The pseudoformula of ascorbic acid is C₆H₈O₆.
Pseudoformula is defined as an empirical formula which is twice or thrice the actual molecular formula of a compound. In this case, ascorbic acid contains 3. 407 mol of carbon, 4. 53 mol of hydrogen, and 3. 406 mol of oxygen.So, the simplest way to get the empirical formula of ascorbic acid is by using the formula weight, which is the sum of the atomic weights of the atoms present in the compound.
The atomic weights of carbon, hydrogen, and oxygen are 12.01, 1.01, and 16.00, respectively.
Calculate the formula weight of ascorbic acid:
(3.407 mol of C × 12.01 g/mol) + (4.53 mol of H × 1.01 g/mol) + (3.406 mol of O × 16.00 g/mol)
= 176.12 g/mol
The empirical formula for ascorbic acid is found by dividing each molar amount by the smallest molar amount, which is 3.406. Doing this will give you a ratio of the elements in the compound, which will be used to construct the empirical formula.
Carbon: 3.407 mol ÷ 3.406 mol = 1.000
Hydrogen: 4.53 mol ÷ 3.406 mol = 1.332
Oxygen: 3.406 mol ÷ 3.406 mol = 1.000
The empirical formula of ascorbic acid is CH1.0000O1.0000, which is not a valid formula. The formula must have whole number subscripts. By multiplying each subscript by the same number to obtain a whole number ratio, the empirical formula of ascorbic acid can be determined.
Multiply the subscripts by 6 to obtain a whole number ratio: C₆H₆O₆
Divide the subscripts by 6 to obtain the simplest whole number ratio: C₁H₁O₁
The simplest whole number ratio is equal to the molecular formula of ascorbic acid: C₆H₈O₆.
The pseudoformula of ascorbic acid is C₆H₈O₆.
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A second concern was discovered when a chemical dosing machine in the bottling line had a minor release during a change out of the chemical dimethyl dicarbonate (DMDC). The employees were evacuated without injury.
These two incidents led the winery to determine the level of risk and potential solutions. A risk assessment team was formed, and an assessment was performed. The team determined that the SO2 and DMDC exposure risks both presented multiple fatality – level risk and required immediate risk treatment.
Management sets the expectations, context, and objectives of the assessment. The risk assessment team was established that included the consultant as facilitator, the winemaker, assistant winemaker, cellar manager, operations manager, bottling
department manager, maintenance manager, and Health and Safety Executive (HSE) manager.
Data were collected regarding the SO2 and DMDC operations, equipment and instruments used, instructions, chemicals, and their SDSs, operator training, procedures, and available incident information. A search for similar events involving SO2 and DMDC were also conducted. Employees were interviewed to learn from their experiences, concerns, and suggestions.
The two procedures were observed to document and understand the sequence of tasks and potential risks associated with tasks. Photographs, tank quantities, room dimensions and configurations, distances to exits, means of egress, and other physical attributes were collected.
After reviewing the information, the potential concerns of fatalities or serious incidents were discussed. Workplace exposures such as pure SO2 releases and DMDC releases which present a potential for fatalities or serious incidents must be given the highest priority and controlled to an acceptable level. As a side note, the consultant explained that unlike like less-serious workplace incident rates, fatality, and serious incident/injury rates have not declined and do require serious attention. FSI exposures that can result in environmental releases, explosions, and disasters have been found to involve some of the following factors (14):
• Unusual and nonroutine work
• Nonproduction tasks
• Facility modification or construction activities
• Shutdowns and startups for repair and maintenance tasks
• Exposure to high-energy sources (e.g. electrical, steam, pneumatic, chemical) • Upsets (situations going from normal to abnormal).
The risk assessment team, consisting of various members from different departments within the winery, conducted a comprehensive assessment of the potential risks associated with sulfur dioxide (SO2) and dimethyl dicarbonate (DMDC) operations.
Data was collected regarding the equipment, chemicals, procedures, training, incident history, and physical attributes of the workplace. The team also interviewed employees to gather their experiences, concerns, and suggestions.
After reviewing the collected information, the team identified the potential risks of fatalities or serious incidents related to workplace exposures of SO2 and DMDC. These risks were considered of utmost priority and required immediate control to ensure an acceptable level of safety. The consultant highlighted the importance of addressing these serious risks, as fatality and serious incident rates have not shown a decline and demand serious attention.
Factors contributing to the potential for environmental releases, explosions, and disasters were identified, including unusual and nonroutine work, nonproduction tasks, facility modification or construction activities, shutdowns and startups for repair and maintenance tasks, exposure to high-energy sources (such as electrical, steam, pneumatic, chemical), and situations transitioning from normal to abnormal (upsets).
Based on the assessment and the identified risks, it is crucial for the winery to implement effective control measures to minimize the potential for fatalities, serious incidents, and environmental disasters. These measures may include improving procedures, enhancing operator training, implementing stricter safety protocols, and ensuring proper handling and storage of chemicals.
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In an experiment, the variable that the experimenter changes is known as the
Independent Variable
True
False
this question answer is true
Answer:
true
Explanation:
it's true....independent variable is what is changed by the experimenter
Give examples of environmental concerns I will be taking into account by a mining company when it created a reclamation plan for a mine site
Answer:
The major potential environmental impacts associated with mining and associated mineral processing operations are related to erosion-prone landscapes, soil and water quality, and air quality.
Explanation:
Plants were native to the mining site and making sure the soil is healthy are the examples of environmental concerns taking into account by a mining company when it created a reclamation plan for a mine site.
What are environmental concerns?Land use may change as a result of mine exploration, construction, operation, and maintenance.
It may also have detrimental effects on the environment, such as deforestation, erosion, contamination of nearby streams and wetlands, alteration of soil profiles, increased noise levels, and dust.
Mining is a naturally invasive operation that can alter the landscape over a much wider region than the actual mining site.
Therefore, years after a mine closes, the impacts of this harm can still be felt, adding to greenhouse gas emissions, causing flora and fauna to die, and eroding the land and environment.
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Calculate the temperature needed to change 75 liters at 27C to 105 liters. The pressure is a constant value throughout this experiment
To calculate the temperature needed to change 75 liters at 27°C to 105 liters, the information about the pressure is required. Please provide the constant pressure value for an accurate calculation.
To calculate the temperature needed to change the volume of a gas from 75 liters at 27°C to 105 liters while keeping the pressure constant, we need to use the ideal gas law. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
Since the pressure is constant throughout the experiment, we can rewrite the ideal gas law as V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Plugging in the values, we have:
V1 = 75 liters
T1 = 27°C + 273.15 (converted to Kelvin)
V2 = 105 liters
T2 = ?
Using the equation V1/T1 = V2/T2, we can rearrange it to solve for T2:
T2 = (V2 * T1) / V1
Substituting the values, we can calculate the temperature T2 needed to change the volume from 75 liters at 27°C to 105 liters.
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Which of the following best describes a hypothesis?
A. An explanation that is supported by scientific data
B. A proven explanation for experimental results
c. A possible answer to a scientific question
D. A question about the natural world
please help i will give brainliest
Which element has the smallest atoms?
A. francium (Fr), located at the left end of the seventh period
B. neon (Ne), located at the right end of the second period
C. lithium (Li), located at the left end of the second period
D. boron (B), located in the middle of the second period
suppose that you add 25.6 g of an unknown molecular compound to 0.250 kg of benzene, which has a k f of 5.12 oc/m. with the added solute, you find that there is a freezing point depression of 3.54 oc compared to pure benzene. what is the molar mass (in g/mol) of the unknown compound?
If we add 25.6 g of an unknown molecular compound to the 0.250 kg of benzene, the molar mass of the unknown compound is 148.8 g/mol.
The Molality of the compound is given as :
ΔT = i Kf m
Where,
ΔT = freezing point depression = 3.54 °C
i = Van't Hoff factor of the Benzene = 1
Kf = constant of the freezing = 5.12 °C/m
m = molality = ?
m = ΔT / i Kf
m = 3.54 / 1 × 5.12
m = 0.69 mol
molality = moles / mass of benzene
moles = 0.172
The molar mass = mass / moles
The molar mass = 25.6 / 0.172
The molar mass = 148.8 g/mol
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A motorcycle starts at rest and moves a distance of 460m .
If it has constant acceleration of 4m55 m/s2 what is its final velocity
Answer:
The final velocity of the motorcycle is 64.7 m/s
Note: Since the value for the acceleration of the motorcycle is not clear, it is assumed to be 4.55 m/s² in the calculation
Explanation:
Using the equation of motion that contains all the given values in the question: v² = u² + 2as
where v is final velocity; u is initial velocity; a is acceleration; s is horizontal distance
v = ?, u = 0 (since the motorcycle starts from rest), a = 4.55 m/s², s = 460 m
v² = 0² + 2 * 4.55 * 460
v² = 4186
take square root of both sides
v = 64.7 m/s
Therefore, the final velocity of the motorcycle is 64.7 m/s
Which d electron count can exhibit both high-spin and low-spin states in an octahedral complex?
d5 electron count can exhibit both high-spin and low-spin states in an octahedral complex
Electron is a free, open-source software framework developed and maintained by GitHub. This framework is designed to build desktop applications using web technologies rendered using a variant of the Chromium browser engine and a backend using the Node.js runtime environment.
Electrons are negatively charged subatomic particles that are free to (not) bond with atoms. Electrons bound to atoms are one of the three main types of particles within atoms. The other two are protons and neutrons. Electrons, protons, and neutrons together form the nucleus.
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Calculate pBa when 75.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+. For the buffered pH of 10, alpha(Y4+) = 0.30. Kf = 7.59 × 107 for BaY2−.
Select one:
a. 4.33
b. 7.06
c. 7.58
d. 4.59
the value of pBa is approximately 1.66.
Volume of 0.1 M EDTA solution = 75.00 mL
Volume of 0.1 M Ba2+ solution = 50.00 mL
alpha(Y4+) = 0.30 (fraction of EDTA that forms the Y4- complex)
Kf = 7.59 × 10^7 (formation constant for BaY2- complex)
First, let's calculate the moles of EDTA and Ba2+ in the solutions:
Moles of EDTA = (Volume of EDTA solution in L) * (Concentration of EDTA)
= (75.00 mL / 1000 mL/L) * 0.1 M
= 0.0075 mol
Moles of Ba2+ = (Volume of Ba2+ solution in L) * (Concentration of Ba2+)
= (50.00 mL / 1000 mL/L) * 0.1 M
= 0.0050 mol
Next, we determine the moles of BaY2- complex formed by reacting Ba2+ with EDTA:
Moles of BaY2- = alpha(Y4+) * Moles of EDTA
= 0.30 * 0.0075 mol
= 0.00225 mol
Since the stoichiometric ratio between Ba2+ and BaY2- is 1:1, the concentration of Ba2+ remaining in solution is equal to the concentration of Ba2+ initially minus the moles of BaY2- formed:
Concentration of Ba2+ remaining = (Moles of Ba2+ - Moles of BaY2+) / (Total volume of solution in L)
= (0.0050 mol - 0.00225 mol) / (0.075 L + 0.050 L)
= 0.00275 mol / 0.125 L
= 0.022 M
Finally, we can calculate pBa by taking the negative logarithm (base 10) of the concentration of Ba2+ remaining:
pBa = -log10(Concentration of Ba2+ remaining)
= -log10(0.022)
≈ 1.66
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The moon has less gravity than we have here on Earth TRUE OR FALSE
Explanation:
obviously true the gravity of moon is 1/6th of the Earth
4) I need Two that apply. How does a cloud form?
Answer:
A cloud occurs when evaporated water condenses into visible water or ice. The three factors required in forming a cloud include: moisture, condensation, and temperature.
Explanation:
Lana drew the diagram below to model asexual reproduction. Based on Lana's diagram, which statement explains the results of asexual reproduction? A. The offspring are not genetically identical to the parent, because each offspring receives only half of the chromosomes from a single parent. B. The offspring are not genetically identical to the parents, because two parents each contribute half of their chromosomes to each offspring. C. The offspring are genetically identical to the parent, because each offspring receives a complete copy of a single parent's chromosomes. D. The offspring are genetically identical to the parents, because two parents each contribute a complete copy of their chromosomes to each offspring.
Based on Lana's diagram, the correct statement that explains the results of asexual reproduction is C. The offspring are genetically identical to the parent, because each offspring receives a complete copy of a single parent's chromosomes.
What happens in asexual reproduction?In the diagram, the parent cell divides into two identical daughter cells, each of which contains a complete copy of the parent cell's genetic material.
This type of reproduction, where a single parent produces offspring that are genetically identical to itself, is called asexual reproduction. It is the process by which many unicellular organisms, such as bacteria and some protists, reproduce.
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the mass in grams of one mole of any pure substance is called its
Answer:
this
Explanation:
Molar mass also called Atomic mass is the atomic weight. You can find this on the periodic table.
-
-
The answer would be
A. the mass in grams of one mole of a substance
When white light strikes this object, the light is completely absorbed, with none of it transmitted or reflected. Which type of object
could this be? (1 point)
O a black piece of paper
O a white sheet of plastic
O a green long-sleeved shirt
O a clear windowpane
The type of object that completely absorbs white light would be a black piece of paper. First option.
What are black bodies?Black bodies are objects that completely absorb all the components of white light.
White light is a mixture of different colors of light, each with a different wavelength and frequency.
When white light strikes a black body, the black body absorbs a large fraction of all the different colors, regardless of their wavelength. This absorption causes the light to be transformed into thermal energy, which increases the temperature of the black body.
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True or false: Denaturation of an enzyme increases the rate at which it may catalyze a chemical reaction.
Consider the balanced equation. N2 + 3H2 Right arrow. 2NH3 What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3? Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100.. 17.8% 33.3% 35.7% 53.4%
Answer:
53.4%
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar mass of H2 = 2 × 1.01 = 2.02 g/mol
Mass of H2 from the balanced equation = 3 × 2.02 = 6.06 g
Molar mass of NH3 = 14 + (3×1.01)
= 14 + 3.03
= 17.03 g/mol
Mass of NH3 from the balanced equation = 2 × 17.03 = 34.06 g
Summary:
From the balanced equation above,
6.06 g of H2 reacted to produce 34.06 g of NH3.
Next, the shall determine the theoretical yield of NH3. This can be obtained as follow:
From the balanced equation above,
6.06 g of H2 reacted to produce 34.06 g of NH3.
Therefore, 26.3 g of H2 will react to produce = (26.3 × 34.06)/6.06 = 147.82 g of NH3.
Thus, the theoretical yield of NH3 is 147.82 g
Finally, we shall determine the percentage yield of NH3. This is illustrated below:
Actual yield = 79 g
Theoretical yield = 147.82 g
Percentage yield =.?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield = 79/147.82 × 100
Percentage yield of NH3 = 53.4%
Answer:
The answer is D(53.4%)
Explanation:
Just took test hopes this help
What percentage of the earth's surface is covered by marine habitats?
Answer: 70 percent
About 97 percent of Earth's water is in the ocean.
The ocean covers more than 70 percent of the surface of our planet. It's hard to imagine, but about 97 percent of the Earth's water can be found in our ocean. Of the tiny percentage that's not in the ocean, about two percent is frozen up in glaciers and ice caps.
Hope this helps...... Stay safe and have a Merry Christmas!!!!!!!!!!! :D
Explanation:
What is the volume of 220g of an object with a density of 55g/cm^3
use dimensional analysis and set up your equation like so:
\(=220g\ x\ \frac{cm^3}{55g}\)
the unit grams would cancel which leaves the unit of cm³, which is the unit for volume in this case.
The volume of 220g of an object with a density of 55g/cm³ is 4 cm³.
A tank contains 100 gal of brine made by dissolving 60 lb of salt in water. Saltwater containing 1 lb of salt per gallon runs in at the rate of 2 gal/min and the well-stirred mixture runs out at the rate of 3 gal/min. Find the amount of salt in the tank after 30 minutes.
The presented statement states that 52.5 lbs. of salt had been consumed after 30 minutes.
What three kind of salt are there?Salts come in a variety of forms, such as a double salt, corrosive salt, or basic salt. A salt that forms when a potent acid and a feeble base interact is referred to as a "acidic salt."
There seem to be initially 100 gal with in tank. The amount of salt is 60 lbs. The tank has a salt content of 60/100, or 0.6 pounds per gal.
One pound per gallon of water is coming in.
Because the combination is continually varying, the concentration of the water that comes out is uncertain.
Each minute, the tank is dropping 1 gal. (3-2=1)
Utilize this information to configure three functions for volume, saturation, and salt content: V(t), S(t), and C(t).
V (t) = 100 - t
s (t) = 60 plus 2t - 3t * C (t)
C (t) = s(t)/V(t) = 60 + 2t - 3t * C(t)/ 100 - t
Solve for C with a little mathematics (t)
C (t) = 60 + 2t / 100 + 2t
Now, we are aware of the tank's salt content at all times.
Evaluate at t = 30
C (30) = 60 + 2(30)/100 - 2(30) = 120/160 = 3/4
s(30) = 60 + 2(30) - 3(30)(3/4) = 120 - 90(3/4) = 52.5
Consequently, 52.5 lbs of salt had been consumed within 30 minutes.
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Use the following words: air, expand, compress, pressure, mass, particle, weight, syringe, balloon balance
WE DEMONSTRATED THAT AIR HAS MASS, BECAUSE WHEN WE _____________________________, WE COLLECTED EVIDENCE THAT ___________________________. WE ALSO ______________________________________________
We collected the evidence that air has mass through balloon balance experiment . We also got the evidence that the Compressed air weighs more than atmospheric air.
All types of gases consist of a certain amount of mass depending upon its amount present in a particular substance or container .Gases are invisible to the human eye but we feel them through our skin because of the pressure they exhibit.
We can prove that the air has mass by performing a simple science experiment that is the balloon balance science experiment.
Take a simple balance machine and tie equal sized inflate balloons on either side of its lever. You will observe both the balloons hang at the same height from the ground. Now, blast any one of the inflated balloon through a syringe and observe that the lever on the deflated balloon side tilts and moves upwards. While the inflated balloon still goes down because it has air which reveals mass. In this way we can prove that air has mass.
Or we can calculate mass of a gas through ideal gas equation PV=nRT
Compressed air exhibits weight more than the surrounding air moving freely outside through balloon balance experiment. Let us see how much weight it reveals.
Compressed air weighs more than atmospheric air and hence a certain amount of compressed air weighs more than the same amount of regular gas.
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an automobile gasoline tank holds 19.0 gal when full. how many pounds of gasoline will it hold if the gasoline has a density of 0.737 g/ml ?
The automobile gasoline tank will hold 59.05 pounds of gasoline.
Given that the volume of the gasoline tank of the automobile is 19.0 gal. and the density of the gasoline is 0.737 g/mL. So, we need to calculate the number of pounds of gasoline the tank can hold.The relationship between volume and density is as follows;Density= Mass/ VolumeRearranging the equation gives; Mass= Density × VolumeSubstitute the given values in the above equation;Mass = 0.737 g/mL × 19.0 gal × 3.78541 L/galWe know that; 1 L = 1 dm³Thus,Mass = 0.737 g/mL × 19.0 gal × 3.78541 dm³/LMass = 53.70 kgThe given mass is in kg and we need to convert it into pounds;1 kg = 2.20462 poundsThus,Mass = 53.70 kg × 2.20462 pounds/kgMass = 118.4 pounds therefore, the automobile gasoline tank will hold 59.05 pounds of gasoline.
Given that the volume of the gasoline tank of the automobile is 19.0 gal. and the density of the gasoline is 0.737 g/mL. So, we need to calculate the number of pounds of gasoline the tank can hold.The relationship between volume and density is as follows;Density= Mass/ VolumeThe above equation can be rearranged as; Mass= Density × Volume Substitute the given values in the above equation;Mass = 0.737 g/mL × 19.0 gal × 3.78541 L/galMass = 53.70 kgThe given mass is in kg and we need to convert it into pounds;1 kg = 2.20462 poundsThus,Mass = 53.70 kg × 2.20462 pounds/kgMass = 118.4 pounds therefore, the automobile gasoline tank will hold 59.05 pounds of gasoline.
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I need help pls help ASAP I'll give brainliest
Answer: A
Explanation:
Answer:
The answer is A
Explanation:
write its IUPAC name?
Answer:
Hello
please make sure that there is any group which is join with alkane
¿How do the products of the reaction to the phenol red test and the splint test? Please help me it's for today.! :((
Answer:
The answer in this question is show you made Sodium Hydroxide and Hydrogen Gas.In order to do the products of the reaction relate to the phenol red test and the splint test you need to show that you made Sodium Hydroxide and Hydrogen Gas. Show that you made Sodium Hydroxide and Hydrogen Gas so that the products of the reaction relate to the phenol red test and the splint test.
When writing 6,020,000 in scientific notation, it is correctly written as:
1. 6.02 x 10 ^6
2. 6.02 x 10 ^-6
3. 0.602 x 10 ^-8
4. 602. x 10 ^6
Answer:
6.02*10^6 is ans
it mean (1)is correct
Air is cooling at night. The frost point (temperature at which RH with respect to ice reaches 100%) is reached at T = -10 degree Celsius. a) What is the RH (normal RH with respect to liquid water) at this point? b) Upon further cooling the air reaches a temperature of T =-11 degree Celsius Kaolinite particles of 200 nm diameter are present. Do you expect ice particles to form? If yes, do they form via deposition nucleation or condensation of droplets followed by freezing? Briefly explain your answer. c) Upon even further cooling the air reaches a temperature of T = -12 degree Celsius. Same question as before: Do you expect ice particles to form now? If yes, do they form via deposition nucleation or condensation of droplets followed by freezing? Briefly explain your answer. Equilibrium vapor pressures may be calculated or taken from the table below. t/°C 0 -1 -2 -3 -4 -5 -6 -7 -8 -9 - 10 -11 -12 -13 T/ Keow /Pa 273.15 611.2 272.15 568.2 271.15 527.9 270.15 490.2 269.15 454.8 268.15 421.8 267.15 390.9 266.15 362.1 265.15 335.1 264.15 310.0 263.15 286.5 262.15 264.7 261.15 244.3 260.15 225.4 259.15 207.8 258.15 191.4 e oi/Pa 611.2 562.7 517.7 476.1 437.5 401.8 368.7 338.2 310.0 283.9 259.9 237.7 217.3 198.5 181.2 165.3 - 14 - 15 Equilibrium vapor pressures with respect to water (eow) and with respect to ice (coi).
The equilibrium vapor pressure with respect to water (eow) is 259.9 Pa. assume that saturation vapor pressure is same as equilibrium vapor pressure.
Therefore, the RH at the frost point is
RH = (eow / saturation vapor pressure) × 100
= (259.9 Pa / 259.9 Pa) × 100
= 100%
b) At T = -11 °C, we need to compare the equilibrium vapor pressure with respect to water (eow) and the equilibrium vapor pressure with respect to ice (coi) to determine if ice particles will form. From the given table, at T = -11 °C, the equilibrium vapor pressure with respect to water (eow) is 237.7 Pa, and the equilibrium vapor pressure with respect to ice (coi) is 165.3 Pa.
The air is supersaturated with respect to ice, and the presence of Kaolinite particles can provide surfaces for water droplets to condense onto, leading to the formation of ice particles.
c) At T = -12 °C, we compare the equilibrium vapor pressure with respect to water (eow) and the equilibrium vapor pressure with respect to ice (coi). From the given table, at T = -12 °C, the equilibrium vapor pressure with respect to water (eow) is 217.3 Pa, and the equilibrium vapor pressure with respect to ice (coi) is 181.2 Pa.
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why are fish lucky that water particles expand as they hit a temperature of 0°c?
Answer:
it is a result of hydrogen bonds present within water molecules.
Explanation:
when the water is transformed to ice at zero degrees Celsius, the water molecules are in crystal lattice in a structure that has a lot of empty space around each molecule.
Determine the moles in 3.45 x1024 formula units (particles) of Sodium Chloride, NaCl?
Answer:
5,73mol
Explanation:
N=n*Na
3.45*10^24=n*6.02*10^23
n=\(\frac{3.45*10^{24}}{6.02*10^{23}}\)
n=5,73mol
The volume and amount of aflium gas remains constant.
Under these conditions, aflium gas has an initial pressure of
635 torr when it is at 67.0°C. What is the new temperature if
the pressure drops to 250 torr?
Answer:
133.85 K
Explanation:
Initial pressure, P₁ = 635 torr
Initial temperature, T₁ = 67.0°C = 340 K
Final pressure, P₂ = 250 torr
We need to find the new temperature. The relation between temperature and pressure is given by :
\(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{250\times 340}{635}\\\\T_2=133.85\ K\)
So, the new temperature is equal to 133.85 K.