You are a high school track & field coach and you have a young female athlete who is running a competitive mile pace (5:20). This would make her competitive at the state and national level. You want to help this athlete to maintain and even increase her competitive potential. What would you do?
Answers
Answer:
By ensuring she is giving speed endurance training
Explanation:
Speed endurance training would be the most ideal in helping an athlete to maintain and even increase her competitive potential.
The speed endurance training would help the athlete maintain and attain a high speed for a longer duration of time.
This would make the athlete get used to running an improved time for much longer distances.
Related Questions
How many grams of Hydronium chromate are produced when 43.4 g of Tin (IV) chromate combines with35.2 g of Hydronium hydrogen phosphate? Use the following balanced equation:2 (H3O)2HPOA + 1 Sn(CrO4)2 ---> 2 (H30)2CrOA + 1 Sm(HPOA)2
Answers
Answer
Mass of (H30)2CrO = 38 g
Explanation
Given:
Mass of Sn(CrO4)2 = 43.4 g
Mass of (H3O)2HPO4 = 35.2 g
Required: The mass of (H30)2CrO4 that will be produced
Solution:
Calculate the possible mass that could be produced by each reactant, so as to determine the limiting reagent. Use stoichiometry.
For Sn(CrO4)2:
\(\begin{gathered} 43.4\text{ g Sn\lparen CrO}_4\text{\rparen}_2\text{ x }\frac{1\text{ mole Sn\lparen CrO}_4)_2}{350.70\text{ g Sn\lparen CrO}_4)_2}\text{ x }\frac{2\text{ mole \lparen H}_3\text{0\rparen}_2\text{CrO}_4}{1\text{ mole Sn\lparen CrO}_4)_2}\text{ x }\frac{153.9\text{ g \lparen H}_3O)_2CrO_4}{1\text{ mol \lparen H}_3O)_2CrO_4} \\ \\ =\text{ 38 g \lparen H}_3\text{O\rparen}_2\text{CrO}_4 \end{gathered}\)For (H3O)2HPO4
\(\begin{gathered} 35.2\text{ g \lparen H}_3\text{O\rparen}_2\text{HPO}_4\text{ x }\frac{1\text{ mole \lparen H}_3\text{O\rparen}_2\text{HPO}_4\text{ }}{133.97\text{ g }(H_3O)_2HPO_4}\text{ x }\frac{2\text{ mole}}{2\text{ mole}}\text{ x }\frac{153.9\text{ g \lparen H}_3\text{O\rparen}_2\text{CrO}_4}{1\text{ mole \lparen H}_3\text{O\rparen}_2\text{CrO}_4} \\ \\ =\text{ 40.43 g \lparen H}_3\text{O\rparen}_2\text{CrO}_4 \end{gathered}\)Sn(CrO4)2 will produce less (H30)2CrO4 therefore, Sn(CrO4)2 is the limiting reagent.
A gas occupies 5L at a temperature of 300k. What is the volume at 405k
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At a temperature of 405 K, the gas will occupy a volume of approximately 6.75 liters.
We can use Charles's Law to find the volume of the gas at the new temperature. Charles's Law states that the volume of a gas is directly proportional to its temperature when the pressure and the amount of gas are constant. Mathematically, this can be represented as:
V1/T1 = V2/T2
Here, V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature. In this problem, V1 is 5 L, T1 is 300 K, and T2 is 405 K. We want to find V2.
To solve for V2, we can rearrange the equation:
V2 = (V1 * T2) / T1
Now, plug in the given values:
V2 = (5 L * 405 K) / 300 K
V2 ≈ 6.75 L
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explain why the crystals, during the first filtration are washed with cold water
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To remove any impure residual crystals, the crystals are washed with cold water during the first filtration.
This process is termed as Recrystallisation, in which some compounds are purified. When a compound is synthesized in solid form, there are some impurities present. So, in order to remove those impurities, we recrystallize it under some specific conditions.
Recrystallisation is also known as Fractional distillation. This process is a time consuming process. We have certain Solubility curves that are used to predict the outcome of the Recrystallization process. This process gives the best results when the impurities are small in amount.
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Answer:
In the case of crystallization, the liquid may contain impurities that can reincorporate into the solid if not removed. To rinse a suction-filtered solid, the vacuum is removed and a small portion of cold solvent is poured over the solid (the " filter cake "). In the case of crystallization, the same solvent from the crystallization is used.
Please answer Q1, Q2, Q3 and Q4 in great detail. Thank you so much
Q1. State the formula for the energy levels of Hydrogen
Q2. What is the wavelength (in nm) for a transition between:
a) n=1⇒n=6?
b) n=25⇒n=26?
Q3. For a gas temperature of 300K, what is the relative density (between the two states) for each of the transitions in Q2? To two decimal points is sufficient.
Q4. The Lambert-Beers law is:
I(x) = I◦ exp(−nσx)
where n is the density of the absorber, σ(λ) is the wavelength-dependent cross section for absorption, x is the position, I◦ is the initial photon flux, I(x) is the photon flux versus position through the absorber.
Derive the Lambert-Beers law. (State and justify any assumptions.)
Answers
Q1. The formula for the energy levels of hydrogen is E = -13.6 eV/n².
Q2. a) The wavelength for the transition between n=1 and n=6 is approximately 93.5 nm. b) The wavelength for the transition between n=25 and n=26 is approximately 29.46 nm.
Q3. For the transitions in Q2, the relative densities are approximately 0.73 and 0.995, respectively.
Q4. The Lambert-Beers law relates the intensity of light transmitted through an absorber to the absorber's density, cross section for absorption, and position within the medium. It is expressed as I(x) = I₀ * exp(-n * σ(λ) * x).
Q1. The formula for the energy levels of hydrogen is given by the Rydberg formula, which is used to calculate the energy of an electron in the hydrogen atom:
E = -13.6 eV/n²
Where:
- E is the energy of the electron in electron volts (eV).
- n is the principal quantum number, which represents the energy level or shell of the electron.
Q2. a) To find the wavelength (in nm) for a transition between n=1 and n=6 in hydrogen, we can use the Balmer series formula:
1/λ = R_H * (1/n₁² - 1/n₂²)
Where:
- λ is the wavelength of the photon emitted or absorbed in meters (m).
- R_H is the Rydberg constant for hydrogen, approximately 1.097 x 10⁷ m⁻¹.
- n₁ and n₂ are the initial and final energy levels, respectively.
Plugging in the values, we have:
1/λ = (1.097 x 10⁷ m⁻¹) * (1/1² - 1/6²)
1/λ = (1.097 x 10⁷ m⁻¹) * (1 - 1/36)
1/λ = (1.097 x 10⁷ m⁻¹) * (35/36)
1/λ = 1.069 x 10⁷ m⁻¹
λ = 9.35 x 10⁻⁸ m = 93.5 nm
Therefore, the wavelength for the transition between n=1 and n=6 in hydrogen is approximately 93.5 nm.
b) Similarly, to find the wavelength (in nm) for a transition between n=25 and n=26 in hydrogen, we can use the same formula:
1/λ = R_H * (1/n₁² - 1/n₂²)
Plugging in the values:
1/λ = (1.097 x 10⁷ m⁻¹) * (1/25² - 1/26²)
1/λ = (1.097 x 10⁷ m⁻¹) * (1/625 - 1/676)
1/λ = (1.097 x 10⁷ m⁻¹) * (51/164000)
1/λ = 3.396 x 10⁴ m⁻¹
λ = 2.946 x 10⁻⁵ m = 29.46 nm
Therefore, the wavelength for the transition between n=25 and n=26 in hydrogen is approximately 29.46 nm.
Q3. To determine the relative density for each of the transitions in Q2, we need to calculate the ratio of the photon flux between the two states. The relative density is given by the equation:
Relative Density = (I(x2) / I(x1))
Where I(x2) and I(x1) are the photon fluxes at positions x2 and x1, respectively.
For a gas temperature of 300K, the relative density is proportional to the Boltzmann distribution of states, which is given by:
Relative Density = exp(-ΔE/kT)
Where ΔE is the energy difference between the two states, k is the Boltzmann constant (approximately 1.38 x 10⁻²³ J/K), and T is the temperature in Kelvin.
a) For the transition between n=1 and n=6, the energy difference is:
ΔE = E₁ - E₂ = (-13.6 eV / 1²) - (-13.6 eV / 6²)
ΔE = -13.6 eV + 0.6 eV = -13.0 eV
Converting the energy difference to joules:
ΔE = -13.0 eV * 1.6 x 10⁻¹⁹ J/eV = -2.08 x 10⁻¹⁸ J
Substituting the values into the relative density equation:
Relative Density = exp(-(-2.08 x 10⁻¹⁸ J) / (1.38 x 10⁻²³ J/K * 300 K))
Relative Density ≈ 0.73
Therefore, for the transition between n=1 and n=6, the relative density is approximately 0.73.
b) For the transition between n=25 and n=26, the energy difference is:
ΔE = E₁ - E₂ = (-13.6 eV / 25²) - (-13.6 eV / 26²)
ΔE ≈ -13.6 eV + 0.0585 eV ≈ -13.5415 eV
Converting the energy difference to joules:
ΔE ≈ -13.5415 eV * 1.6 x 10⁻¹⁹ J/eV ≈ -2.1664 x 10⁻¹⁸ J
Substituting the values into the relative density equation:
Relative Density = exp(-(-2.1664 x 10⁻¹⁸ J) / (1.38 x 10⁻²³ J/K * 300 K))
Relative Density ≈ 0.995
Therefore, for the transition between n=25 and n=26, the relative density is approximately 0.995.
Q4. Derivation of the Lambert-Beers law:
To derive the Lambert-Beers law, we consider a thin slice of the absorber with thickness dx. The intensity of light passing through this slice decreases due to absorption.
The change in intensity, dI, within the slice can be expressed as the product of the intensity at that position, I(x), and the fraction of light absorbed within the slice, nσ(λ)dx:
dI = -I(x) * nσ(λ)dx
The negative sign indicates the decrease in intensity due to absorption.
Integrating this equation from x = 0 to x = x (the total thickness of the absorber), we have:
∫[0,x] dI = -∫[0,x] I(x) * nσ(λ)dx
The left-hand side represents the total change in intensity, which is equal to I₀ - I(x) since the initial intensity is I₀.
∫[0,x] dI = I₀ - I(x)
Substituting this into the equation:
I₀ - I(x) = -∫[0,x] I(x) * nσ(λ)dx
Rearranging the equation:
I(x) = I₀ * exp(-nσ(λ)x)
This is the Lambert-Beers law, which shows the exponential decrease in intensity (photon flux) as light passes through an absorber. The law quantifies the dependence of intensity on the density of the absorber, the absorption cross section, and the position within the absorber.
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What’s different about ionic bonding and covalent bonding ? (Help I need the answer ASAP)
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is it possible to use the same colored central atom to make a model for all of these molecules? why?
Answers
Yes, it is possible to use the same colored central atom to make a model for all of these molecules. This is because the central atom in all of these molecules is the same, so it does not matter what color it is. The other atoms attached to the central atom will determine the shape of the molecule.
All the molecules have the same central atom because they are all hydrocarbons with the general formula CnH2n+2. The only difference between them is the number of carbon atoms that are present.For example, methane (CH4), ethane (C2H6), propane (C3H8), and butane (C4H10) are all hydrocarbons, and they all have Carbon as their central atom. The number of hydrogen atoms in each molecule varies based on the number of carbon atoms present.For instance, Methane (CH4) has one carbon atom and four hydrogen atoms, Ethane (C2H6) has two carbon atoms and six hydrogen atoms, Propane (C3H8) has three carbon atoms and eight hydrogen atoms, and Butane (C4H10) has four carbon atoms and ten hydrogen atoms. Therefore, you can use the same central atom, Carbon (C), to create a model for all of these molecules.
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a 50.00-ml sample of aqueous ca(oh)2 requires 34.66 ml of 0.885 m nitric acid for neutralization. calculate the concentration (molarity) of the original solution of calcium hydroxide.
Answers
The concentration (molarity) of the original solution of calcium hydroxide is 305.771 M.
To calculate the concentration (molarity) of the original solution of calcium hydroxide (Ca(OH)2), we can use the stoichiometry of the neutralization reaction between calcium hydroxide and nitric acid.
The balanced chemical equation for the reaction is:
Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(l)
From the balanced equation, we can see that one mole of calcium hydroxide reacts with two moles of nitric acid. Therefore, the number of moles of calcium hydroxide can be calculated by multiplying the volume of nitric acid used (34.66 ml) by its concentration (0.885 M) and dividing by the stoichiometric coefficient (2).
Moles of Ca(OH)2 = (34.66 ml) × (0.885 M) / 2 = 15.28855 mmol
Next, we convert the volume of the calcium hydroxide solution (50.00 ml) to liters and calculate the molarity.
Molarity of Ca(OH)2 = (15.28855 mmol) / (50.00 ml / 1000) = 305.771 M
Therefore, the concentration (molarity) of the original solution of calcium hydroxide is 305.771 M.
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What is the pH of a solution with a concentration of 4.2 x 10−5 M H3O+? (4 points)
Select one:
a. 2.31
b. 4.38
c. 5.62
d. 6.87
What is the pH of a solution with a 4.60 x 10−4 M hydroxide ion concentration? (4 points)
Select one:
a. 0.663
b. 3.33
c. 10.1
d. 10.7
Answers
Answer:
1) b. 4.38
2) d. 10.7
Explanation:
1) The problem tells us that [H₃O⁺] = 4.2x10⁻⁵ M, and keeping in mind that [H₃O⁺]= [H⁺], we can calculate the pH of the solution:
pH = -log[H⁺] = 4.382) First we calculate the pOH of the solution:
pOH = -log[OH⁻] = -log(4.6x10⁻⁴) = 3.33Then we calculate the pH using the following formula:
pH = 14 - pOH = 14 - 3.33pH = 10.7The pH of a solution with a concentration of 4.2x10⁻⁵ M H₃O⁺ ion is 4.38 and of a solution with a concentration of 4.6x10⁻⁴ OH⁻ ion is 10.7.
What is pH and pOH?pH of any solution is define as the negative log of the concentration of H⁺ ions and pOH is define as the negative log of the concentration of OH⁻ ions.
Given that concentration of H⁺ ions in solution = 4.2x10⁻⁵ M
pH for this solution is calculated as:
pH = -log(4.2x10⁻⁵) = 4.38
Also given that concentration of OH⁻ ions in solution = 4.6x10⁻⁴
pOH = -log(4.6x10⁻⁴) = 3.33
We know that, pH + pOH = 14.
pH = 14 - 3.33 = 10.7
Hence, value of pH & pOH is 4.38 and 10.7 respectively.
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can you determine the contents of the unlabeled jar through observation alone?
Answers
Answer:
This question appears incomplete but generally the answer is NO
Explanation:
Observation of the contents of an unlabeled jar depends solely on some of the physical properties of the content which is not good enough when identifying the contents of an unlabeled jar. This is because several substances have the same physical properties. For example, water and several acidic solutions appear the same in terms of colour and viscosity. Hence, a jar of water and a jar of hydrochloric acid cannot be differentiated by mere observation but with the use of litmus paper.
And a solution of hydrochloric acid and sulphuric acid cannot be differentiated by observation alone except a jar ammonia is brought close to the two jars (the one with hydrochloric acid will form a white fume with ammonia); hence relying on the chemical properties of both substances.
NOTE: Observation is done through sight, smell, feel, sound and taste without relying on another procedure/analysis to determine them.
TRUE OR FALSE
Astronomers use spectroscopes to identify elements in stars because each element produces
unique emission spectrum
Answers
Answer:
True!
Explanation:
A spectrum is simply a chart or a graph that shows the intensity of light being emitted over a range of energies. Have you ever seen a spectrum before? Probably. Nature makes beautiful ones we call rainbows. Sunlight sent through raindrops is spread out to display its various colors (the different colors are just the way our eyes perceive radiation with slightly different energies).
Hope i helped!
Can someone help me with this chemistry question?
Answers
Answer:
B no. is the answer because
Explanation:
acids increase the hydrogen ion concentration, which provides more ionsan acid comes in contact with a metal, the acid steals free electrons away. That loosens the bonds holding metal atoms in place, so the metal dissolves. That's a chemical change, so the ability of acids to dissolve metals is a chemical propertyWhen a strong acid is neutralized by a strong base there are no excess hydrogen ions left in the solution. The solution is said to be neutral as it is neither acidic nor alkaline. The pH of such a solution is close to a value of 7; the exact pH value is dependent on the temperature of the solution.So this can probably be the the answer
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Mark me Brainliest （⌒▽⌒）#beyourselfwhat is the best description of the unknown acid? a weak monoprotic acid a weak diprotic acid a strong diprotic acid a strong monoprotic acid what is the first pka of the unknown acid? report the answer to one decimal point. what is the second pka of the unknown acid? report the answer to one decimal point.
Answers
The unknown acid is a weak: diprotic acid. The first pKa of the unknown acid is: 3.0, and the second pKa is: 5.0.
What is the definition of an acid?A substance that releases hydrogen ions in water is known as an acid. It has a pH of less than 7, which is the acidic range. Acids are the opposite of bases in chemistry, which release hydroxide ions when dissolved in water.The unknown acid is a weak diprotic acid.
What is pKa?pKa is the negative logarithm of the acid dissociation constant (Ka). It is a measure of the acidity of a molecule, with lower values indicating stronger acids. A pKa value of 3.0 corresponds to a weak acid.
What is the definition of diprotic acid?A diprotic acid is a substance that has two hydrogen ions to donate when dissolved in water. The chemical formula for a diprotic acid is often written in the following way: H2A, with the acid donating two hydrogen ions to the water. A weak diprotic acid is an acid that does not fully dissociate in water.
What is the first pKa of the unknown acid?The first pKa of the unknown acid is 3.0.
What is the second pKa of the unknown acid?The second pKa of the unknown acid is 5.0.
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How many moles of calcium are present in 204 grams of calcium?
Answers
R. F.M (Relative Formula Mass ) of Ca=40
40g of Ca=1 mole
204g of Ca=(1*204)/40
=5.1 moles
Pressure is defined as _______.1
Answers
Answer:the perpendicular force per unit area
pressure, in the physical sciences, the perpendicular force per unit area, or the stress at a point within a confined fluid.
Explanation:
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chromatography of food dyes lab why is it important to mark the solvent level on the chromatography paper as soon as you remove it from the petri dish
Answers
It is important to mark the solvent level on the chromatography paper as soon as you remove it from the petri dish in a chromatography of food dyes lab because if the solvent level is not marked as soon as possible, the solvent front can evaporate causing the results to be inaccurate.
Chromatography is a laboratory technique for separating a mixture into its individual components. The mixture is dissolved in a solvent and then placed in contact with a stationary phase. The components of the mixture are then separated based on their individual interactions with the stationary phase and the solvent. Chromatography of food dyes is a lab that is used to separate different food dyes that are present in a sample.
The sample is placed on chromatography paper which is then placed in a petri dish containing a solvent. As the solvent moves up the chromatography paper, the different dyes in the sample are separated based on their individual interactions with the paper and the solvent.
In a chromatography of food dyes lab, it is important to mark the solvent level on the chromatography paper as soon as it is removed from the petri dish because the solvent front can evaporate causing the results to be inaccurate. If the solvent front evaporates, the distance traveled by the different dyes will be shorter, making it appear as though they are less separated than they actually are.
By marking the solvent level as soon as possible, the distance traveled by the different dyes can be accurately measured, and the results will be more accurate.
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The reason why it is important to mark the solvent level on the chromatography paper as soon as you remove it from the petri dish is that the solvent level must be measured to calculate the Rf value. The Rf value is a way to quantify how far a particular compound travels in chromatography.
It is calculated as the distance traveled by the compound divided by the distance traveled by the solvent.The chromatography of food dyes lab is a experiment that aims to identify the dyes used in food products by using paper chromatography. The procedure includes: Cut a strip of chromatography paper and mark the solvent level using a pencil as soon as you remove it from the petri dish; prepare the chromatography solvent by mixing rubbing alcohol with water; then, spot the dyes on the chromatography paper using toothpicks or capillary tubes.
Afterwards, place the paper in the petri dish containing the solvent, making sure that the dyes do not touch the solvent, and cover it. Allow the solvent to travel up the paper until it reaches the solvent level mark. Once the solvent level has reached the mark, remove the paper from the petri dish and allow it to dry before analyzing the results.
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what is the Lewis structure for iodine
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A sphere of radius 0.457 m, temperature 32.2 ∘
C, and emissivity 0.924 is located in an environment of temperature 82.9 ∘
C. At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere's net rate of energy exchange? (a) Number (b) Number Units Units
Answers
a) The sphere emits thermal radiation at a rate of 139.75 Watts.
b) The sphere absorbs thermal radiation at a rate of 37.66 Watts.
c) The sphere's net rate of energy exchange is 102.09 Watts.
What are the rates of thermal radiation emission, absorption, and net energy exchange for the sphere?To calculate the rates of thermal radiation emission and absorption, we can use the Stefan-Boltzmann law, which states that the rate of thermal radiation emitted or absorbed by an object is proportional to its surface area, temperature, and the Stefan-Boltzmann constant.
a) The rate of thermal radiation emitted by the sphere can be calculated using the formula:
Emitting Rate = emissivity * surface area * Stefan-Boltzmann constant * (\(temperature^4 - environment\ temperature^4\))
Plugging in the given values:
Emitting Rate = \(0.924 * (4\pi * (0.457)^2) * 5.67 \times 10^{-8} * ((32.2 + 273.15)^4 - (82.9 + 273.15)^4)\)
Emitting Rate ≈ 139.75 Watts
b) The rate of thermal radiation absorbed by the sphere can be calculated in a similar way but using the environment temperature as the object's temperature:
Absorbing Rate = emissivity * surface area * Stefan-Boltzmann constant * (\(environment\ temperature^4 - temperature^4\))
Plugging in the given values:
Absorbing Rate = \(0.924 * (4\pi * (0.457)^2) * 5.67 \times 10^{-8} * ((82.9 + 273.15)^4 - (32.2 + 273.15)^4)\)
Absorbing Rate ≈ 37.66 Watts
c) The net rate of energy exchange is the difference between the emitting rate and the absorbing rate:
Net Rate = Emitting Rate - Absorbing Rate
Net Rate = 139.75 Watts - 37.66 Watts
Net Rate ≈ 102.09 Watts
Therefore, the sphere emits thermal radiation at a rate of 139.75 Watts, absorbs thermal radiation at a rate of 37.66 Watts, and has a net rate of energy exchange of 102.09 Watts.
Note: The units for all the rates are Watts.
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Please answer by number.
Answers
Answer:
2)compound
3)element
4)nonmetal
5)solute
3. element
4. nonmetal
5. solute
Would a saturated fatty acid hold more potential energy or less potential energy than an unsaturated fatty acid
Answers
Answer:
Yes, it is a fact that saturated fats contain more gross energy due to higher H+ concentration. However, unsaturated fats are more digestible and have higher absorption rate, and therefore the metabolizable energy that unsaturated fats provide is higher than that provided by saturated fats.
Explanation:
Problem 11.39 A sample of helium gas has a volume of 6.50L at a pressure of 845 mmHg and a temperature of 25°C. What is the final pressure of the gas, in atmospheres, when the volume and temperature of the gas sample are changed to the following, if the amount of gas does not change? Part A 1600 mL and 320 K Express your answer with the appropriate units. "! A p. Value = Om ? Units Submit Request Answer Part 3 2.50 L and 12 Express your answer with the appropriate units. Units P-value Submit B est Answer
Answers
Part A: The final pressure of the gas, when the volume and temperature are changed to 1600 mL and 320 K respectively, is approximately 4.57 atm.
Part B: The final pressure of the gas, when the volume and temperature are changed to 2.50 L and 12 K respectively, is approximately 0.19 atm.
To solve this problem, we can use the combined gas law, which states:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = Initial pressure
V1 = Initial volume
T1 = Initial temperature
P2 = Final pressure (what we want to find)
V2 = Final volume
T2 = Final temperature
Let's solve each part of the problem:
Part A:
Given:
Initial volume (V1) = 6.50 L
Initial pressure (P1) = 845 mmHg (we'll convert it to atmospheres later)
Initial temperature (T1) = 25°C = 25 + 273.15 = 298.15 K
Final volume (V2) = 1600 mL = 1600/1000 = 1.6 L
Final temperature (T2) = 320 K
Now let's plug these values into the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
(P2 * 1.6) / 320 = (845 * 6.50) / 298.15
P2 = (845 * 6.50 * 320) / (1.6 * 298.15)
P2 ≈ 3477.59 mmHg
To convert the pressure to atmospheres, we divide by the conversion factor of 760 mmHg/1 atm:
P2 (in atmospheres) = 3477.59 mmHg / 760 mmHg/atm
P2 ≈ 4.57 atm
So the final pressure of the gas, when the volume and temperature are changed to 1600 mL and 320 K respectively, is approximately 4.57 atm.
Part B:
Given:
Initial volume (V1) = 6.50 L
Initial pressure (P1) = 845 mmHg (we'll convert it to atmospheres later)
Initial temperature (T1) = 25°C = 25 + 273.15 = 298.15 K
Final volume (V2) = 2.50 L
Final temperature (T2) = 12 K
Using the same combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
(P2 * 2.50) / 12 = (845 * 6.50) / 298.15
P2 = (845 * 6.50 * 2.50) / (12 * 298.15)
P2 ≈ 144.41 mmHg
Converting to atmospheres:
P2 (in atmospheres) = 144.41 mmHg / 760 mmHg/atm
P2 ≈ 0.19 atm
Therefore, the final pressure of the gas, when the volume and temperature are changed to 2.50 L and 12 K respectively, is approximately 0.19 atm.
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what process below explain the difference between oxymercuration/demercuration and hydroboration/oxidation? group of answer choices heteroatom-containing intermediate conversion of an alkene to an alcohol markovnikov regiochemistry hydration
Answers
The process that explains the difference between oxymercuration/demercuration and hydroboration/oxidation is regiochemistry.
Oxymercuration/demercuration is a chemical reaction in which an alkene is converted to an alcohol. The reaction is carried out in the presence of mercury acetate. The reaction proceeds in two steps. In the first step, the alkene is treated with mercury (II) acetate and a proton source such as water to form an organomercurial intermediate. The intermediate is then hydrolyzed to form the alcohol. The hydroboration/oxidation is a chemical reaction that is used to convert alkenes to alcohols. This reaction proceeds in two steps. In the first step, the alkene is treated with borane (BH3) to form an organoboron intermediate. The intermediate is then oxidized to form the alcohol. In oxymercuration/demercuration reaction, Markovnikov regiochemistry is observed while in hydroboration/oxidation reaction, anti-Markovnikov regiochemistry is observed. This difference in regiochemistry is due to the difference in the reaction mechanism and the intermediates involved. Therefore, the process that explains the difference between oxymercuration/demercuration and hydroboration/oxidation is regiochemistry.
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What’s in the gaps...
Answers
Answer:
Faster
number
reactants
increases
Explanation:
The rate of reaction refers to how fast or slow a reaction occurs. The rate of reaction is dependent on factors such as surface area, concentration, temperature, nature of reactants, catalyst etc.
The concentration refers to the amount of reactants present. When the acid is made more concentrated, its amount is increased. This increases the number of collisions between the acid particles and the marble chips ultimately leading to increase in the rate of reaction according to Arrhenius theory.
Suppose that 5-ml containers of helium (He), neon (Ne), and argon (Ar) are at pressures of 1 atm,
2 atm, and 3 atm, respectively. The He and Ne are then added to the container of Ar.
a. What is the partial pressure of He in the container after the three gases are mixed?
b. What is the total pressure in the container after the three gases are mixed?
Answers
The total pressure in the container after the three gases are mixed is 6 atm and the partial pressure of He is 1 atm.
What is Dalton's Law of Partial Pressure ?According to Dalton's Law of Partial Pressure , the pressure exerted by a mixture of gas is equal to the sum of the partial pressure of the gases in the mixture.
P(total) = p₁ +p₂+p₃+.....
As the volume of the containers are same and the
n = 1*5/RT for Helium
n = 2*5 /RT for Neon
n= 3*5/RT for Argon
Mole fraction = moles of the element/Total moles
Moles fraction of Helium is
\(\rm \dfrac{1*5}{1*5+2*5+3*5}\)
= 1/6
Total Pressure in the container = 1 + 2 +3 = 6 atm
The partial pressure of Helium will be
=Mole fraction * Total pressure
=(1/6)*6 atm
= 1 atm , as the volume is same the temperature is same , the pressure will also be same.
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What would be one way to avoid static electricity? А) Dry cotton fabrics and synthetic fabrics separately. B) Run a metal clothes hanger across your dry hair. C Use a blow dryer to dry your hair. D Wear shoes with rubber soles.
Answers
Answer: The answer would be A.
Explanation:
Answer: B
Explanation:
Every atom of the __ carbon has six protons
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"Asbestos needs to be removed, whether or not it will be
disturbed.
True or False"
Answers
False, Asbestos is a naturally occurring mineral fiber that was commonly used in various industries due to its heat resistance, strength, and insulating properties.
Asbestos does not necessarily need to be removed if it will not be disturbed or pose a risk to human health. Asbestos-containing materials that are in good condition and undisturbed are generally considered safe. However, if asbestos-containing materials are damaged, deteriorating, or will be disturbed during renovation or demolition activities, it is necessary to take appropriate precautions, which may include professional removal or encapsulation, to prevent the release of asbestos fibers into the air. The decision to remove asbestos should be based on an assessment of its condition, potential for disturbance, and adherence to local regulations and guidelines.
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which element is more reactive out of oxygen and sulphur
Answers
Answer:
Oxygen
Explanation:
Oxygen is more reactive because it has small atomic size which has high tendency to attract more electrons.
to what tempature must a sample of helim gas be cooled from 119
Answers
The sample of helium gas must be cooled to approximately -220°C to reduce its volume from 5.9 L to 0.2 L at constant pressure.
According to the ideal gas law, the relationship between the volume (V), temperature (T), and pressure (P) of a gas can be expressed as PV = nRT, where n is the number of moles of the gas and R is the ideal gas constant. In this case, the pressure is constant, so we can simplify the equation to V/T = constant.
To find the final temperature required to reduce the volume from 5.9 L to 0.2 L, we can set up the following ratio:
(V1 / T1) = (V2 / T2)
Where V1 is the initial volume (5.9 L), T1 is the initial temperature (119°C + 273.15 = 392.15 K), V2 is the final volume (0.2 L), and T2 is the final temperature that we need to find.
Rearranging the equation, we have:
T2 = (V2 / V1) * T1
= (0.2 L / 5.9 L) * 392.15 K
≈ 13.28 K
Converting 13.28 K back to Celsius, we get:
T2 ≈ -259.87°C
Therefore, the sample of helium gas must be cooled to approximately -220°C (or -259.87°C) to reduce its volume from 5.9 L to 0.2 L at constant pressure.
The question should be:
To what temperature must a sample of helium gas be cooled from 119°C to reduce its volume from 5.9 L to 0.2 L at constant pressure?
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What is the major non-covalent interaction that dictates the interaction between chloromethane and water?.
Answers
These sights are an instance of hydrogen bonds, vulnerable interactions that shape among a hydrogen with a partial fine price and a extra electronegative atom, which includes oxygen.
Hydrogen bonding takes place among the −OH organization of methanol and the −OH organization of water. This is the more potent non-covalent appeal due to the fact the appeal arises from everlasting dipoles due to the distinction in electronegativity of oxygen and hydrogen atoms. Intermolecular forces are non-covalent interactions that arise among one of a kind molecules, instead of among one of a kind atoms of the equal molecule. These interactions are the end result of dipole-dipole interactions shaped thru temporary in homo genieties withinside the electrons inside a molecule. worried the vulnerable sharing of an electron pair among a hydrogen atom and any other atom.
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Describe what happens to water particles as you increase/decrease the temperature/pressure? (hint: movement/speed, attraction, density, volume, state of matter)
Answers
Explanation:
as the temperature of a solid liquid or a gas increase the particles move rapidly as the temperature falls the particles slow down if a liquid is pulled sufficiently it forms a solid.
Answer: As the temperature of a solid, liquid or gas increases, the particles move more rapidly. As the temperature falls, the particles slow down.