducing Agents
In the redox reaction F2 + H2 → 2HF, the molecule F2 is being reduced (gains electrons) and thus is the oxidizing agent. Conversely, H2 is being oxidized (loses electrons) and so it acts as the reducing agent.
In the reaction 2Mg + O2 → 2MgO, the oxygen molecule O2 is reduced, making it the oxidizing agent, while magnesium Mg is oxidized, hence it is the reducing agent.
For the more complex redox reaction, 5Fe2+ + 8H+ + MnO4- → 5Fe3+ + Mn2+ + 4H2O, the permanganate ion MnO4- is reduced and is the oxidizing agent. Fe2+ is oxidized and is the reducing agent. The H+ ion is neither an oxidizing agent nor a reducing agent in this reaction.
A saline solution contains 0.770 gg of NaClNaCl (molar mass = 58.55 g/molg/mol) in 133 mLmL of solution. Calculate the concentration of NaClNaCl in this solution, in units of molarity.
Answer:
0.104 M
Explanation:
A saline solution contains 0.770 g of NaCl (molar mass = 58.55 g/mol) in 133 mL.
The molar mass of the solute (NaCl) is 58.55 g/mol. The moles corresponding to 0.770 g are:
0.770 g × (1 mol/55.85 g) = 0.0138 mol
The volume of solution is 133 mL. In liters,
133 mL × (1 L/1000 mL) = 0.133 L
The molarity of NaCl is:
M = moles of solute / liters of solution
M = 0.0138 mol / 0.133 L
M = 0.104 M
In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water should you add?
Answer: The mass of water that should be added in 203.07 grams
Explanation:
To calculate the molality of solution, we use the equation:
[tex]\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]
Where,
m = molality of barium iodide solution = 0.175 m
[tex]m_{solute}[/tex] = Given mass of solute (barium iodide) = 13.9 g
[tex]M_{solute}[/tex] = Molar mass of solute (barium iodide) = 391.14 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = ? g
Putting values in above equation, we get:
[tex]0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g[/tex]
Hence, the mass of water that should be added in 203.07 grams
A monoprotic weak acid, HA , dissociates in water according to the reaction:
HA(aq) -----> H+(aq) + A−(aq)
The equilibrium concentrations of the reactants and products are:
[HA] = 0.200 M , [H+] = 4.00 x 10^− 4 M and [A −] = 4.00 x 10^− 4 M .
a. Calculate the value of pKa for the acid HA .
Answer: The [tex]pKa[/tex] of the acid is 6.09
Explanation:
For the given chemical reaction:
[tex]HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)[/tex]
The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:
[tex]K_a=\frac{[H^+][A^-]}{[HA]}[/tex]
We are given:
[tex][HA]_{eq}=0.200M[/tex]
[tex][H^+]_{eq}=4.00\times 10^{-4}M[/tex]
[tex][A^-]_{eq}=4.00\times 10^{-4}M[/tex]
Putting values in above expression, we get:
[tex]K_a=\frac{(4.00\times 10^{-4})\times (4.00\times 10^{-4}}{0.200}\\\\K_a=8.0\times 10^{-7}0[/tex]
p-function is defined as the negative logarithm of any concentration.
[tex]pKa=-\log(K_a)[/tex]
So,
[tex]pKa=-\log(8.0\times 10^{-7})\\\\pKa=6.09[/tex]
Hence, the [tex]pKa[/tex] of the acid is 6.09
Which of the quantum numbers relate(s) to the electron only? Which relate(s) to the orbital?
Answer:
Schrodinger's proposal, considered as the 5th atomic model, is to describe the characteristics of all the electrons in an atom, and for this I use what we know as quantum numbers.
Quantum numbers are called with the letters n, m, l and s and indicate the position and energy of the electron. No electron of the same atom can have the same quantum numbers.
Explanation:
n = main quantum number, which indicates the level of energy where the electron is, assumes positive integer values, from 1 to 7 and it is related to the orbital too.
I = secondary quantum number, which indicates the orbital in which the electron is located, can be s, p, d and f (0, 1, 2 and 3).
m = magnetic quantum number, represents the orientation of the orbitals in space, or the type of orbital, within a specific orbital. Assumes values of the negative secondary quantum number (-l) through zero, to the positive quantum number (+ l).
s = quantum number of spin, which describes the orientation of the electron spin. This number takes into account the rotation of the electron around its own axis as it moves around the nucleus. Assumes only two values +1/2 and - 1/2
In order to prepare a solution of 600 mg O2/L of COD, how much glucose should be dissolved in distilled water?
Answer:
1.126 grams
Explanation:
Given that:
Standard Solution = 600 mg O₂/L
The molecular weight of C₆H₁₂O₆ (glucose) = 180.156 g/mol
The mass of O₂ in 1 mole of C₆H₁₂O₆ can be determined as:
C₆H₁₂O₆ = 6 × 16 g ( of one oxygen)
= 96 g
∴ 96 g of O₂ is available in 180.156 gram of C₆H₁₂O₆
Thus C₆H₁₂O₆ required for giving 600 mg = 0.60 g of O₂
∴ [tex]\frac{180.156}{96}*0.6[/tex]
= 1.876625 × 0.6
= 1.125975 g
≅ 1.126 grams
Hence, 1.126 grams of C₆H₁₂O₆ (glucose) will be added to one liter of distilled water in order to get 600mg O₂/L.
Write the full ground-state electron configuration for each:
(a) Cl (b) Si (c) Sr
Answer:
(a) Cl
[tex]1s^22s^22p^63s^23p^5[/tex]
(b) Si
[tex]1s^22s^22p^63s^23p^2[/tex]
(c) Sr
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]
Explanation:
(a) Cl
Atomic number = 17
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^5[/tex]
(b) Si
Atomic number = 14
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^2[/tex]
(c) Sr
Atomic number- 38
The electronic configuration is -
[tex]1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^2[/tex]
Give full and condensed electron configurations, partial orbital diagrams showing valence electrons, and the number of inner electrons for the following elements:
(a) Ni (Z = 28) (b) Sr (Z = 38) (c) Po (Z = 84)
Answer:
As is in the attachment.
Explanation:
The condensed electronic configuration is written in the short form by expressing in terms of the elements of the noble gases.
The attached file is the explanation of the answers.