The number of moles of H₂O in the thimble of water is approximately 0.00664 mol. The answer is option B) 6.6 x 10⁻³
The number of molecules in a substance is related to Avogadro's number (6.022 x 10²³ molecules/mol) by the formula:
number of molecules = number of moles x Avogadro's number
We are given that there are 4.0 x 10²¹ molecules of water. To find the number of moles of water, we need to rearrange the formula and solve for moles:
number of moles = number of molecules / Avogadro's number
number of moles = 4.0 x 10²¹ molecules / 6.022 x 10²³ molecules/mol
number of moles = 0.00664 mol
The number of moles of H₂O in the thimble of water is approximately 0.00664 mol. The answer is option B) 6.6 x 10⁻³
The quantity of substance in a system is represented by a measurement unit called a mole. A material is said to have one mole if there are exactly as many atoms, molecules, or ions in one mole of it as there are in 12 grams of carbon-12.
It is practical to quantify substance concentrations in chemical reactions and other chemical processes using moles. It enables chemists to deal with quantifiable amounts of substances, like grams, and to quickly convert these amounts to a standard unit of measurement, like moles.
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draw and explain the energy profile diagram along with various possible conformations of cyclohexane
Answer:
we're is the diagram I don't see it
Gallium is a solid metal at room temperature but melts at 29.9 °C. If you hold gallium in your hand, it melts from body heat. How much heat must 2.5 g of gallium absorb from your hand to raise its temperature from 25.0 °C to 29.9 °C? The specific heat capacity of gallium is 0.372 J/g
In order for 2.5 g of gallium to transform from a solid state at 25.0 °C to a liquid state at 29.9 °C, it would require absorbing 19.56 J of heat from your hand.
Why is gallium utilised in high temperature applications?Only gallium has a low melting point of 29.7°C and a high boiling point of 1500–2000°C. Together with these peculiar characteristics, it also exhibits undercooling (20 °C or below), which would make it a perfect thermometric liquid if not for its propensity to wet quartz and glass surfaces.
Q1 = m × c × ΔT
Q1 = 2.5 g × 0.372 J/g·°C × (29.9 °C - 25.0 °C)
Q1 = 5.58 J
Q2 = m × ΔH_fusion
Q2 = 2.5 g × 5.59 J/g
Q2 = 13.98 J
Q_total = Q1 + Q2
Q_total = 5.58 J + 13.98 J
Q_total = 19.56 J
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Given the thermochemical equations
X2+3Y2⟶2XY3Δ1=−370 kJ
X2+2Z2⟶2XZ2Δ2=−130 kJ
2Y2+Z2⟶2Y2ZΔ3=−220 kJ
Calculate the change in enthalpy for the reaction.
4XY3+7Z2⟶6Y2Z+4XZ2
The change in enthalpy for the given reaction is +330 kJ.
To calculate the change in enthalpy for the reaction:
4XY3 + 7Z2 ⟶ 6Y2Z + 4XZ2
we need to use the Hess's Law, which states that if a chemical reaction can be expressed as the sum of several stepwise reactions, the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.
We can write the reaction in terms of the given thermochemical equations as follows:
4XY3 ⟶ 2X2 + 6Y2 (reverse of equation 1, with ΔH = +370 kJ)
2X2 + 4XZ2 ⟶ 8XY3 (multiply equation 2 by 2, with ΔH = -2×130 kJ = -260 kJ)
2Y2 + Z2 ⟶ 2Y2Z (reverse of equation 3, with ΔH = +220 kJ)
Adding these three equations gives:
4XY3 + 7Z2 ⟶ 6Y2Z + 4XZ2 (with ΔH = 370 kJ - 260 kJ + 220 kJ = +330 kJ).
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name two acids used in the manufacture of fertilizers
Two acids commonly used in the manufacture of fertilizers are sulfuric acid and phosphoric acid
According to Beers Law, A-&bc; what should the slope and the intercept be in calibration curve for 'plot
If the route length is constant, the intercept of the calibration curve should be zero, and the slope should be proportional to the molar absorptivity times the path length.
What is the calibration graph for Beer's law's slope?Beer's law, which links absorbance to concentration, is represented by a linear function. The molar attenuation coefficient multiplied by the cuvette width, or pathlength—1 centimetre in this lab—gives you the slope of your calibration curve. To find the concentration, rearrange the linear solution.
According to Beer's Law, a solution's absorbance (A) is inversely proportionate to its concentration.
A = εcl
The molar absorptivity () times the route length (l) are represented by the calibration curve's slope (m):
m = εl
If the route length is constant, the calibration curve's intercept (b) should be zero because there shouldn't be any absorbance at zero concentration:
b = 0
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Identify the system and surroundings!! And c, d, and e
The candle and the wax represent the structure. Everything that is not a part of the flame or the wax, such as the air, items in the room, and anything else, is considered to be the surroundings. As the liquid wax solidifies, heat is transferred from the wax to the environment.
What kind of response occurs when candle wax melts?Wax fire causes a chemical change while wax melting causes a physical change: Wax changes from a solid to a liquid on its own when it melts. Only the physical state of a substance alters in the described procedure.
When candle wax melts, where does it go?The New York Times claims that the majority of a candle's material truly evaporates into the air. Actually, the wax rises as it begins to melt and pool around the cotton flame of the candle.
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Calculate the number of atoms in a 7.08 x 103 g sample of aluminum.
Answer:
There are 4.59×1024 4.59 × 10 24 atoms of Al in 7.63 moles of Al.
ASAP PLEASE!!!
1. Claim: How are elements arranged on the periodic table in terms of valence
electrons? (2 points)
2. Evidence: Use the Element symbol provided to create a Bohr/ Orbital Model for
each. Use the PhET simulation to work through each. Complete the table below.
Include a picture of each that you either snip from the simulation or draw. We
The periodic table is arranged in such a way that elements with similar valence electron configurations are placed in the same group or column.
How are elements arranged on the periodic table in terms of valence?Valence electrons are the outermost electrons in an atom, and they play a critical role in determining the chemical properties of an element.
The elements in each column of the periodic table have the same number of valence electrons, which gives them similar chemical properties
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Why do sex-linked traits follow different patterns of inheritance than other
traits?
Answer: Sex-linked traits are different from other traits because they are located on the sex chromosomes. Since males and females have different numbers of these chromosomes, the inheritance of sex-linked traits is different between the two sexes. This means that certain traits are more likely to be expressed in males than in females, and that females can inherit these traits from both parents while males can only inherit them from their mother.
Explanation: Sex-linked traits follow different patterns of inheritance than other traits because they are located on the sex chromosomes (X and Y) rather than on the autosomes (non-sex chromosomes). Since males have one X and one Y chromosome, and females have two X chromosomes, the inheritance of sex-linked traits is affected by the sex of the individual and the number of copies of the gene present.
The X chromosome contains many more genes than the Y chromosome and therefore, most sex-linked traits are inherited in a dominant or recessive manner on the X chromosome. In females, the presence of two copies of the X chromosome allows for a greater range of genetic variability, as both copies can potentially express different alleles. In males, however, the presence of only one X chromosome means that any alleles on that chromosome will be expressed, regardless of whether they are dominant or recessive. This is why sex-linked traits are more commonly expressed in males than in females.
Additionally, since males only inherit one X chromosome from their mother, they can only inherit X-linked traits from her. Females, on the other hand, inherit one X chromosome from each parent, which means they can inherit X-linked traits from both their mother and father. This can affect the frequency and distribution of certain sex-linked traits in a population.
Overall, the unique inheritance patterns of sex-linked traits are a consequence of their location on the sex chromosomes and the differences in chromosome inheritance between males and females.
Consider the following intermediate chemical equations.
2 equations: first: upper C (s) plus one-half upper O subscript 2 (g) right arrow upper C upper O (g). Second: upper C upper O (g) plus one-half upper O subscript 2 (g) right arrow upper C upper O subscript 2 (g).
When you form the final chemical equation, what should you do with CO?
The CO gas produced in the first equation is used in the second equation to produce CO2 in the final equation.
In the intermediate equations, solid carbon (C) and molecular oxygen (O2) are transformed into gaseous carbon monoxide (CO), which is then reacted with more oxygen to produce carbon dioxide (CO2).
The final chemical equation can be created by combining the intermediate equations and cancelling out the intermediate reactant and product (CO and O2) to obtain the overall balanced equation for the reaction:
C(g) + O2(s) = CO2 (g)
Thus, the CO generated in the first equation is consumed in the second equation and does not show up in the third and final equation. The two intermediate reactions' combined outcome is represented by the final equation, which only includes the reactants (C and O2) and product (CO2).
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how many moles of helium gas are present in a 11.2l container at 298k and 1.35 atm the gas constant r =0.0821 atm/k mol
Answer
The first step is to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Rearranging the equation to solve for n, we get:
n = PV/RT
Substituting the given values, we get:
n = (1.35 atm) x (11.2 L) / [(0.0821 atm·L/mol·K) x (298 K)]
n = 0.553 mol
Therefore, there are 0.553 moles of helium gas present in the 11.2L container at 298K and 1.35 atm.
Question 1
Imagine yourself in the shoes of Dimitri Mendeleev. You are provided with two sets of cards that list properties of various
elements. These cards resemble the cards used by Mendeleev when he grouped elements. One set of cards lists the names
of known elements and their properties, while the other set of cards lists the properties of a few unknown elements. These
sets are shown below.
Known Elements Set
K
Physical State: solid
Density: 0.86 g/cm³
Conductivity: good
Physical State: solid
Density: 4.93 g/cm³
Conductivity: very poor
Solubility (H₂O): reacts rapidly Solubility (H₂O): negligible
Melting Point: 63°C
Melting Point: 113.5°C
Ge
Physical State: solid
Density: 5.32 g/cm³
Conductivity: fair
Solubility (H₂O): none
Melting Point: 937°C
CI
Ba
Physical State: solid
Density: 3.6 g/cm³
Conductivity: good
Au
Rb
Physical State: solid
Density: 19.3 g/cm³
Conductivity: excellent
Solubility (H₂O): None
Melting Point: 1064°℃
Physical State: gas
Density: 0.00178 g/cm³
Conductivity: none
Solubility (H₂O): reacts strongly Solubility (H₂O): negligible
Melting Point: 710°C
Melting Point: -189.2°C
Ag
Ar
A
What is the pH of an aqueous solution with a hydrogen ion concentration of [H+]=3.1×10−9 M?
the pH of the aqueous solution is 8.51 with a hydrogen ion concentration of [H+]=3.1×[tex]10^-9[/tex]M
The pH of the aqueous solution can be calculated using the formula:
pH = -log[H+]
where [H+] is the hydrogen ion concentration of the solution.
Substituting the given value, we get:
pH = -log(3.1×[tex]10^-9[/tex])
pH = 8.51
An aqueous solution is one in which water serves as the solvent. It is utilised in a variety of applications, including analytical chemistry, biochemistry, and industrial chemistry. It is the most prevalent kind of solution used in chemical reactions. Water serves as both the solvent and the solute in an aqueous solution, where the solute is often a solid, liquid, or gas. Due to its high polarity and capacity to make hydrogen bonds with other molecules, water is an excellent solvent that can dissolve a variety of materials, including polar molecules and ionic compounds. Acid-base reactions, redox reactions, and precipitation reactions are just a few of the numerous chemical processes that take place in aqueous solutions. A variety of variables can have an impact on an aqueous solution's characteristics.
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Select the two true statements about natural selection. Natural selection makes less advantageous variations become more advantageous over many generations. A population's environment affects the outcome of natural selection. Natural selection always makes a population gain new advantageous variations. Natural selection can change which variations are more common in a population over time. Submit
Answer:
The two true statements about natural selection are:
A population's environment affects the outcome of natural selection.
Natural selection can change which variations are more common in a population over time.
Natural selection acts on the existing variations within a population and favors those that confer a survival or reproductive advantage in a particular environment. Over time, the advantageous variations become more prevalent, and less advantageous ones may become less common or disappear from the population. However, natural selection does not necessarily create new variations; rather, it acts on the genetic variation that already exists within a population.
If electrons jump from a lower energy shell to a higher energy shell, they are said to be ____
Answer:
they are said to be excited.
Explanation:
excitation is a fundamental concept in atomic physics that helps explain many of the properties and behaviors of matter on both the macroscopic and microscopic scales.
ASAP PLEASE!!!3. Reasoning: Explain how the evidence supports your claim. Explain how the
evidence from your data table shows the trends for valence electrons for both
groups and periods on the periodic table. (4 points)
The data table supports the idea that valence electrons affect the chemical characteristics of elements and may be used to forecast chemical reactions by showing how the amount of valence electrons follows different patterns on the periodic table.
How is the number of valence electrons represented in the periodic table?The number of valence electrons in groups 1-2 and 13–18 rises by one from one element to the next throughout each row, or period, of the periodic table.
What are valence electrons and valence valence?The ability of an atom to make covalent bonds with other atoms is known as its "valency." Valence electrons, on the other hand, are the quantity of electrons required in a compound's entire outer shell in order for bonds to form.
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How many percent by mass of mercury are there in a sample of tap water with a mass of 750 g containing 2.2g of Hg?
Answer:
Divide the mass of the water lost by the mass of hydrate and multiply by 100. The theoretical (actual) percent hydration (percent water) can be calculated from the formula of the hydrate by dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and multiplying by 100.
A 0.231 M solution of acetate has a pOH of 4.90. What is the Kb of acetate?
Explanation:
o solve this problem, we need to use the relation between pOH, pH, and the dissociation constant of the conjugate base of the weak acid, which is given by:
Kb = Kw / Ka
where Kb is the dissociation constant of the conjugate base, Ka is the dissociation constant of the weak acid, and Kw is the ion product constant of water, which is 1.0 x 10^-14 at 25°C.
First, we need to find the pH of the solution, since we know the pOH:
pH + pOH = 14
pH = 14 - 4.90 = 9.10
The weak acid in this case is the acetic acid (CH3COOH), which dissociates in water according to the equation:
CH3COOH + H2O ↔ CH3COO- + H3O+
The dissociation constant of acetic acid (Ka) is 1.8 x 10^-5 at 25°C. We can use this value and the relation between Ka and Kb to find Kb:
Kb = Kw / Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10
Therefore, the Kb of acetate is 5.56 x 10^-10.
Which of these prostheses is used to support blood flow through an artery?
OA. A nanotube
OB. A passive prosthesis
OC. A stent
OD. A pacemaker
For each of the following equilibria, write the equilibrium constant expression for Kc.
1. BaSO4(s) <---->Ba2+(aq) + SO42-(aq)
2. CH3COOH (aq) + H2O (l) <--->CH3COO- (aq) + H3O+ (aq)
Equilibria are chemical reactions that happen with a change in the concentration of the reactants or products. If the forward reaction is favored, a greater concentration of the product will be formed than the reactant(s), and the reaction is said to be favored. If the reverse reaction is favored, a greater concentration of the reactant(s) will be formed and thus the reaction is said to be favored. An equilibrium constant (constant K), is a number that describes the ratio of products to reactants at an equilibrium. K is also referred to as the equilibrium coefficient.
Answer:
1. Kc = [Ba2+][SO42-]
2. Kc = [CH3COO-][H3O+]/[CH3COOH]
Explanation:
The equilibrium constant expression for Kc is the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients.
1. For the first equilibrium, BaSO4(s) <---->Ba2+(aq) + SO42-(aq), the equilibrium constant expression for Kc is: Kc = [Ba2+][SO42-].
2. For the second equilibrium, CH3COOH (aq) + H2O (l) <—>CH3COO- (aq) + H3O+ (aq), the equilibrium constant expression for Kc is: Kc = [CH3COO-][H3O+]/[CH3COOH].
Note that the concentration of water is not included in the expression because it is a pure liquid and its concentration is considered constant.
Keq= 798 for the reaction:
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
In a particular mixture at equilibrium, [SO2]= 4.20 M and [SO3]=11.0 M.
Calculate the equilibrium concentration of O2 ([O2]) this mixture.
It's easy
Explanation:
We can use the equilibrium constant expression to calculate the equilibrium concentration of O2:
Kc = [SO3]^2 / ([SO2]^2 [O2])
At equilibrium, the value of Kc is constant, so we can use the equilibrium concentrations of SO2 and SO3 to solve for [O2]:
Kc = [SO3]^2 / ([SO2]^2 [O2])
Kc = (11.0 M)^2 / ((4.20 M)^2 [O2])
Simplifying:
[O2] = (11.0 M)^2 / (Kc (4.20 M)^2)
The value of Kc for this reaction is 4.67 × 10^1, as determined by experiment.
[O2] = (11.0 M)^2 / (4.67 × 10^1 (4.20 M)^2)
[O2] = 0.153 M
Therefore, the equilibrium concentration of O2 in this mixture is 0.153 M.
How are hybrid and electric cars related to air pollution
Predicting Products: Mg + K2SO4. (2 and 4 are coefficients)
The reaction between magnesium (Mg) and potassium sulfate (K2SO4) is a double displacement reaction that occurs in an aqueous solution. The products that will form depend on the solubility of the resulting compounds.
When Mg and K2SO4 react, magnesium sulfate (MgSO4) and potassium (K) will be produced. This is because the magnesium will displace the potassium ion in the potassium sulfate compound, resulting in the formation of magnesium sulfate and potassium metal.
The balanced chemical equation for this reaction is:
2Mg + K2SO4 → MgSO4 + 2K
It is important to note that this reaction will only occur if the magnesium is more reactive than the potassium in the solution. If the opposite were true, no reaction would occur.
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Q. The order of acidic strength of hydrogen halides is: acid HF < HCl CH-1 > Br¹>1-1 (b) F-1 < CH-1 < Br¹ Br¹> -1 (d) F-1> CI-1 < Br¹> -1. Hint: A strong acid has a weak conjugate base and vice versa.
Answer:
im just here grabbing points dont take me as rude or anything
Explanation:
yours appreciatively bye
Can you explain in detail how anion exchage occur in soil.
Answer:
With the adsorption of cations like zinc as Zn (OH)+ or ZnCl+ or both, the anion exchange is known to increase. The solid phase has an impact on the anions' concentration in the soil solution. Anions are negatively adsorbed as a result of the exchange complex's overall negative charge.
I NEED THIS DONE TODAY !!!!!!!!Electromagnetic Spectrum Lab Report
Destructions: In this virtual lab, you will use a virtual spectrometer to analyze astronomical
bodies in space. Record your hypothesis and spectrometric recular in the lab report below. You
will submit your completed report to your butructor.
Name and Title:
Include your name, instru
1
and name of lab.
Objectives (1):
In your own words, what is the purpose of this lab?
Hypothesis:
In this section, please include the predictions you developed during your lab activity. These
statements reflect your predicted outcomes for the experiment.
Procedure:
The materials and procedures are listed in your virtual lab. You do not need to repeat them here.
However, you should note if you experienced any errors or other factors that might affect your
outcome. Using your summary questions at the end of your virtual lab activity, please clearly
define the dependent and independent variables of the experiment.
Data:
Record the elements present in each unknown astronomical object. Be sure to indicate "yes" or
"no" for each element.
Hydrogen Helium Lithium Sodiam Carbon
Moon One
Moon Two
Planet One
Planet Two
Nitrogen
Conclusion:
Your conclusion will inchade a summary of the lab results and an interpretation of the results.
Please answer all questions in complete sentences using your own words.
1. Using two to three sentences, summarize what you investigated and observed in this lab
2. Astronomers use a wide variety of technology to explore space and the electromagnetic
spectrum; why do you believe it is essential to use many types of equipment when
studying space?
3. If carbon was the most common element found in the moons and planets, what element is
missing that would make them splat to Earth? Explain why. (Hint: Think about the
carbon cycle)
4.
We know that the electromagnetic spectrum uses wavelengths and frequencies to
determine a lot about outer space. How does it help us find out the make-up of stars?
5. Why might it be useful to determine the elements that a planet or moon is made up of?
PLEASE MAKE SURE YOU ANSWER THE HYPOTHESIS AND PROCEDURE QUESTION!!!!
Below contains the complete lab report on electromagnetic spectrum
The Lab ReportName: [Your Name]
Title: Electromagnetic Spectrum Lab Report
Instructor: [Instructor's Name]
Objectives:
The purpose of this lab is to analyze the elemental composition of different astronomical bodies using a virtual spectrometer and understand the importance of the electromagnetic spectrum in astronomical research.
Hypothesis:
I predict that the moons and planets will have varying compositions of elements, with hydrogen and helium being more common in gaseous bodies and heavier elements like carbon and nitrogen more common in rocky bodies.
Dependent variable: Presence of elements in astronomical bodies
Independent variable: Astronomical bodies (Moon One, Moon Two, Planet One, Planet Two)
Data:
[Please input your data for each object as per your virtual lab results]
Conclusion:
In this lab, I investigated the elemental composition of four different astronomical bodies using a virtual spectrometer and observed the presence or absence of various elements.
It is essential to use many types of equipment when studying space because different instruments can detect and analyze different aspects of the electromagnetic spectrum, providing a comprehensive understanding of the universe.
To make these moons and planets similar to Earth, oxygen would need to be present as it is a vital component of the carbon cycle and essential for life as we know it.
The electromagnetic spectrum helps us find out the makeup of stars by analyzing the emitted light, which contains information about the elements and their abundance within the star.
Determining the elements that a planet or moon is made up of helps us understand their formation, potential for life, and possible resources for future exploration or colonization.
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Formation of the Solar System Lab Report
Instructions: In this virtual lab, you will investigate the law of universal gravitation by
manipulating the size of the ster and the positions of planets within Solar System X. Record your
hypothesis and results in the lab report below. You will submit your completed report.
Name and Tide:
Include your name, instructor's name, date, and name of lab
Awi Ulivar, Mrs. Harmon, 3/30/21, Formation of the solar system lab report
Objectives():
is your own words, what is the purpose of this lab?
Hypothesia:
In this section, please inchade the if the statements you developed during your lab activity
These statements reflect your predicted outcomes for the experiment.
If the mass of the sun is Is at least
If the mass of the sun is 2x, at least one planet will fall into the habitable rone if I place a planet
and all planets will orbit the sun successfully.
arbits
If the
e planet will fall into the habitable zone if I place a planet
and all planets will orbit the sun successfully.
the sun i
at least one planci will fall into the habitable zone if I place a plant
and all planets will orbit the sun successfully.
Procedure:
The materials and procedures are listed in your virtual lah. You do not need to repeat them here.
However, you should note if you experienced any errors or other factors that might affect your
Using the summary quvons at the end of your virtual lab activity, please clearly define the
dependent and independent variables of the experiment
Data:
Record your observation statements from Space Academy.
When the mass of the sun is larger, Farth moves around the sun at a
pace.
When the mass of the sun is smaller, Farth moves around the sun at a
pace.
When Farth is closer to the sun, its orbit becomes
When Farth is farther from the sun, its orbit becomes
Example:
'smas I
MAT'S THIS
I Trial One
MAT'S HILL
1-Tial Two
For each trial, record the orbit manber of each planet from the sun. Be sure to indicate the
amber of planets in the habitable zone after each trial. Create a different configuration of
planets for each trial. An example has been supplied for you.
MAY
2x-Trial One
way's mass
2- Tial Two
mask
34-Trial One
WAY's mass
J-Tial Two
Orbit
Number
Planet One
Orbit
Number
Planet
Two
(faster, slower)
3
(faster, slower)
(faster, slower)
(faster, slower)
Orbit
Orbit
Number Number
Planet Planet
Three
Four
Number of Number of
planets in planets left
the
habitable
PODE
successful
orbit
Conclusion:
Your conclusion will include a ummary of the lab results and an interpretation of the results
Please awwer all questions in complete sentences using your own work
1. Using two to three sentences, semmarize what you investigated and observed in this lab
2. You completed three terra forming trials. Describe the how the sun's mass affects planets
in a solar system, Use data you receeded to support your conclusions
3. In this simulation the masses of the planets were all the same do you think of the masses of the planets were different it would affect the results why or why not?
4. How does this simulation demonstrate the law of universal gravitation
5. It is year 2085 and the world population has grown at an alarming rate as a space explorer you have been sent on a terraforming mission into space your mission to search for a habitable planet for humans to colonize in addition to planet earth you found a planet you believe would be habitable and now need to report beach your findings describe the new planet and why it would be perfect for maintaining human life.
According to the stimulation, if the bulk of the plants is the same, there will be no change.
What makes planet masses different?Planets have varying masses because they are formed of diverse materials, and their mass dictates their thickness and thinness.
The weight of an item is determined by its mass and the strength with which gravity pulls on it. The strength of gravity is proportional to the distance between two objects. As a result, the same thing weights differently on various planets.
The mass indicates the influence of gravity as well as the density of the atmosphere. If the masses are the same, all planets will be the same size, and many will be incapable of supporting life.
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A sample of helium occupies a volume of 160cm3 at 100 KPa and 25°c. what volume will it occupy if the pressure is adjusted to 80 KPa and the temperature remains unchanged?
Answer:
Explanation:Explore this page
About the gas laws calculator
This is an ideal gas law calculator which incorporates the Boyle's law , Charles's law, Avogadro's law and Gay Lussac's law into one easy to use tool you can use as a:
Boyle's Law-
[tex]\:\:\:\:\:\:\:\:\:\:\:\star\:\sf \underline{ P_1 \: V_1=P_2 \: V_2}\\[/tex]
(Pressure is inversely proportional to the volume)
Where-
[tex]\sf V_1[/tex] = Initial volume[tex]\sf V_2[/tex] = Final volume[tex]\sf P_1[/tex] = Initial pressure[tex]\sf P_2[/tex] = Final pressureAs per question, we are given that -
[tex]\sf V_1[/tex] = 160 cm³[tex]\sf P_1[/tex] = 100KPa[tex]\sf P_2[/tex] = 80KPaNow that we have all the required values and we are asked to find out that volume which will be occupied if the pressure is adjusted to 80 KPa and the temperature remains unchanged. For that we can put the values and solve for the final volume of helium-
[tex]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\star\:\sf \underline{ P_1 \: V_1=P_2 \: V_2}[/tex]
[tex]\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf 100 \times 160 = 80 \times V_2\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2 = \dfrac{100 \times 160}{80}\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\longrightarrow \sf V_2 =100\times \cancel{\dfrac{ 160}{80}}\\[/tex]
[tex]\:\:\:\: \:\:\:\:\:\:\longrightarrow \sf V_2 = 100 \times 2\\[/tex]
[tex] \:\:\:\:\:\:\:\:\:\:\longrightarrow \sf \underline{V_2 = 200 \:cm^3 }\\[/tex]
Therefore, 200 cm³ will be occupied if the pressure is adjusted to 80 KPa and the temperature remains unchanged.
How many grams of hydrogen chloride can be produced from 0.490 g of hydrogen and 50.0 g chlorine? The balanced equation is:
H₂(g) + Cl₂(g) → 2 HCI(g)
Hydrogen is the limiting reactant and chlorine is in excess since there is less hydrogen chloride that can be made from it (4.56 g) than there is chlorine (51.7 g) that can. As a result, 4.56 g of hydrogen chloride can be generated.
From 0.490 g of hydrogen and 50.0 g of chlorine, how many grams of hydrogen chloride may be produced?Let's first determine how much hydrogen chloride can be made from 0.490 g of hydrogen. We'll convert from moles of hydrogen to moles of hydrogen chloride using the balanced chemical equation:
1 mole of H2 yields 2 moles of HCl.
4.56 g HCl is equal to 0.490 g H2 times (1 mole H2 / 2.016 g H2), 2 moles HCl to 1 mole H2, and 36.46 g HCl to 1 mole H2.
The result is that 4.56 g of hydrogen chloride may be made from 0.490 g of hydrogen.
Let's now determine how much hydrogen chloride 50 g of chlorine can yield:
Cl2 creates 2 moles of HCl from 1 mole.
50.0 g Cl2 multiplied by (1 mole Cl2/70.90 g Cl2), (2 moles HCl/Mole Cl2), and (36.46 g/Mole Cl2) results in 51.7 g HCl.
The result is that 50.0 g of chlorine can make 51.7 g of hydrogen chloride.
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Must be Correct 50 POINTS
The chemical formula for the product.
(a)Orbital diagram for Li:
1s² 2s¹
Orbital diagram for S:
1s² 2s² 2p⁶ 3s² 3p⁴
Lewis structure for Li:
Li: [Li]+
Lewis structure for S:
:S:::S:
Combination of Li and S:
Li₂S
(b)
Orbital diagram for Ca:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
Orbital diagram for Cl:
1s² 2s² 2p⁶ 3s² 3p⁵
Lewis structure for Ca:
Ca: [Ca]²⁺
Lewis structure for Cl:
:Cl:
Combination of Ca and Cl:
CaCl₂
(c)
Orbital diagram for K:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
Orbital diagram for Cl:
1s² 2s² 2p⁶ 3s² 3p⁵
Lewis structure for K:
K: [K]+
Lewis structure for Cl:
:Cl:
Combination of K and Cl:
KCl
(d) Orbital diagram for Na:
1s² 2s² 2p⁶ 3s¹
Orbital diagram for N:
1s² 2s² 2p³
Lewis structure for Na:
Na: [Na]+
Lewis structure for N:
:N:::N:
Combination of Na and N:
Na₃N
An orbital diagram is a visual depiction of the electrons located in an atom's or molecule's orbitals. Each electron is represented by an arrow, while each orbital is illustrated by a line.
The two electrons in each orbital's two lines are drawn in pairs to represent their opposing spins. Lewis structures, on the other hand, are schematics that display the interactions between the atoms in a molecule as well as any potential lone pairs of electrons.
Each atom's valence electrons are shown as dots, and the connections between atoms are shown as lines. The kind of bond that can be created between two elements depends on the number of valence electrons.
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