Answer:
The difference in the magnetic orientation influences the thermal stability of the allotropes of iron.
Explanation:
It is known that the allotropes of iron exist in three phases: α - phase, β- phase, and γ-phase. However, two prominent structures are the α - phase and γ-phase. Now, let us look at the two phrases:
α - phase
This structure is a body-centered cube. It means that the unit cell structure resembles a cube. The lattice points are in the face of the cube. This subsequently affects the magnetic structure of the iron allotrope.
γ-phase
This allotrope has a lattice structure. It simply means that the structure has lattice points on the face of the cube. The structure generally affects the magnetic properties of the transitional metal; hence the stability of the γ-phase compared to α-phase.
Allotropes stable at high temperatures have higher enthalpies than those at low temperatures because they require more energy to maintain their structural bonds at these elevated temperatures. Hess's law and observations like the enthalpy differences in the thermite reaction further support this understanding in the context of energy changes and phases of matter.
Explanation:The enthalpies of allotropes that are stable at high temperatures are higher than those stable at low temperatures because more energy is required to maintain the structure and bonding in the high-temperature allotrope. For example, in the case of iron, γ-Fe (gamma iron) has a higher enthalpy than α-Fe (alpha iron). This is due to the difference in bonding and structure at different temperatures. As temperature increases, thermal energy overcomes stronger bonds, resulting in allotropes with higher enthalpies at these temperatures.
Hess's law can illustrate this concept further. Considering the thermite reaction, the heat produced during the reaction of aluminum with iron(III) oxide indicates an exothermic reaction that causes the iron to melt. In general, transformations like changing phases from solid to liquid require energy, and allotropes that must retain more complex, less stable structures at higher temperatures inherently have higher enthalpies.
Moreover, substances with high melting and boiling points usually have strong bonds and interactions to maintain those phases, which means their reactions typically involve greater changes in enthalpy. This is why the enthalpy of vaporization is much greater than the enthalpy of fusion, and the same principle can be applied to allotropes stable at different temperatures. Allotropes stable at higher temperatures have structures that require more energy to maintain, hence their higher enthalpies.
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Why do the deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer?
Answer:
The deprotonated organic acid (RCO2-) and protonated organic base (RNH3+) go into the aqueous layer because Organic compounds that are neither acids or bases do not react with either NaOH or HCl and, therefore remain more soluble in the organic solvent and are not extracted.
Explanation:
The density of platinum is 21.45 g/cm3. What is the volume of a platinum sample with a mass of 11.2 g?
Final answer:
To find the volume of a platinum sample with a mass of 11.2 g, divide the mass by the density of platinum, which is 21.45 g/cm3, resulting in a volume of approximately 0.5221 [tex]cm^3[/tex].
Explanation:
To determine the volume of a platinum sample with a mass of 11.2 g, given the density of platinum is 21.45 g/cm3, we can use the formula for density which is:
Density = Mass / Volume.
By rearranging the formula to solve for volume, we get:
Volume = Mass / Density.
Substituting the given values:
Volume = 11.2 g / 21.45 g/cm3 = 0.5221 [tex]cm^3[/tex].
Therefore, the volume of the platinum sample is approximately 0.5221 [tex]cm^3[/tex].
Make a plot of arsenate species vs. pH for a 25 mM arsenate solution. Vary pH from 0-14. Neglect activity corrections. A groundwater contaminated with arsenate has a pH range of 7.0 to 8.5. What is/are the dominant arsenate species in this range?
For main-group elements, are outer electron configurations similar or different within a group? Within a period? Explain.
Answer:
Same within the groups...
Different within the periods...
Explanation:
The outer configuration is same with a group because of their arrangement in the periodic table. If we see group 1a of alkalies, all of them have 1 electron in their outer shell and this order goes on increasing up-to noble gases that have 8 electrons in their outer shell each.
This arrangement is called periodicity of the periodic table and elements are arranged according to the periodic law.
Along a period, these properties of outer configuration goes on increasing than within the group.
A sodium atom has a mass number of 23. Its atomic number is 11. How many electrons does a neutral sodium atom have?
Answer:
it contains 11 electrons
Explanation:
Calculate the shortest wavelength of light capable of dissociating the Br–I bond in one molecule of iodine monobromide if the bond energy, or bond dissociation energy, is 179 kJ/mol .
Answer: 6.7*10^-7 m
Explanation:
The full explanation is shown in the image attached. The energy of the photon is obtained by dividing the bond energy by the Avogadro's number. Using the Plank's equation, we cash obtain the frequency or wavelength of radiation required by substituting into the given equation appropriately.
Final answer:
To calculate the shortest wavelength of light capable of dissociating the Br-I bond, the bond energy given (179 kJ/mol) is first converted to Joules per photon. Then, applying the equation E = hc / λ with Planck's constant and the speed of light gives the shortest wavelength needed to break the bond.
Explanation:
The question asks to calculate the shortest wavelength of light capable of dissociating the Br–I bond in iodine monobromide, given that the bond energy is 179 kJ/mol. To solve this, we apply the equation relating the energy of a photon (E) to its wavelength (λ), using Planck's constant (h) and the speed of light (c).
The energy of the photon needed can be calculated using the equation E = hc / λ, where h = 6.626 x 10-34 J·s (Planck's constant) and c = 3.00 x 108 m/s (speed of light). First, convert the bond dissociation energy from kJ/mol to Joules (J) by multiplying by 1000 and dividing by Avogadro's number (6.022 x 1023 mol-1), giving the energy required per molecule.
Finally, rearrange the equation to solve for λ: λ = hc / E. Plugging in the values, we find the shortest wavelength of light capable of breaking the Br–I bond. Remember, since the question gives the bond dissociation energy in kJ/mol, conversion to Joules is necessary for the calculation to proceed correctly.
Solid silver chloride, AgCI(s), decomposes to its elements when exposed to sunlight Write a balanced equation for this reaction with correct formulas for the products. What is the coefficient in front of AgCI(s) when the equation is properly balanced? 01 03 02 04
Answer: The coefficient in front of AgCl when the equation is properly balanced is 2.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.
Decomposition of silver chloride is represented as:
[tex]2AgCl(s)\rightarrow 2Ag(s)+Cl_2(g)[/tex]
Thus the coefficient in front of AgCl when the equation is properly balanced is 2.
Answer:
the coefficient in front of AgCl(s) when the equation is properly balanced is 2.
Explanation:
The decomposition of solid silver chloride (AgCl) into its elements when exposed to sunlight can be represented by the following balanced chemical equation:
2AgCl(s) → 2Ag(s) + Cl2(g)
In this balanced equation:
The coefficient in front of AgCl(s) is 2.
The coefficient in front of Ag(s) is 2.
The coefficient in front of Cl2(g) is also 2.
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A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction. Calculate the concentrations of H2SO4 and NaOH that remain in solution after the reaction is complete, and the concentration of the salt that is formed during the reaction.
Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
For NaOH:
Initial molarity of NaOH solution = 0.195 M
Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol[/tex]
For sulfuric acid:
Initial molarity of sulfuric acid solution = 0.248 M
Volume of solution = 20.0 mL = 0.020 L
Putting values in equation 1, we get:
[tex]0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol[/tex]
The chemical equation for the reaction of NaOH and sulfuric acid follows:
[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of sulfuric acid
So, [tex]2.925\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol[/tex] of sulfuric acid
As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, NaOH is considered as a limiting reagent because it limits the formation of product.
Excess moles of sulfuric acid = [tex](4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol[/tex]
By Stoichiometry of the reaction:
2 moles of KOH produces 1 mole of sodium sulfate
So, [tex]2.925\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol[/tex] of sodium sulfate
For sodium sulfate:Moles of sodium sulfate = [tex]1.462\times 10^{-3}moles[/tex]
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M[/tex]
For sulfuric acid:Moles of excess sulfuric acid = [tex]3.498\times 10^{-3}mol[/tex]
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M[/tex]
For NaOH:Moles of NaOH remained = 0 moles
Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L
Putting values in equation 1, we get:
[tex]\text{Molarity of NaOH}=\frac{0}{0.050}=0M[/tex]
Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.
After the reaction between solutions of H2SO4 and NaOH is complete, the remaining concentrations are 0.0317 M H2SO4, 0 M NaOH, and 0.0836 M Na2SO4 are formed.
Explanation:The balanced chemical equation for the reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is: H2SO4 + 2NaOH -> Na2SO4 + 2H2O. This equation shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide to produce one mole of sodium sulfate (Na2SO4) and two moles of water.
Using this stoichiometry, the moles of H2SO4 and NaOH in the solutions can be calculated with moles = concentration x volume (in L). That gives us 0.020 L * 0.248 mol/L = 0.00496 mol for H2SO4 and 0.015 L * 0.195 mol/L = 0.002925 mol for NaOH.
Since one mole of H2SO4 reacts with two moles of NaOH, all of the NaOH will react, and you will have 0.00496 mol - 2* 0.002925 mol = 0.00111 mol of H2SO4 left. The concentration of H2SO4 then is 0.00111 mol / 0.035 L (sum of volumes) = 0.0317 M.
NaOH is completely reacted, so its concentration is 0 M. From the reaction stoichiometry, 0.002925 mol of Na2SO4 is formed, so its concentration is 0.002925 mol / 0.035 L = 0.0836 M.
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The tarnish that forms on objects made of silver is solid silver sulfide; it can be removed by reacting it with aluminum metal to produce silver metal and solid aluminum sulfide. How many moles of the excess reactant remain unreacted when the reaction is over if 5 moles of silver sulfide react with 8 moles of aluminum metal? Hint: Write a balanced chemical equation first. Enter to 1 decimal place.
When 5 moles of silver sulfide react with 8 moles of aluminum metal, there is an excess of aluminum. After the reaction is complete, 3.3 moles of the excess aluminum remain unreacted.
Explanation:The balanced chemical equation for the reaction between solid silver sulfide (Ag2S) and aluminum metal (Al) is:
3Ag2S + 2Al → 6Ag + Al2S3
Based on this equation, we can see that every 3 moles of Ag2S react with 2 moles of Al to produce 6 moles of Ag and 1 mole of Al2S3.
In the given question, 5 moles of Ag2S react with 8 moles of Al. Therefore, we have an excess of Al. To determine the moles of excess Al remaining unreacted, we can set up a ratio:
(8 moles Al reacted) / (2 moles Al required to react with 3 moles Ag2S) = x moles Ag2S / 5 moles Ag2S
Simplifying this ratio, we find:
x = (8 moles Al / 2) × (5 moles Ag2S / 3 moles Al)
x = 20/6 = 3.3 moles
Therefore, 3.3 moles of the excess reactant (Al) remain unreacted when the reaction is over.
Final answer:
After writing a balanced chemical equation for the reaction between silver sulfide and aluminum, we determine that 4.7 moles of aluminum remain unreacted when 5 moles of silver sulfide react with 8 moles of aluminum.
Explanation:
To determine the number of moles of the excess reactant that remain unreacted, we first need to write a balanced chemical equation for the reaction between silver sulfide and aluminum metal. Here is the balanced equation:
3 Ag₂S (s) + 2 Al (s) → 6 Ag (s) + Al₂S₃(s)
Using the balanced equation, we see that 3 moles of silver sulfide react with 2 moles of aluminum. Therefore, if we had 5 moles of silver sulfide, we would need 2/3 × 5 = 10/3 moles of aluminum to react completely with the silver sulfide.
Since 8 moles of aluminum were originally present, we subtract the amount of aluminum that reacted to find the excess:
8 moles Al - 10/3 moles Al = 14/3 moles Al
Thus, 14/3 moles or 4.7 moles of aluminum remain unreacted.
A 7.73 mass % aqueous solution of ethylene glycol (HOCH2CH2OH) has a density of 1.19 g/mL. Calculate the molarity of the solution. Give your answer to 2 decimal places.
Answer:
Molarity for the solution is 1.48 M
Explanation:
Molarity involves the moles of solute that are contained in 1L of solution. It is a sort of concentration. The most usual.
7.73 by mass involves the mass of solute that are contained in 100 g of solution, so this one of ethylene glycol contains 7.73 g of it.
Let's determine the moles of solute
7.73 g / molar mass of ethylene glycol = moles
7.73 g / 62 g/mol = 0.124 mol
If the mass of solution is 100 g, we can determine the volume with density.
Density = Mass / Volume
Volume = Mass / Density
Volume = 100 g / 1.19 g/mL = 84.03 mL
In conclusion, 0.124 moles are contained in 84.03 mL
Molarity is mol/L, so let's convert the volume in L
84.03 mL . 1L / 1000 mL = 0.08403 L
0.124 mol / 0.08403 L = 1.48 M
How many electrons in an atom can have each of the following quantum number or sublevel designations?
(a) n = 2, l = 1
(b) 3d
(c) 4s
Answer: a:6 electrons, b:10electrons, c: 2electrons
Explanation:
1.Principle quantum number(n)
n is the number of shell, it is a positive integer. The maximum number of electrons in a shell is 2(n2)
2.Azimuthal quantum number(l)
l is the number of subshell in the principal shell
The name of subshell are s, p, d, f
The number of nodes in each subshell is
s is l=0, p is l=1, d is l=2, f is l=3.
Each shell can have 2 x l + 1 sublevels, and each sublevel can accommodate maximum of 2 electrons and have two electrons.
A. n=2, l=1 then 2 x 1 + 1= 3 subshell, 3*2= 6 electrons in the p subshell
b. 3d, d is l = 2 then 2 x 2 + 1 = 5 sublevels, 5*2=10 electrons in the d subshell
c. 4s, s is l=0 then 2 x 0 + 1 = 1 sublevel, 1*2=2 electrons in the s subshell.
Two classrooms, labeled A and B, are of equal volume and are connected by an open door. Classroom A is warmer than classroom B (maybe it has a sunny window). Derive a formula that relates the masses of air in each room MA and MB to the temperatures TA and TB. Which room contains a greater mass of air
Answer:
Room B contains the greater mass of air
Explanation:
According to the given data, the volume of both rooms is the same, so let's consider both volumes as V. An open door connects both the room, which infer that pressure for both the rooms is also the same, let it be P.
Considering the above conditions, the general gas equation for the air in both rooms can be given as:
[tex]PV=n_{A}RT_{A}\\ \\ PV=n_{B}RT_{B}[/tex]
Here, n represents the moles of air, R is the gas constant, and T is the temperature. Taking the ratio of both the above equations we get
[tex]\frac{PV}{PV}= \frac{n_{A}RT_{A}}{n_{B}RT_{B}}\\[/tex]
moles of the gas are mass per molecular mass, as the molecular mass is the same in both the cases so n can be replaced by m, as follows
[tex]\frac{PV}{PV}= \frac{m_{A}RT_{A}}{m_{B}RT_{B}}\\[/tex]
By simplifying we get,
[tex]\frac{m_{A}}{m_{B}}=\frac{T_{B}}{T_{A}}[/tex]
According to the given conditions, TA is greater than TB. So from the derived relation, it can be seen that the mass of air in room B is greater than the mass of air in room A.
Calculate the volume in milliliters of a sodium carbonate solution that contains of sodium carbonate . Be sure your answer has the correct number of significant digits.
Answer:
25.0 g of sodium carbonate are present in 220 ml of the solution.
Explanation:
Hi there!
I have found the complete question on the web:
Calculate the volume in milliliters of a 1.07 M sodium carbonate solution that contains 25.0g of sodium carbonate Na2CO3. Be sure your answer has the correct number of significant digits.
First, let's find how many moles of sodium carbonate have a mass of 25.0 g.
The molar mass of Na₂CO₃ is 106 g/mol.
So, if 106 g of sodium carbonate is equal to 1 mol, then 25.0 g will be:
25.0 g · (1 mol / 106 g) = 0.236 mol Na₂CO₃
The solution contains 1.07 mol sodium carbonate in 1 liter.
So, if 1.07 mol sodium carbonate is present in 1 l solution, then, 0.236 mol will be present in:
0.236 mol · (1000 ml / 1.07 mo) = 221 ml (220 ml without any intermediate rounding).
25.0 g of sodium carbonate are present in 220 ml of the solution.
Final answer:
To calculate the volume in milliliters of a sodium carbonate solution, you need to know the concentration of the solution.
Explanation:
To calculate the volume in milliliters of a sodium carbonate solution, we need to know the concentration of the solution. Without that information, we are unable to provide an accurate calculation. Please provide the concentration of the sodium carbonate solution in the question.
Sucralfate (molecular weight: 2087 g/mol) is a major component of Carafate®, which is prescribed for the treatment of gastrointestinal ulcers. The recommended dose for a duodenal ulcer is 1.0g taken orally in 10.0 mL (2 teaspoons) of a solution four times a day. Calculate the molarity of the solution.a. 0.0005 Mb. 0.048 Mc. 0.02087 Md. 0.010 M
Answer:
Option b. 0.048 M
Explanation:
We have the molecular weight and the mass, from sulcralfate.
Let's convert the mass in g, to moles
1 g . 1 mol / 2087 g = 4.79×10⁻⁴ moles.
Molarity is mol /L
Let's convert the volume of solution in L
10 mL . 1L/1000 mL = 0.01 L
4.79×10⁻⁴ mol / 0.01 L = 0.048 mol/L
Consider the following types of electromagnetic radiation:
(1) Microwave
(2) Ultraviolet
(3) Radio waves
(4) Infrared
(5) X-ray
(6) Visible
(a) Arrange them in order of increasing wavelength.
(b) Arrange them in order of increasing frequency.
(c) Arrange them in order of increasing energy.
Electromagnetic radiation can be arranged by wavelength, frequency, or energy. In increasing order, by wavelength, it is: X-ray, Ultraviolet, Visible, Infrared, Microwave, Radio waves. In increasing order, by frequency and energy, it is: Radio waves, Microwave, Infrared, Visible, Ultraviolet, X-ray.
Explanation:The types of electromagnetic radiation listed can be arranged by wavelength, frequency, and energy. The relationship among these three properties is that as wavelength increases, frequency and energy decrease, and vice versa.
In order of increasing wavelength, the arrangement is: X-ray, Ultraviolet, Visible, Infrared, Microwave, Radio waves.In order of increasing frequency, the arrangement is the inverse: Radio waves, Microwave, Infrared, Visible, Ultraviolet, X-ray. In order of increasing energy, the arrangement is the same as frequency, because energy and frequency are directly proportional: Radio waves, Microwave, Infrared, Visible, Ultraviolet, X-ray.Learn more about Electromagnetic Radiation here:https://brainly.com/question/2334706
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Whenever dry nitrogen from a portable cylinder is used in service and installation practice, what item is of most importance in consideration of safety?
Answer:
a relief valve is inserted in the downstream line from the pressure regulator.
Explanation:
Whenever dry nitrogen from a portable cylinder is used in service and installation practice the item of most importance in consideration of safety is a relief valve is inserted in the downstream line from the pressure regulator.
The most important item to consider for safety when using dry nitrogen from a portable cylinder in service and installation practice is a pressure regulator, which controls the flow and release of the gas. It is also crucial to inspect the system for leaks regularly.
Explanation:When using dry nitrogen from a portable cylinder in service and installation practice, the most important item to consider for safety is a pressure regulator. A pressure regulator is used to control the flow and release of nitrogen gas from the cylinder. It ensures that the pressure does not exceed safe limits and reduces the risk of an explosion or other accidents.
Additionally, it is crucial to always check for leaks in the system before using dry nitrogen. Leaks can result in the accumulation of nitrogen gas, which can displace oxygen in the air and lead to asphyxiation. Therefore, regularly inspecting the system and using appropriate leak detection methods, such as soapy water or a leak detection solution, is essential for safety.
In summary, when working with dry nitrogen from a portable cylinder, the two most important safety considerations are the use of a pressure regulator and regular inspection for leaks.
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A 1.85 mass % aqueous solution urea (CO(NH2)2)has a density of 1.05 g/mL. Calculate the molality of the solution. Give your answer to 2 decimal places.
Answer:
Molality for this solution is 0.31 m
Explanation:
Molality → Mol of solute / kg of solvent
Mass of solute = 1.85 g (CO(NH₂)₂
% by mass, means mass of solute in 100 g of solution
Mass of solution = Mass of solute + Mass of solvent
As 1.85 g is the mass of solute, the mass of solvent would be 98.15 g to complete the 100g of solution.
Let's convert the mass of solvent (g to kg)
1 g = 1×10⁻³ kg
98.2 g . 1×10⁻³ kg / 1g = 0.0982 kg
Let's convert the mass of solute to moles (mass / molar mass)
1.85 g / 60 g/mol = 0.0308 mol
Molality = 0.0308 mol / 0.0982 kg → 0.31 m
To find the molality of the solution, we calculate the number of moles of urea (0.031 mol) and the mass of water (0.10315 kg). The molality is then computed as moles of urea divided by the mass of water, which comes out to be approximately 0.30 m.
To calculate the molality of the solution, we need to first know the number of moles of the solute (urea) and the mass of the solvent (water) in kilograms.
The mass % of the solution is given as 1.85%, meaning that we have 1.85g of urea in 100g of the solution. Considering that the molar mass of urea (CO(NH2)2) is about 60.06 g/mol, the number of moles of urea can be calculated as follows: Moles of urea = mass / molar mass = 1.85g / 60.06 g/mol ≈ 0.031 mol.
Since the density of the solution is given as 1.05 g/mL, the mass of the solution for 100 mL (or 100g since density = mass/volume) would be 100 mL * 1.05 g/mL = 105g. In this 105g of the solution, 1.85g is urea, so the mass of water (solvent) would be 105g - 1.85g = 103.15g or 0.10315 kg (since 1g = 0.001 kg).
Molality is the number of solute's moles divided by the mass of solvent (in kg). Thus, the molality of the solution would be: Molality = moles of urea / mass of water in kg = 0.031 mol / 0.10315 kg ≈ 0.30 mol/kg or 0.30 m.
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The energy changes for many unusual reactions can be determined using Hess’s law.
(a) Calculate ΔE for the conversion of F⁻(g) into F⁺(g).
(b) Calculate ΔE for the conversion of Na⁺(g) into Na⁻(g).
Answer:
(a) F⁻(g) ⇒ F⁺(g) ΔE=2009 KJ/mol
(b) Na⁺(g) ⇒ Na⁻(g) ΔE=-548.6 KJ/mol
Explanation:
Hess's Law
"Energy changes for a reaction is independent of steps or route involved."
(a) F⁻(g) ⇒ F⁺(g)Now the conversion F⁻ to F⁺ is a two step reaction. In first step F⁻ loses an electron e⁻ to become neutral atom.
F⁻ ⇒ F + e⁻ ΔH=328 KJ/mol (i)
Kindly note this energy is derived from electron affinity, which is the energy changes while adding an electron to neutral atom and its same while removing an electron from ion.
In second step,
Another electron is removed from F.
F ⇒ F⁺ + e⁻ ΔH= 1681 KJ/mol (ii)
This energy is 1st ionization energy for F, which is the energy required to remove first electron from outer most shell of neutral atom.
Now the total energy change can be found by applying Hess's law and adding equations (i) and (ii);
F⁻ ⇒ F + e⁻ ΔH=328 KJ/mol
F ⇒ F⁺ + e⁻ ΔH= 1681 KJ/mol
F⁻ ⇒ F⁺ + 2e⁻ ΔH= 2009 KJ/mol
(b) Na⁺(g) ⇒ Na⁻(g)This reaction involves two steps in first Na⁺(g) gains electron and become neutral atom while in second step Na(g) accepts another electron to become Na⁻(g).
Na⁺(g) + e⁻⇒ Na(g) ΔH=-495.8 KJ/mol (i)
Na(g) + e⁻ ⇒ Na⁻(g) ΔH=-52.8 KJ/mol (ii) (By Applying Hess's Law)
Na⁺(g) + 2e⁻ ⇒ Na⁻(g) ΔH=-548.6 KJ/mol
Kindly note as there is no molar change in above mentioned reactions therefore enthalpy changes(ΔH) and internal energy changes(ΔE) are same.
As a science major, you are assigned to the "Atomic Dorm" at college. The first floor has one suite with one bedroom that has a bunk bed for two students. The second floor has two suites, one like the one on the first floor, and the other with three bedrooms, each with a bunk bed for two students. The third floor has three suites, two like the ones on the second floor, and a third suite that has five bedrooms, each with a bunk bed for two students. Entering students choose room and bunk on a first-come, first-serve basis by criteria in the following order of importance:
(1) They want to be on the lowest available floor.
(2) They want to be in the smallest available dorm room.
(3) They want to be in a lower bunk if available.
(a) Which bunk does the 1st student choose?
(b) How many students are in top bunks when the 17th student chooses?
(c) Which bunk does the 21st student choose?
(d) How many students are in bottom bunks when the 25th student chooses?
Explanation:
A.
The first student will be on the lower bunk on the first floor because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available.
B.
7 students are in the TOP bunks because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available. Therefore, all the rooms up till the third floor (Remember, third floor has 3 suites), so the first floor is filled - 1 person on the top bunk, 2 floor is filled- 4 persons and the third floor; the first suite is filled - 1 person and the second suite is a little partially filled- 1 person.
C.
Following the criteria 1, 2 and 3, the 21st student occupies the third suite on the third floor because all the floors (1 and 2) are occupied so the third suite on the third floor is still vacant.
D.
From the criteria there are therefore 10 persons at the TOP bunk. All the rooms up till the third floor are filled, so the first floor is filled - 1 person on the top bunk, second floor is filled (2 suites) - 4 persons and the third floor; the first suite and second suite is filled - 4 persons; the thirs suite has 6 persons present so 1 person is at the top bunk.
tyrosine kinase inhibitor binds and inhibits BTK. As a result of the experiment, you are able to elute BTK from the column, but in a mixture of other tyrosine kinases. Why are tyrosine kinases other than BTK present in the eluate?
Answer: It is because tyrosine kinases and BTK have similar solubilities
Explanation:
In column chromatography, components of a mixture are seperated based on their relative solubilities in two non-mixing phases.
In essence, tyrosine kinases and BTK are present in the eluate due to their similar solubility rates that arise from the similar chemical structure both possess (otherwise it would be impossible for the inhibitor meant for Tyrosine kinase to bind and also inhibits BTK)
Thus, the similar solubilities of both groups is the reason they could elute out of the column without being adsorped.
Calculate the molalities of the following aqueous solutions:________. (a) 1.22 M sugar (C12H22O11) solution (density of solution = 1.12 g/ml), (b) 0.87 M NaOH solution (density = 1.04 g/ml), (c) 5.24 M NaHCO3 solution (density = 1.19 g/ml).
a) The molality of the 1.22 M sugar solution is approximately [tex]\(1.089 \, \text{mol/kg}\).[/tex]
b) The molality of the 0.87 M NaOH solution is approximately [tex]\(0.8365 \, \text{mol/kg}\).[/tex]
c) The molality of the 5.24 M NaHCO3 solution is approximately [tex]\(4.4034 \, \text{mol/kg}\).[/tex]
(a) 1.22 M sugar (C12H22O11) solution:
Given:
- Molarity (\(M\)) of sugar solution = 1.22 M
- Density of solution = 1.12 g/ml
We'll assume that the density given is for the solution and not just for water.
1. Calculate the mass of 1 L (1000 ml) of solution:
[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]
[tex]\[ \text{mass of solution} = 1.12 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]
[tex]\[ \text{mass of solution} = 1120 \, \text{g} \][/tex]
2. Convert the mass of the solution to kilograms:
[tex]\[ \text{mass of solution in kg} = \frac{{1120 \, \text{g}}}{{1000}} = 1.12 \, \text{kg} \][/tex]
3. Calculate the moles of sugar (C12H22O11):
[tex]\[ \text{moles of sugar} = \text{Molarity} \times \text{volume in L} \][/tex]
[tex]\[ \text{moles of sugar} = 1.22 \, \text{mol/L} \times 1 \, \text{L} = 1.22 \, \text{mol} \][/tex]
4. Calculate the molality ([tex]\(m\)[/tex]):
[tex]\[ m = \frac{{\text{moles of sugar}}}{{\text{mass of solvent in kg}}} \][/tex]
Let's perform the calculation.
For the 1.22 M sugar solution:
1. Calculate the mass of 1 L (1000 ml) of solution:
[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]
[tex]\[ \text{mass of solution} = 1.12 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]
[tex]\[ \text{mass of solution} = 1120 \, \text{g} \][/tex]
2. Convert the mass of the solution to kilograms:
[tex]\[ \text{mass of solution in kg} = \frac{{1120 \, \text{g}}}{{1000}} = 1.12 \, \text{kg} \][/tex]
3. Calculate the moles of sugar (C12H22O11):
[tex]\[ \text{moles of sugar} = \text{Molarity} \times \text{volume in L} \][/tex]
[tex]\[ \text{moles of sugar} = 1.22 \, \text{mol/L} \times 1 \, \text{L} = 1.22 \, \text{mol} \][/tex]
4. Calculate the molality ([tex]\(m\)[/tex]):
[tex]\[ m = \frac{{\text{moles of sugar}}}{{\text{mass of solvent in kg}}} \][/tex]
[tex]\[ m = \frac{{1.22 \, \text{mol}}}{{1.12 \, \text{kg}}} \][/tex]
[tex]\[ m \approx 1.089 \, \text{mol/kg} \][/tex]
So, the molality of the 1.22 M sugar solution is approximately [tex]\(1.089 \, \text{mol/kg}\).[/tex]
(b) 0.87 M NaOH solution:
Given:
- Molarity ([tex]\(M\)[/tex]) of NaOH solution = 0.87 M
- Density of solution = 1.04 g/ml
We'll follow the same steps as above to calculate the molality.
Let's calculate for the NaOH solution.
For the 0.87 M NaOH solution:
1. Calculate the mass of 1 L (1000 ml) of solution:
[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]
[tex]\[ \text{mass of solution} = 1.04 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]
[tex]\[ \text{mass of solution} = 1040 \, \text{g} \][/tex]
2. Convert the mass of the solution to kilograms:
[tex]\[ \text{mass of solution in kg} = \frac{{1040 \, \text{g}}}{{1000}} = 1.04 \, \text{kg} \][/tex]
3. Calculate the moles of NaOH:
[tex]\[ \text{moles of NaOH} = \text{Molarity} \times \text{volume in L} \][/tex]
[tex]\[ \text{moles of NaOH} = 0.87 \, \text{mol/L} \times 1 \, \text{L} = 0.87 \, \text{mol} \][/tex]
4. Calculate the molality (\(m\)):
[tex]\[ m = \frac{{\text{moles of NaOH}}}{{\text{mass of solvent in kg}}} \][/tex]
[tex]\[ m = \frac{{0.87 \, \text{mol}}}{{1.04 \, \text{kg}}} \][/tex]
[tex]\[ m = \frac{{0.87 \, \text{mol}}}{{1.04 \, \text{kg}}} \][/tex]
[tex]\[ m \approx 0.8365 \, \text{mol/kg} \][/tex]
So, the molality of the 0.87 M NaOH solution is approximately [tex]\(0.8365 \, \text{mol/kg}\).[/tex]
(c) 5.24 M NaHCO3 solution:
Given:
- Molarity (\(M\)) of NaHCO3 solution = 5.24 M
- Density of solution = 1.19 g/ml
Let's follow the same steps as above to calculate the molality for the NaHCO3 solution.
For the 5.24 M NaHCO3 solution:
1. Calculate the mass of 1 L (1000 ml) of solution:
[tex]\[ \text{mass of solution} = \text{density} \times \text{volume} \][/tex]
[tex]\[ \text{mass of solution} = 1.19 \, \text{g/ml} \times 1000 \, \text{ml} \][/tex]
[tex]\[ \text{mass of solution} = 1190 \, \text{g} \][/tex]
2. Convert the mass of the solution to kilograms:
[tex]\[ \text{mass of solution in kg} = \frac{{1190 \, \text{g}}}{{1000}} = 1.19 \, \text{kg} \][/tex]
3. Calculate the moles of NaHCO3:
[tex]\[ \text{moles of NaHCO3} = \text{Molarity} \times \text{volume in L} \][/tex]
[tex]\[ \text{moles of NaHCO3} = 5.24 \, \text{mol/L} \times 1 \, \text{L} = 5.24 \, \text{mol} \][/tex]
4. Calculate the molality (\(m\)):
[tex]\[ m = \frac{{\text{moles of NaHCO3}}}{{\text{mass of solvent in kg}}} \][/tex]
[tex]\[ m = \frac{{5.24 \, \text{mol}}}{{1.19 \, \text{kg}}} \][/tex]
[tex]\[ m = \frac{{5.24 \, \text{mol}}}{{1.19 \, \text{kg}}} \][/tex]
[tex]\[ m \approx 4.4034 \, \text{mol/kg} \][/tex]
So, the molality of the 5.24 M NaHCO3 solution is approximately [tex]\(4.4034 \, \text{mol/kg}\).[/tex]
The volume density of atoms for a bcc lattice is 5 x 1026 m-3. Assume that the atoms are hard spheres with each atom touching its nearest neighbors. Determine the lattice constant and effective radius of the atom.
Explanation:
It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.
This means that volume occupied by 2 atoms is equal to volume of the unit cell.
So, according to the volume density
[tex]5 \times 10^{26} atoms = 1 [tex]m^{3}[/tex]
2 atoms = [tex]\frac{1 m^{3}}{5 \times 10^{26} atoms} \times 2 atoms[/tex]
= [tex]4 \times 10^{-27} m^{3}[/tex]
Formula for volume of a cube is [tex]a^{3}[/tex]. Therefore,
Volume of the cube = [tex]4 \times 10^{-27} m^{3}[/tex]
As lattice constant (a) = [tex](4 \times 10^{-27} m^{3})^{\frac{1}{3}}[/tex]
= [tex]1.59 \times 10^{-9} m[/tex]
Therefore, the value of lattice constant is [tex]1.59 \times 10^{-9} m[/tex].
And, for bcc unit cell the value of radius is as follows.
r = [tex]\frac{\sqrt{3}}{4}a[/tex]
Hence, effective radius of the atom is calculated as follows.
r = [tex]\frac{\sqrt{3}}{4}a[/tex]
= [tex]\frac{\sqrt{3}}{4} \times 1.59 \times 10^{-9} m[/tex]
= [tex]6.9 \times 10^{-10} m[/tex]
Hence, the value of effective radius of the atom is [tex]6.9 \times 10^{-10} m[/tex].
Rank the following bonds from strongest to weakest and provide the bond energy: the bond between hydrogen and oxygen in a water molecule; the bond between sodium and chloride in the NaCl molecule; the bond between atoms in a metal; the van der Waals bond between adjacent hydrogen atoms.
The ranking of the bonds from strongest to weakest is: bond between atoms in a metal, bond between sodium and chloride in NaCl, bond between hydrogen and oxygen in water, and the van der Waals bond between adjacent hydrogen atoms. The bond energies for each bond are different, with the bond between atoms in a metal having the highest bond energy, followed by the bond between sodium and chloride in NaCl, the bond between hydrogen and oxygen in water, and the van der Waals bond between adjacent hydrogen atoms having the lowest bond energy.
Explanation:The ranking of the bonds from strongest to weakest is as follows:
The bond between atoms in a metal (strongest) The bond between sodium and chloride in the NaCl molecule The bond between hydrogen and oxygen in a water molecule The van der Waals bond between adjacent hydrogen atoms (weakest)
The bond energy, also known as bond dissociation energy, is the energy required to break a bond. Here are the bond energies for each bond:
The bond between atoms in a metal: high bond energy (specific values vary) The bond between sodium and chloride in the NaCl molecule: high bond energy (~788 kJ/mol) The bond between hydrogen and oxygen in a water molecule: moderate bond energy (~464 kJ/mol) The van der Waals bond between adjacent hydrogen atoms: low bond energy (negligible) Learn more about bond strength here:
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The strongest bond is the bond between atoms in a metal, followed by the bond between sodium and chloride in NaCl. The bond between hydrogen and oxygen in water is weaker than the previous two, and the van der Waals bond between adjacent hydrogen atoms is the weakest. The bond energy is high for the bonds in a metal and NaCl, moderate for the bond in water, and low for the van der Waals bond between hydrogen atoms.
Explanation:The bonds can be ranked from strongest to weakest as follows:
Bond between atoms in a metal: This bond is the strongest among the given options.Bond between sodium and chloride in the NaCl molecule: This bond is stronger than the bond between hydrogen and oxygen in a water molecule.Bond between hydrogen and oxygen in a water molecule: This bond is weaker than the bond between sodium and chloride.Van der Waals bond between adjacent hydrogen atoms: This bond is the weakest among the given options.The bond energy for each bond is:
Bond between atoms in a metal: High bond energyBond between sodium and chloride in the NaCl molecule: High bond energyBond between hydrogen and oxygen in a water molecule: Moderate bond energyVan der Waals bond between adjacent hydrogen atoms: Low bond energyLearn more about bond strength here:https://brainly.com/question/30226871
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If one wished to obtain 0.15 moles of glacial acetic acid, how many milliliters of glacial acetic acid should be obtained? Enter only the number with two significant figures.
Answer:
8.6 mL
Explanation:
Glacial acetic acid (or anhydrous acetic acid) can be thought of as pure acetic acid, and has a density of 1.05 g/cm³ (or g/mL).
To answer this problem first we convert moles of acetic acid (CH₃CO₂H) to grams, using its molar mass (60 g/mol):
0.15 mol acetic acid * 60 g/mol = 9.0 g acetic acid.Now we convert grams of acetic acid to mL, using its density:
9.0 g ÷ 1.05 g/mL = 8.6 mLAt a certain temperature this reaction follows first-order kinetics with a rate constant of 0.184 s^-1 ? Suppose a vessel contains Cl_2 O_5 at a concentration of 1.16 M. Calculate the concentration of Cl_2 O_5 in the vessel 5.70 seconds later.
To calculate the concentration of Cl2O5 in the vessel 5.70 seconds later, use the first-order rate law equation and the given rate constant and initial concentration.
Explanation:To calculate the concentration of Cl2O5 in the vessel 5.70 seconds later, we can use the first-order rate law equation.
The rate law equation for a first-order reaction is: ln([A]t/[A]0) = -kt
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
In this case, we know the rate constant (k) is 0.184 s^-1 and the initial concentration ([A]0) is 1.16 M. We need to find the concentration at 5.70 seconds ([A]t).
Plugging in the values: ln([A]t/1.16) = -0.184 * 5.70
Solving for [A]t, we find that the concentration of Cl2O5 in the vessel 5.70 seconds later is approximately 0.64 M.
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Given data: Rate constant (k) = 0.184 s⁻¹
Initial concentration (Cl₂O₅) = 1.16 M
Time (t) = 5.70 s
To find the concentration of Cl₂O₅ after 5.70 seconds
We know that the first-order integrated rate law equation is:
ln([A]t/[A]0) = -where [A]t and [A]0 are the concentrations of reactant at time 't' and initial time '0' respectively. We can rearrange this equation to find the concentration of the reactant at any time 't':[A]t = [A]0 × e^(-kt)Putting the values in the equation, we get:
[Cl₂O₅]5.70 = 1.16 M × e^(-0.184 s⁻¹ × 5.70 s)
[Cl₂O₅]5.70 = 0.501
Therefore, the concentration of Cl₂O₅ in the vessel 5.70 seconds later is 0.501 M.
There are four types of Chemical Warfare Agents (CWA). Which CWA prevents blood from providing oxygen to tissues and organs?
Answer:
Blood agent
Explanation:
chemical warfare agents are of four major types:
nerve, blister, choking and blood agents
Nerve agents - it attacks the central nervous system and can enter the body through inhalation
Blister agents - it attacks skins and it is rapidly absorbed by the skin cells
chocking agents - as name denoted, it is related to the respiratory system.
Blood agent- it attacks the circulatory system of the body. it prevents the circulation of blood into the tissue and organs
For an ideal gas, which pairs of variables are inversely proportional to each other (if all other factors remain constant)?
Answer:
Pressure and volume
Explanation:
The ideal gas equation is given below:
PV = nRT
P = nRT /V
If nRT is constant,
P will be inversely proportional to V(P & 1/V)
Therefore, pressure and volume are inversely proportional if all factors remains constant
In an ideal gas, pressure and volume are inversely proportional to each other if all other factors remain constant. This is known as Boyle's Law.
Explanation:In an ideal gas, pressure and volume are inversely proportional to each other if all other factors remain constant.
This is known as Boyle's Law, which states that when the volume of a given amount of gas is decreased, the pressure increases, and vice versa.
For example, if the volume of a gas is reduced by half, the pressure will double, and if the volume is doubled, the pressure will be halved.
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In an aqueous solution containing Ni (II) and Ni (IV) salts, which cation would you expect to be the more strongly hydrated? Why?
Answer:
Well in comparison of Ni (II) and Ni (IV), Ni (IV) is a stronger vation with the +4 charge so it will attract the more oxygen ions in the aqueous solution. That is why Ni (IV) will be more strongly hydrated.
Explanation:
Final answer:
In an aqueous solution, the Ni (IV) cation is more strongly hydrated than the Ni (II) cation due to its higher charge density, which attracts more water molecules into its hydration sphere.
Explanation:
In an aqueous solution containing Ni (II) and Ni (IV) salts, the Ni (IV) cation is expected to be more strongly hydrated. This outcome is attributed primarily to the charge density. The Ni (IV) has a higher charge (+4) compared to Ni (II) which has a +2 charge. In terms of hydration, water molecules, which act as dipoles, are more strongly attracted to ions with a higher charge density. This means that the Ni (IV) ion, with its higher charge, attracts and binds water molecules more strongly than the Ni (II) ion.
This strong attraction results in a greater degree of hydration for the Ni (IV) ion as it pulls more water molecules into its hydration sphere. This process is crucial in understanding the properties of solutions, especially in predicting the behavior of ions in biological and chemical systems.
A graduated cylinder contains 15.0 mLmL of water. What is the new water level after 34.0 gg of silver metal with a density of 10.5 g/mLg/mL is submerged in the water?
Answer: The new water level when silver metal is submerged is 18.24 mL
Explanation:
We are given:
Volume of graduated cylinder = 15.0 mL
To calculate volume of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of silver metal = 10.5 g/mL
Mass of silver metal = 34.0 g
Putting values in above equation, we get:
[tex]10.5g/mL=\frac{34.0g}{\text{Volume of silver metal}}\\\\\text{Volume of silver metal}=\frac{34.0g}{10.5g/mL}=3.24mL[/tex]
New water level = Volume of silver metal + Volume of graduated cylinder
New water level = (3.24 + 15.0) mL = 18.24 mL
Hence, the new water level when silver metal is submerged is 18.24 mL
In a certain acidic solution at 25 ∘C, [H+] is 100 times greater than [OH −]. What is the value for [OH −] for the solution?
Answer: The value of [tex][OH^-][/tex] for the solution is [tex]10^{-6}M[/tex]
Explanation:
To calculate the concentration of hydroxide ion for the solution, we use the equation:
[tex][H^+]\times [OH^-]=10^{-14}[/tex]
We are given:
[tex][H^+]=100\times [OH^-][/tex]
Putting values in above equation, we get:
[tex]100\times [OH^-]\times [OH^-]=10^{-14}[/tex]
[tex][OH^-]^2=\frac{10^{-14}}{100}[/tex]
[tex][OH^-]=\sqrt{10^{-12}}[/tex]
[tex][OH^-]=10^{-6}M[/tex]
Hence, the value of [tex][OH^-][/tex] for the solution is [tex]10^{-6}M[/tex]