Answer:Hydrogen is placed such because it exhibits some similar characteristics of both group1 and group VII elements.
Explanation:
The reason why hydrogen is similar to group 1 metals:
#It has same valence electron and inorder achieve octet state it can lose that electron and forms H+ ion
#It acts as a good reducing agent similar to group1 metals
#It can also halides
Similarity to halogens:
#hydrogen can also gain one electron to gain noble gas configuration. It can combine with other non metals to form molecules with covalent bonding.
#It exists as diatomin molecule,H2
#Have the same electronegativity nature
#its reaction with other metal
Calculate the force of attraction between a Ca2 and an O2- ion the centers of which are separated by a distance of 1.25 nm. Express your answer in N in engineering notation to three significant figures. Do not include the units in your answer.
Answer:
5.9 x 10^-9 N.
Explanation:
Below is an attachment containing the solution.
The force of attraction between a Ca2+ and an O2- ion can be calculated using Coulomb's Law. Coulomb's Law states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Explanation:The force of attraction between a Ca2+ and an O2- ion can be calculated using Coulomb's Law. Coulomb's Law states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation is given as:
[tex]F = (k * q1 * q2) / r^2[/tex]
Where F is the force of attraction, k is the proportionality constant[tex](2.31 × 10^16 J pm),[/tex] q1 and q2 are the charges of the ions, and r is the distance between their centers.
In this case, Ca2+ has a charge of +2 and O2- has a charge of -2. The distance between their centers is 1.25 nm, which is equivalent to 1250 pm. Plugging in the values:
[tex]F = (2.31 × 10^16 J pm * 2 * -2) / (1250 pm)^2[/tex]
Simplifying the equation:
[tex]F = -1.48 x 10^10 N[/tex]
Therefore, the force of attraction between the Ca2+ and O2- ions is[tex]-1.48 x 10^10 N (attractive force).[/tex]
Calculate the molality of a 9.2 M solution of perchloric acid given that the density of this solution is 1.54 g/mL.
Answer: The molality of perchloric acid in the solution is 14.95 m
Explanation:
We are given:
Molarity of perchloric acid solution = 9.2 M
This means that 9.2 moles of perchloric acid are contained in 1 L or 1000 mL of solution
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of perchloric acid = 9.2 moles
Molar mass of perchloric acid = 100.5 g/mol
Putting values in above equation, we get:
[tex]9.2mol=\frac{\text{Mass of perchloric acid}}{100.5g/mol}\\\\\text{Mass of perchloric acid}=(9.2mol\times 100.5g/mol)=924.6g[/tex]
To calculate mass of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.54 g/mL
Volume of solution = 1 L = 1000 mL
Putting values in above equation, we get:
[tex]1.54 g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.54g/mL\times 1000mL)=1540g[/tex]
We are given:
Mass of solute (perchloric acid) = 924.6 grams
Mass of solution = 1540 grams
Mass of solvent = Mass of solution - mass of solute = [1540 - 924.6] g = 615.4 g
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,
[tex]m_{solute}[/tex] = Given mass of solute (perchloric acid) = 924.6 g
[tex]M_{solute}[/tex] = Molar mass of solute (perchloric acid) = 100.5 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent = 615.4 g
Putting values in above equation, we get:
[tex]\text{Molality of perchloric acid}=\frac{924.6\times 1000}{100.5\times 615.4}\\\\\text{Molality of perchloric acid}=14.95m[/tex]
Hence, the molality of perchloric acid in the solution is 14.95 m
Considering the definition of molality, the molality of a 9.2 M solution of perchloric acid is 14.95 [tex]\frac{moles}{kg}[/tex].
Definition of molalityMolality is a measure of concentration that is defined as the ratio of the number of moles of any dissolved solute to kilograms of solvent.
The Molality of a solution is determined by the expression:
[tex]Molality=\frac{number of moles of solute}{kilograms of solvent}[/tex]
Molality is expressed in units [tex]\frac{moles}{kg}[/tex].
Number of moles of soluteBeing the molarity is the number of moles of solute that are dissolved in a certain volume a molarity of 9.2 M means that 9.2 moles of perchloric acid are contained in 1 L or 1000 mL of solution.
Then, the number of moles of solute is 9.2 moles of perchloric acid.
Kilograms of solventIn first place, density is the ratio of the weight (mass) of a substance to the volume it occupies. So, a density of 1.54 [tex]\frac{g}{mL}[/tex]means that you have 1.54 grams per 1 mL of solution or 1540 grams per 1 L (1000 mL) of solution. So, the mass of solution is 1540 grams in 1 L of solution
The mass of solution is the sum between the mass of solute and the mass of solvent.
Being the moles of perchloric acid 9.2 moles and the molar mass of perchloric acid 100.5 [tex]\frac{g}{mole}[/tex], the mass of the solute perchloric acid is calculated as:
mass of solute= number of moles of solute× molar mass
mass of solute= 9.2 moles× 100.5 [tex]\frac{g}{mole}[/tex]
mass of solute= 924.6 grams
Considering 1 L of solution, then, the mass of solvent is calculated as:
mass of solution= mass of solute +the mass of solvent
1540 grams = 924.6 grams +the mass of solvent
1540 grams - 924.6 grams= mass of solvent
615.4 grams= 0.6154 kg= mass of solvent
Finally, the mass of solvent is 0.6154 kg.
Molality of a 9.2 M solution of perchloric acidSo, being the number of moles of solute 9.2 moles and mass of solvent 0.6154 kg, the molality is calculated as:
[tex]Molality=\frac{9.2 moles}{0.6154 kg}[/tex]
Molality= 14.95 [tex]\frac{moles}{kg}[/tex]
Finally, the molality of a 9.2 M solution of perchloric acid is 14.95 [tex]\frac{moles}{kg}[/tex].
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Define shielding and effective nuclear charge. What is the connection between the two?
Answer:
The relation between the shielding and effective nuclear charge is given as
[tex]Z_{eff} = Z -S[/tex]
where s denote shielding
z_{eff} denote effective nuclear charge
Z - atomic number
Explanation:
shielding is referred to as the repulsion of an outermost electron to the pull of electron from valence shell. Higher the electron in valence shell higher will be the shielding effects.
Effective nuclear charge is the amount of net positive charge that valence electron has.
The relation between the shielding and the effective nuclear charge is given as
[tex]Z_{eff} = Z -S[/tex]
wheres denote shielding
z_{eff} denote effective nuclear charge
Z - atomic number
Use valence bond theory to describe the bonding in hydrogen sulfide (H2S). What are the predicted bond angles in hydrogen sulfide?
Answer:
The bond angle between the p orbital is assumed to be 90 degrees But according to VSEPR theory, as molecular geometry is a little bit bent thus the angle is less than 109.5 degrees.
Explanation:
IN hydrogen sulfide, H_2 S, both hydrogen atoms consist of para-magnetic 1s orbital while sulfur consists of diamagnetic 3s and 2 para magnetic 3p orbital.
The bond angle between the p orbital is assumed to be 90 degrees But according to VSEPR theory, as molecular geometry is a little bit bent thus the angle is less than 109.5 degrees.
In hydrogen sulfide (H2S), the bonding is explained using valence bond theory. The bond angle in H2S is approximately 92 degrees.
Explanation:In hydrogen sulfide (H2S), the bonding can be explained using valence bond theory. According to this theory, covalent bonds are formed by the overlap of atomic orbitals. In H2S, the sulfur atom forms a covalent bond with each hydrogen atom through overlap of its sp3 hybrid orbital and the hydrogen 1s orbital.
The predicted bond angle in H2S is approximately 92 degrees. This angle is less than the ideal tetrahedral angle of 109.5 degrees due to the presence of two lone pairs of electrons on the sulfur atom, which exert greater repulsion than the bonding pairs. This results in a distorted tetrahedral shape with a smaller bond angle.
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In a titration experiment, 31.4 mL of 1.120 M HCOOH is neutralized by 16.3 mL of Ba(OH)2. What is the concentration of the Ba(OH)2 solution?
Final answer:
To find the concentration of the Ba(OH)₂ solution that neutralizes HCOOH, calculate the moles of HCOOH used and apply these to the volume of Ba(OH)₂ solution. The concentration is found to be 2.157 M.
Explanation:
The question asks about calculating the concentration of a Ba(OH)₂ solution used to neutralize a known volume and concentration of HCOOH. The balanced chemical equation for the reaction between formic acid (HCOOH) and barium hydroxide (Ba(OH)₂) is:
HCOOH + Ba(OH)₂ → BaCO₃ + 2H₂O
First, calculate the moles of HCOOH used in the titration:
Moles of HCOOH = volume (L) × concentration (M) = 0.0314 L × 1.120 M = 0.03517 moles.According to the balanced equation, 1 mole of HCOOH reacts with 1 mole of Ba(OH)₂, so moles of Ba(OH)₂ = moles of HCOOH = 0.03517 moles.To find the concentration of Ba(OH)₂, use the formula: Concentration = moles/volume (L).Volume of Ba(OH)₂ solution = 16.3 mL = 0.0163 L.Concentration of Ba(OH)₂ = 0.03517 moles / 0.0163 L = 2.157 M.Therefore, the concentration of the Ba(OH)₂ solution is 2.157 M.
The concentration of the [tex]Ba(OH)\(_2\)[/tex]solution is approximately 1.078 M, when rounded to four decimal places.
To find the concentration of the [tex]Ba(OH)\(_2\)[/tex] solution, we need to use the concept of molarity and the stoichiometry of the neutralization reaction. The balanced chemical equation for the reaction between formic acid (HCOOH) and barium hydroxide [tex](Ba(OH)\(_2\))[/tex] is:
[tex]\[ 2HCOOH + Ba(OH)_2 \rightarrow Ba(HCOO)_2 + 2H_2O \][/tex]
From the equation, we see that 2 moles of HCOOH react with 1 mole of [tex]Ba(OH)\(_2\).[/tex]
First, we calculate the number of moles of HCOOH that reacted:
[tex][ \text{moles of HCOOH} = \text{volume of HCOOH} \times \text{concentration of HCOOH} \][/tex]
[tex]\[ \text{moles of HCOOH} = 31.4 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times 1.120 \text{ M} \][/tex]
[tex]\[ \text{moles of HCOOH} = 0.0314 \text{ L} \times 1.120 \text{ M} \][/tex]
[tex]\[ \text{moles of HCOOH} = 0.035128 \text{ moles} \][/tex]
Now, using the stoichiometry of the reaction (2 moles of HCOOH to 1 mole of [tex]Ba(OH)\(_2\)[/tex] we find the moles of [tex]Ba(OH)\(_2\)[/tex] that reacted:
[tex]\[ \text{moles of Ba(OH)}_2 = \frac{1}{2} \times \text{moles of HCOOH} \][/tex]
[tex]\[ \text{moles of Ba(OH)}_2 = \frac{1}{2} \times 0.035128 \text{ moles} \][/tex]
[tex]\[ \text{moles of Ba(OH)}_2 = 0.017564 \text{ moles} \][/tex]
Next, we calculate the concentration of [tex]Ba(OH)\(_2\)[/tex] using the definition of molarity:
[tex]\[ \text{concentration of Ba(OH)}_2 = \frac{\text{moles of Ba(OH)}_2}{\text{volume of Ba(OH)}_2 \text{ solution in liters}} \][/tex]
[tex]\[ \text{concentration of Ba(OH)}_2 = \frac{0.017564 \text{ moles}}{16.3 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}}} \][/tex]
[tex]\[ \text{concentration of Ba(OH)}_2 = \frac{0.017564 \text{ moles}}{0.0163 \text{ L}} \][/tex]
[tex]\[ \text{concentration of Ba(OH)}_2 = 1.0775 \text{ M} \][/tex]
Therefore, the concentration of the [tex]Ba(OH)\(_2\)[/tex]solution is approximately 1.078 M, when rounded to four decimal places.
The correct answer is: [tex]\[ \boxed{1.078 \text{ M}} \][/tex]
(a) which of these gases would you expect to have the largest van der waals constant a? H2, HF, F2******** (b) which of these gases would you expect to have the largest van der waals constant b? H2, HCL, CL2
In the given gases, HF should have the largest van der Waals constant a because of its stronger intermolecular forces, and Cl2 should have the largest van der Waals constant b due to its larger molecular size.
Explanation:The van der Waals constants (a) and (b) are indicative of the strength of the forces between molecules, and the physical size of the molecules in a given gas, respectively.
(a) In the selection of H2, HF, F2, we would expect the molecule with the strongest intermolecular forces to have the largest van der Waals constant a. That would be HF because when comparing these gases, HF has permanent dipole-dipole interaction which is stronger than the London dispersion forces in H2 and F2.
(b) For the gases H2, HCl, Cl2, we would expect the molecule with the largest physical size to have the largest van der Waals constant b. In this case, Cl2 is the largest molecule and thus would have the largest van der Waals constant b under normal conditions.
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In terms of the molecular properties of water, why is the temperature in San Antonio hotter in the summer then in Galveston?
Explanation:
San Antonio's climate is more humid than Galveston. More water in the air has more specific heat
capacity i.e. humid air retains more heat at energy. So San Antonio is warmer than Galveston in summer.
San Antonio has far more heat hot springs that elevate the temperature, San Antonio's
humidity is higher, water in the air absorbs heat and makes people feel hotter.
Final answer:
Water has higher specific heat than other materials, which leads to higher summer temperatures in San Antonio compared to Galveston due to the molecular properties of water.
Explanation:
Water has higher specific heat than other materials, which means it absorbs or releases more heat for the same change in temperature. The temperature in San Antonio is hotter in the summer than in Galveston because of the molecular properties of water. The bodies of water surrounding San Antonio, such as rivers or lakes, have a larger heat capacity compared to the ocean surrounding Galveston. This means that it takes more energy to raise the temperature of the bodies of water in San Antonio, resulting in higher summer temperatures.
When measuring water using the graduated pipette, the mass of beaker and the water is 23.670 g. If the empty beaker weighs 13.712 g. What is the volume of water measured assuming a density of 0.9982071 g/mL.
Answer:
The volume of water measured is 10mL
Explanation:
Given;
Mass of mass of beaker and the water = 23.670 g
Mass of empty beaker = 13.712 g
Then, mass of water only = Total mass of of beaker and the water minus Mass of empty beaker
mass of water only = 23.670 g - 13.712 g = 9.958 g
Density = mass/volume
Given density of water = 0.9982071 g/mL
Density of water = Mass of water/ Volume of water
Then, Volume of water = Mass of water/Density of water
Volume of water = 9.958 g/0.9982071 g/mL
Volume of water = 9.975886 mL ≅ 10mL
Therefore, The volume of water measured is 10mL
CH3CH2OH(l) 3O2(g) Classify each chemical reaction: reaction type of reaction (check all that apply) combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition combination precipitation single replacement combustion double replacement acid-base decomposition
Answer:
Reaction I is a COMBINATION REACTION - A reaction that involves the mixing of two or more elements to form a single product.
Reaction II is COMBUSTION REACTION - A reaction that involves Oxygen to produce carbon(iv)oxide and water vapor
Reaction III is DOUBLE DISPLACEMENT REACTION - A reaction that involves the exchange of radicals.
Reaction IV is a COMBUSTION REACTION
Explanation:
Reaction I is a COMBINATION REACTION - A reaction that involves the mixing of two or more elements to form a single product.
Reaction II is COMBUSTION REACTION - A reaction that involves Oxygen to produce carbon(iv)oxide and water vapor
Reaction III is DOUBLE DISPLACEMENT REACTION - A reaction that involves the exchange of radicals.
Reaction IV is a COMBUSTION REACTION
Attached is the reactions I - 1V
The given chemical equation CH3CH2OH(l) + 3O2(g) represents a combustion reaction where CH3CH2OH reacts with oxygen to produce carbon dioxide and water.
Explanation:Based on the given chemical equation, CH3CH2OH(l) + 3O2(g), the reaction is a combustion reaction. Combustion reactions involve the rapid combination of a fuel (in this case, CH3CH2OH) with oxygen (O2) to produce heat, light, and new products. In a combustion reaction, a fuel is oxidized and reacts with oxygen to form carbon dioxide and water.
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Sam records the mass of his evaporating dish as 9.687 g. He records the mass of the evaporating dish and the sample of hydrate as 18.407 g. After heating the sample in the evaporating dish to constant weight, the mass of them combined is 14.007 g. How many moles of water were removed from the sample by the heating process?
Answer:
0.2544 moles of water were removed from the sample by the heating process.
Explanation:
Mass of empty evaporating dish = 9.687 g
Mass of evaporating dish and sample of hydrate = 18.407 g
Mass of hydrate = 18.407 g - 9.687 g = 8.72 g
Mass of the evaporating dish and hydrate after heating = 14.007 g
14.007 g = mass of dish + mass of dehydrate
Mass of dehydrate = 14.007 g - 9.867 g = 4.14 g
Mass of water evaporated = hydrated sample - dehydrated sample
= 8.72 g - 4.14 g = 4.58 g
Moles of water evaporated :
[tex]\frac{4.58 g}{18 g/mol}=0.2544 mol[/tex]
0.2544 moles of water were removed from the sample by the heating process.
The mass of the water removed from the sample by the heating process is determined by subtracting the mass of the evaporating dish and sample after heating from before heating. This mass is then divided by the molar mass of water (18.015 g/mol) to calculate the number of moles of water removed.
Explanation:To calculate the moles of water removed from the sample by the heating process, first we have to determine the mass of the water removed. This would be equal to the mass of the evaporating dish and sample of hydrate before heating minus the mass after heating, i.e., 18.407 g - 14.007 g = 4.4 g.
Then, we calculate the number of moles, using the molar mass of water as a conversion factor from mass to moles. The molar mass of water is 18.015 g/mol. Therefore, the number of moles of water removed would be 4.4 g / 18.015 g/mol = 0.244 mol.
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The osmotic pressure of a solution containing 7.0 g of a sugar per liter is 23 torr at 25 °C. Assuming ideal solution behavior, what is the molar mass of this sugar?
Answer:
5650 g/mol
Explanation:
The osmotic pressure (π) is the pressure needed to prevent that the osmose occur in a system. Osmose is the process that a solvent goes through a membrane where the solution is more concentrated.
This property can be calculated by:
π = M*R*T
Where M is the molarity of the solution (mol/L), R is the ideal gas gas constant (62.3 torr.L/mol.K), and T is the temperature (25°C = 298 K), so:
23 = M*62.3*298
M = 1.24x10⁻³ mol/L
So, if the concentration is mass is 7.0 g/L, the molar mass (MM) of the sugar is its concentration is mass divided by the molarity:
MM = 7/1.24x10⁻³
MM = 5650 g/mol
5650 g/mole is the molar mass of the given sugar, whose osmotic pressure of solution is is the molar mass of this sugar.
What is osmotic pressure?It is that pressure which is applied on the solution to stop the flow of pure solvent from low concentration to high concentration through semipermeable membrane, and it is calculated as follow:
π = CRT, where
π = osmotic pressure = 23 torr (given)
C= concentration = to find?
R = ideal gas gas constant = 62.3 torr.L/mol.K
T = temperature = 25 degree C = 298 K
Putting all these value in the above equation, we get
23 = C × 62.3 × 298
C = 1.24x10⁻³ mol/L
Here the concentration is given in the form of molarity, which defines as no. of moles of solute present in per liter of solvent. And no. of moles is calculated as :
n = W/M, where
W = given mass = 7.0 g (given)
M = molar mass = to find?
n = no. of moles = 1.24x10⁻³ mol/L (calculated above)
Putting all values in above equation we get,
M = 7/1.24x10⁻³ = 5650 g/mol.
Hence, 5650 g/mol is the molar mass of this given sugar.
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is taking place in the gas phase in an ideal, isothermal PFR at constant total pressure. The reaction temperature and pressure are 400 K and 1 atm, respectively. At these conditions, the gas mixture obeys the ideal gas law. The rate equation isAAAACKkCr+=−1The volume of the reactor is 1000 L, and the molar flowrate of A is 500 mol/h. The mole fraction of A in the feed stream is 0.50. At 400 K, k = 45 h-1 and KA = 50 L/mol-A. What is the fractional conversion of A in the stream leaving the PFR?
Answer:
Explanation:
PFR - Plug Flow Rate
The steps and detailed integration and appropriate substitution is as shown in the attached file.
The fractional conversion of chemical A in the PFR under the given conditions is approximately 1, indicating that the conversion is almost complete.
Explanation:The question pertains to the calculation of the fractional conversion of a reactant, A, in a Plug Flow Reactor (PFR). According to the given conditions, we know that the rate equation is -rA = kCa and the value of k is 45 h⁻¹. Given that the volume of the reactor is 1000 L and the molar flowrate of A is 500 mol/h, we can calculate the residence time τ as V/F = 1000 L / 500 mol/h = 2h.
For a first order irreversible reaction in a PFR, the conversion, X, is given by X = 1 - exp(-k*τ). Plugging in the given value of k and the calculated value of τ, we find that X = 1 - exp(-45 h⁻¹ * 2 h) which gives X approximately equal to 1. Hence, the fractional conversion of A is almost complete.
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Question 20 It takes 614./kJmol to break a carbon-carbon double bond. Calculate the maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon. Be sure your answer has the c
The maximum wavelength of light that can break a carbon-carbon double bond by absorbing a single photon is calculated to be 1940 nm. This is done by converting the energy required to break the bond to J/particle, and then using this to find the wavelength using the equation E=h*c/λ.
Explanation:To find the maximum wavelength of light that can break a carbon-carbon double bond by absorbing a single photon, it is necessary to convert the energy required to break the bond from kJ/mol to energy per photon and then use that to calculate the wavelength. Using the relation between energy and wavelength given by the formula E=h×c/λ, where E is Energy, h is Planck's constant (6.626 x 10⁻³⁴ Js), c is the speed of light and λ is the wavelength.
First, convert the energy required to break the bond to J/particle by converting kJ to J (1kJ = 1000J) and then dividing by Avogadro's number (6.022 x 10²³). Thus, E = 614.4 kJ/mol × 1000 J/kJ / 6.022 x 10²³ particles/mol = 1.02 x 10⁻¹⁹ J/particle.
Then, rearrange the formula to solve for λ. We get λ = h × c/E. Substituting values (h = 6.626 x 10⁻³⁴ Js, c = 3.00 x 10⁸ m/s, and E = 1.02 x 10⁻¹⁹ J) gives λ = 1.94 x 10⁻⁶ meters or 1940 nm.
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The maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon is 341 nm.
Explanation:To calculate the maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon, we need to use the formula E = hc/λ, where E is the energy of the photons, h is Planck's constant (6.63 x 10^-34 J·s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength of light in meters.
First, we need to convert the bond energy from kJ/mol to J/molecule by multiplying it by Avogadro's number (6.02 x 10^23). So, the bond energy is 614 x 10^3 J/mol. Next, we can rearrange the formula to solve for λ:
λ = hc/E = (6.63 x 10^-34 J·s)(3.0 x 10^8 m/s)/(614 x 10^3 J/mol)
Calculating this expression gives us a value of λ = 3.41 x 10^-7 m, or 341 nm. Therefore, the maximum wavelength of light for which a carbon-carbon double bond could be broken by absorbing a single photon is 341 nm.
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The human eye is a complex sensing device for visible light. The optic nerve needs a minimum of 2.0 x 10⁻¹⁷ J of energy to trigger a series of impulses that eventually reach the brain.
(a) How many photons of red light (700. nm) are needed?
(b) How many photons of blue light (475 nm)?
Answer:
No. of Photons (red) = 7.05 No. of Photons
No. of Photons (blue) = 4.78 No. of Photons
Explanation:
v (red) = 3 x 108/700 x 10-9 = 4.28 x 10 power 15 Hz
v (blue) = 3 x 108/475 x 10-9 = 6.31 x 10 power 15 Hz
(a) No. of Photons (red) = E/hv = 2.0 x 10 power -17 / 6.626 x 10 power -34 x 4.28 x 10 power 15
No. of Photons (red) = 7.05 No. of Photons
(b) No. of Photons (blue) = E/hv = 2.0 x 10 power -17 / 6.626 x 10 power -34 x 6.31 x 10 power 15
No. of Photons (blue) = 4.78 No. of Photons
An aqueous solution is 36% by mass silver nitrate, AgNO3, and has a density of 1.41 g/mL. What is the molality of silver nitrate in the solution?
To find the molality of the silver nitrate solution, we can use the mass percent and density of the solution. The molality is found to be 2.99 mol/kg.
Explanation:To find the molality of the silver nitrate solution, we need to first determine the moles of silver nitrate in the solution. We can use the mass percent and density to do this.
First, let's assume we have 100 g of the solution.
This means we have 36 g of silver nitrate in the solution (since it is 36% by mass).
The molar mass of AgNO3 is 169.88 g/mol, so we can convert the mass of the silver nitrate to moles: 36 g / 169.88 g/mol = 0.2124 mol.
The density of the solution is given as 1.41 g/mL, so 100 g of the solution would have a volume of 100 g / 1.41 g/mL = 70.92 mL.
Now, we can calculate the molality using the moles of AgNO3 and the mass of the water:
Molality = moles of solute / mass of solvent (in kg)
Molality = 0.2124 mol / 0.07092 kg
= 2.99 mol/kg
Chemists studied the formation of phosgene by sealing 0.96 atm of carbon monoxide and 1.15 atm of chlorine in a reactor at a certain temperature. The pressure dropped smoothly to 1.47 atm as the system reached equilbrium. Calculate Kp (in atm-1) for CO(g) + Cl2(g) ↔ COCl2(g)
Answer:
3.9
Explanation:
Let's consider the following reaction at equilibrium.
CO(g) + Cl₂(g) ↔ COCl₂(g)
We can find the pressures at equilibrium using an ICE chart.
CO(g) + Cl₂(g) ↔ COCl₂(g)
I 0.96 1.15 0
C -x -x +x
E 0.96-x 1.15-x x
The sum of the partial pressures is equal to the total pressure.
pCO + pCl₂ + pCOCl₂ = 1.47
(0.96-x) + (1.15-x) + x = 1.47
2.11 - x = 1.47
x = 0.64
The pressures at equilibrium are:
pCO = 0.96 - x = 0.32 atm
pCl₂ = 1.15 - x = 0.51 atm
pCOCl₂ = x = 0.64 atm
The pressure equilibrium constant (Kp) is:
Kp = pCOCl₂ / pCO × pCl₂
Kp = 0.64 / 0.32 × 0.51
Kp = 3.9
Final answer:
The equilibrium constant, Kp, for the formation of phosgene from carbon monoxide and chlorine is calculated to be 2.94 atm⁻¹.
Explanation:
In this reaction, the formation of phosgene from carbon monoxide and chlorine is represented as:
CO(g) + Cl₂(g) → COCl₂(g)
The equilibrium constant, Kp, can be calculated using the changes in pressures of the reactants and products at equilibrium:
Change in pressure for CO: -0.46 atm
Change in pressure for Cl2: -0.15 atm
Pressure of CO at equilibrium: 0.50 atm
Pressure of Cl₂ at equilibrium: 1.00 atm
Therefore, the expression for Kp would be: Kp = (PCOCl₂) / (PCO × PCl₂) = (1.47) / (0.50 × 1.00) = 2.94 atm⁻¹
Hydrogenation reactions, in which H2 and an "unsaturated" organic compound combine, are used in the food, fuel, and polymer industries. In the simplest case, ethene (C2H4) and H2 form ethane (C2H6).
(a) If 138 kJ is given off per mole of C2H4 reacting, how much heat (in kJ) is released when 15.9 kg of C2H6 forms? Enter a positive number since released implies a negative number. Enter to 0 decimal places.
Answer:
73140 kJ.
Explanation:
Equation of the hydrogenation reaction:
C2H4 + H2 --> C2H6
Molar mass = (2*12) + (6*1)
= 30 g/mol
Mass of C2H4 = 15.9 kg
= 15900 g
Number of moles = mass/molar mass
= 15900/30
= 530 mol of C2H6
By stoichiometry, 1 mole of C2H4 is hydrogenated to give 1 mole of C2H6
Number of moles of C2H4 = 530 mole
Q = n * q
= 530 * 138 kJ
= 73140 kJ.
Final answer:
To calculate the total heat released when 15.9 kg of ethane (C₂H₆) is formed, we convert the mass of C₂H₆ to moles and multiply by the heat released per mole of ethene (C₂H₄). The total heat released is 73015 kJ.
Explanation:
The student is asking about the heat released during a hydrogenation reaction where ethene (C₂H₄) and hydrogen gas (H2) are converted into ethane (C₂H₆). The reaction is exothermic, meaning that heat is released when it occurs. Given that 138 kJ is released per mole of C₂H₄ reacting, to find the total heat released when 15.9 kg of C₂H₆ is formed, we need to follow these steps:
First, convert the mass of C₂H₆ to moles using its molar mass (30.07 g/mol for C₂H₆).Since we are starting with the product, C₂H₆, we calculate the moles of C₂H₄ that would have reacted by assuming a 1:1 molar ratio (since each mole of C₂H₆ formed comes from a mole of C₂H₄ reacting).Multiply the moles of C₂H₄ by the heat released per mole to find the total heat released.Here's the calculation:
Mass of C₂H₆ (15.9 kg) = 15,900 gMoles of C₂H₆ = 15,900 g / 30.07 g/mol = 528.95 molHeat released = 528.95 mol × 138 kJ/mol = 73015.1 kJTherefore, the total heat released when 15.9 kg of ethane is formed is 73015 kJ (to zero decimal places).
Draw the four structures of the compounds with molecular formula C5H10O that contain a carbon-carbon double bond, an unbranched carbon chain, and a hydroxyl group one carbon from the end of the chain.
Explanation:
Molecular formula of the compound =
The carbon chain in the molecule will look like :
1. [tex]H_2C=CH-CH_2-CH_2-CH_2-OH[/tex]
2. [tex]H_3C-CH=CH-CH_2-CH_2-OH[/tex]
3. [tex]H_3C-CH-CH=CH-CH_2-OH[/tex]
4. [tex]H_3C-CH_2-CH_2-CH=CH-OH[/tex]
No branching is present, so that means the valency of the carbon will be fulfilled by 10 hydrogen atoms and 1 oxygen atom
After doing an experiment, a chemist determines the Rf value of a compound to be 4. He also notes that the solvent travelled 4 cm on the plate. What can you conclude about this experiment
Answer:
We can conclude that the Rf of that compound has a ratio of 4. It means that the solute has a ratio value of 4 times than that of solvent. As we can see that it has traveled 4 cm , this data is useful in determination of the compound in a mixture when compared with Rf values of other compounds.
when 45.8 g of alanine (C3H7NO2) are dissolved in 1200. g of a certain mystery liquid X, the freezing point of the solution is 4.10 ℃ less than the freezing point of pure X. Calculate the mass of sodium chloride that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i-1.82 for sodium chloride in X. Be sure your answer has a unit symbol, if necessary, and is rounded to the correct number of significant digits x n?
Answer:
16.5 g
Explanation:
When a nonvolatile compound is dissolved in a pure solvent, the freezing point of the solvent is reduced, because the interaction solvent-solute requires more energy to be joined, and so, it freezes. This property is called cryoscopy, and the temperature change (ΔT) can be calculated by:
ΔT = Kc*W*i
Where Kc is the cryoscopy constant of the solute X, W is the molality of the solution, and i is the van't Hoff factor, which determines the percent of the solute that is dissolved. For organic molecules, such as alanine, i = 1.
The molality is the number of moles of the solute divided by the mass (in kg) of the solvent (1200 g = 1.2 kg). The molar mass of alanine is 89.09 g/mol, and the number of moles of it is the mass divided by the molar mass:
n = 45.8/89.09
n = 0.5141 mol
W = 0.5141/1.2 = 0.4284 mol/kg
So, Kc of X is:
4.10 = Kc*0.4284*1
Kc = 9.57 °C.kg/mol
So, if now sodium chloride is added to X, and the variation temperature is the same, and i = 1.82:
4.10 = 9.57*W*1.82
W = 0.2354 mol/kg
The number of moles of the solute is then:
W = n/1.2
0.2354 = n/1.2
n = 0.2825 mol
The molar mass of sodium chloride is 58.44 g/mol, thus the mass is the molar mass multiplied by the number of moles:
m = 58.44*0.2825
m = 16.5 g
Suppose you want to test the results of a transformation by growing Escherichia coli cells in LB medium containing ampicillin as the antibiotic for selection. Ampicillin at a concentration of 100 µg/mL will kill cells that do not contain an ampicillin resistance gene, but will allow the growth of cells that have been transformed with this gene. The concentrated stock of ampicillin is 100 mg/mL. How many microliters of the ampicillin stock should you add to 50 mL of LB for a bacterial culture?
Answer:
50.0 μL
Explanation:
When a dilution is done, the mass of the solute (in this case the ampicillin) remains constant, following the Lavoiser's law that the mass is conserved. The mass is the concentration (C) multiplied by the volume (V), so if 1 is the stock solution, and 2 is the bacterial culture after the addition of the antibiotic:
m1 = m2
C1*V1 = C2*V2
C1 = 100 mg/mL = 100000 μg/mL (1 mg = 1,000μg)
C2 = 100 μg/mL
V2 = 50 mL + V1 = 50000μL + V1 (V1 in μL)
100000*V1 = 100*(50000 + V1)
1000V1 = 50000 + V1
999V1 = 50000
V1 = 50.0 μL
How many orbitals in an atom can have each of the following designations: (a) 5f; (b) 4p; (c) 5d; (d) n = 2?
Answer: (a) seven orbitals, (b). 3 orbitals, (c). 5 orbitals and (d). 4 orbitals.
Explanation:
In order to solve this question we need to know how to explain the behaviour of electrons in atoms,and what we need to know is what is called the quantum numbers. There are four different kinds of quantum numbers and they are;
(1). Principal quantum numbers: the principal quantum number is denoted by the letter 'n'. It is used to describe the orbitals' energy. It has the values of n=1,2,3,4,...
(2). The spin quantum numbers: the spin quantum numbers is denoted by m(s). The 's' in the parenthesis is in subscript. It has the values of +1/2 and -1/2.
(3). Azimuthal quantum numbers: this is denoted by ℓ and it is used to explain orbital angular momentum and orbital shape. It has the values of ℓ= 0,1,2,3,....n-1.
Note that => ℓ = 0; we have a s-subshell,sphere shape.
ℓ = 1; p-subshell, dumb bell shape.
ℓ=2; d- subshell, double dumb bell shape.
ℓ= 3; f - subshell, multiple lobes.
(4). Magnetic quantum number: it is denoted by m(l) where the 'l' in the parenthesis is in subscript.
===> NOTE: there are (2ℓ + 1 ) orbitals in a subshell, also, there are n^2 number of orbitals in a shell.
Having known all that above, let us jump right in to the solution.
(a). From above we can see that; there are (2ℓ + 1 ) orbitals' in a subshell, also, f= ℓ= 3.
Therefore, the number of orbitals in 5f = 2 ℓ + 1 = (2×3) + 1 = 6+1 = 7 orbitals for 5f.
(b). 4p, the numbers of orbitals in 4p is; p= ℓ= 1=> 4 ℓ + 1 = (2×1) + 1 = 2+1 = 3 orbitals for 4p.
(c). 5d, the numbers of orbitals' in 5d is; d= ℓ= 2 = (2×2) + 1 = 4 + 1 = 5 orbitals for 5d.
(d). For n= 2, the numbers of orbital is ; n^2. Where the n given is 2. Therefore, 2^2= 2×2 = 4 orbitals in n=2.
An orbital refers to a region in space where electrons can be found.
An orbital refers to a region in space where there is a high probability of finding an electron. Orbitals that posses the same amount of energy are called degenerate orbitals.
The number of orbitals in an atom that can have the following designations are shown below;
5f - seven orbitals can have this designation because the f orbital is seven fold degenerate.4p - three orbitals can have this designation because the p orbital is three fold degenerate5d - five orbitals can have this designation because the d orbital is five fold degeneraten = 2 - the total number of orbitals in an energy level is given by n^2. Hence there are four orbitals that has the designation n =2Learn more: https://brainly.com/question/1527403
Write the 1 g equivalents using the following prefixes: (a) mega−, (b) kilo−, (c) deci−, (d) centi−, (e) milli−, (f) micro−, (g) nano−, (h) pico−
Final answer:
The 1 g equivalents using the given prefixes are: mega−: 1 Mg, kilo−: 1 kg, deci−: 1 dg, centi−: 1 cg, milli−: 1 mg, micro−: 1 μg, nano−: 1 ng, pico−: 1 pg.
Explanation:
The 1 g equivalents using the given prefixes are as follows:
(a) mega−: 1 Mg (megagram)
(b) kilo−: 1 kg (kilogram)
(c) deci−: 1 dg (decigram)
(d) centi−: 1 cg (centigram)
(e) milli−: 1 mg (milligram)
(f) micro−: 1 μg (microgram)
(g) nano−: 1 ng (nanogram)
(h) pico−: 1 pg (picogram)
1 message Which process results in the increase in entropy of the universe? the cooling of a hot cup of coffee in room temperature air the melting of snow above 0 °C all of the above the evaporation of water from a desk at room temperature
Answer:
Melting of snow
Evaporation of water from desk
Explanation:
Processes that increase the entropy of the universe are those processes that have an increased disorderliness. We should note that there are three principal states of matter which are the liquid, gas and solid. The gaseous state is the most disorderly while the solid is the least disorderly.
Now. We can see that the cooling of a hot cup of coffee is a process that needs or leads to a loss in temperature which obviously decreases disorderliness of the universe.
The melting of snow however is a process that leads to an increase in the disorderliness of the universe. It entails moving from the solid state to the liquid state. It tends to move to a more disordered state indicating an increase in the entropy of the universe.
The evaporation of water from the desk is quite similar to that above. Hence since we are moving from the liquid to the gaseous state via evaporation, we can state that the entropy of the universe has increased since we have moved from a state with a lesser degree of disorderliness to a state that is more disordered I.e from liquid to gaseous state.
The processes that result in the increase in entropy of the universe would be the melting of snow above 0 °C and the evaporation of water from a desk at room temperature.
Entropy refers to the degree of disorderliness of a system. The degree of disorderliness increases when the temperature of a system increases or when there is a phase change in the order solid > liquid > gas.
Thus, when snow melts, a phase change occurs from solid to liquid leading to an increase in entropy. Also. when water evaporates from a desk, it changes phase from liquid to gas and thereby leads to an increase in entropy.
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Choose the best option for the precursor to the tosylate intermediate Design a synthesis of cis-2-methylcyclopentyl acetate from trans-2-methylcyclopentanol.
Answer:
To synthesize cis-2-methylcyclopentyl from methycyclopentanol, you need to replace the acetate hydroxyl group with acetate by inverting the configuration.
Explanation:
To understand the process, you need to understand the nucleophilic mechanism taking place in the process. This is the first stage of the process. Hydroxide is a poor leaving group, to it must be converted to a good leaving group. To effect the change, it is necessary to use p-toluenesuphate.
p-toluenesuphate is favored because this can be prepared by a reaction that alters none of the bonds attached to the stereogenic center.
The reaction of p-toluensulfonate with potassium acetate in acetic acid effects the conversion to give the final product: cis-2-methylcyclopentyl.
Final answer:
The question involves synthesizing cis-2-methylcyclopentyl acetate from trans-2-methylcyclopentanol, requiring a reaction sequence that includes the use of a tosylate intermediate and control of the stereochemistry to retain the cis configuration.
Explanation:
The question asks how to synthesize cis-2-methylcyclopentyl acetate from trans-2-methylcyclopentanol. To achieve the transformation from trans to cis stereochemistry, one might need to carry out a series of reactions, including esterification and the use of a tosylate intermediate to facilitate inversion of stereocenter or through other stereospecific mechanisms. An appropriate synthetic route could potentially involve the protection of hydroxyl group, formation of the tosylate, followed by a nucleophilic substitution that inverts the stereochemistry, and concluding with the esterification to yield the desired acetate.
It is important to note that the 'tosylate intermediate' is a common intermediate in organic synthesis allowing for an alcohol to be converted into a better leaving group for subsequent substitution reactions. Stereoselectivity is crucial as the goal is to preserve the cis stereochemistry in the final product.
A 2.25 g sample of an unknown gas at 63 ∘ C and 1.10 atm is stored in a 1.15 L flask. What is the density of the gas? density: g / L What is the molar mass of the gas? molar mass:
Answer:
1. Density = 1.96g/L
2. Molar Mass of the gas = 42.61g/mol
Explanation:
1. Mass = 2.25g
Volume = 1.15L
Density = Mass /volume = 2.25/1.15 = 1.96g/L
2. V = 1.15L
P 1.10atm
R = 0.082atm.L/mol /K
T = 63°C = 63 + 273 = 336K
n =?
PV = nRT
n = PV/RT = ( 1.1 x 1.15)/(0.082 x 336)
n = 0.046mol
Mass = 1.96g
n = Mass /Molar Mass
Molar Mass = Mass / n = 1.96/0.046
Molar Mass = 42.61g/mol
75 mL of water is added to a 360 mL solution of acetic acid with a concentration of 0.87 M. Determine the molarity of the new solution.
Answer:
The molarity of the new solution is 0.72 M
Explanation:
Step 1: Data given
Volume of the original solution = 360 mL =.360 L
Molarity = 0.87 M
We add 75 mL = 0.075 L
Step 2: Calculate moles
Moles = molarity * volume
Moles = 0.87 M * 0.360 L
Moles = 0.3132 moles
Step 3: Calculate new molarity
The number of moles stays constant
Molarity = moles / volume
Molarity = 0.3132 moles / (0.36+0.075)
Molarity = 0.3132 moles / 0.435 L
Molarity = 0.72 M
The molarity of the new solution is 0.72 M
To determine the molarity of the new solution, calculate the amount of acetic acid remaining after dilution and then use the formula Molarity = Moles / Volume.
Explanation:To determine the molarity of the new solution, we need to calculate the amount of acetic acid that remains in the solution after adding water. The number of moles of acetic acid before dilution can be determined using the formula:
Moles = Concentration x Volume
Next, we need to calculate the final volume of the solution by adding the volume of water (75 mL) to the initial volume of the acetic acid solution (360 mL). Using this final volume and the number of moles of acetic acid, we can determine the molarity of the new solution using the formula:
Molarity = Moles / Volume
Substituting the values into the equation, the molarity can be calculated.
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A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.
0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.
If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?
A. less than 20∘C
B. 20∘C
C. greater than 20∘C
Answer:
Option C is correct
Explanation:
Increase in the mass of aluminium block would increase the heat capacity of block isothermaly before immersed in water at 0° so heat available for transfer is higher so equilibrium temperature of system would increase.
The thermite reaction, used for welding iron, is the reaction of Fe3O4 with Al. 8 Al (s) + 3 Fe3O4 (s) ⟶ 4 Al2O3 (s) + 9 Fe (s) Δ H° = -3350. kJ/mol rxn. Because this large amount of heat cannot be rapidly dissipated to the surroundings, the reacting mass may reach temperatures near 3000. °C. How much heat (in kJ) is released by the reaction of 19.3 g of Al with 63.2 g of Fe3O4?
The reaction of 19.3 g of aluminum with 63.2 g of ferric oxide is exothermic, releasing 2395.25 kJ of heat. Aluminum is the limiting reagent in the reaction.
The limiting reagent is the reactant that is used up completely in the reaction. To determine the limiting reagent, compare the ratio of the moles of each reactant to the stoichiometric coefficients in the balanced chemical equation. The reactant with the smaller ratio is the limiting reagent.
1: Calculate the number of moles of each reactant.
Moles of aluminum = 19.3 g / 27 g/mol = 0.715 mol
Moles of ferric oxide = 63.2 g / 231.5 g/mol = 0.272 mol
2: Determine the limiting reagent.
The ratio of moles of aluminum to moles of ferric oxide is 0.715 mol / 0.272 mol = 2.63. The stoichiometric ratio of aluminum to ferric oxide in the balanced chemical equation is 8:3. Since 2.63 is less than 8/3, aluminum is the limiting reagent.
3: Calculate the amount of heat released.
q = -3350 kJ/mol * 0.715 mol = -2395.25 kJ
4: Interpret the answer.
The amount of heat released is -2395.25 kJ. This means that the reaction releases 2395.25 kJ of heat.
The amount of heat released by the reaction of 19.3 g of Al with 63.2 g of Fe_3O_4 is -2395.25 kJ.
The orange color of carrots and orange peel is due mostly to β-carotene, an organic compound insoluble in water but soluble in benzene and chloroform. Describe an experiment to determine the concentration of β-carotene in the oil from orange peel.
Answer:
First Method: Vacuum Distillation and Chromatographic separation of the remains that were precipitated out from the peel.
Second Method: Extraction of components from orange peels by help of precipitation procedures that are mostly done In Situ. Those components can be recovered using the saponification process. Then these are examined under UV light spectroscopy. Now, the existence and extent of carotenoids can be determined by checking the levels of anti-oxidants.