Answer: The [tex]\Delta H^o_{rxn}[/tex] for the reaction is -162.5 kJ/mol
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
[tex]O_3(g)+Cl(g)\rightarrow ClO(g)+O_2[/tex] [tex]\Delta H^o_{rxn}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]ClO(g)+O_3(g)\rightarrow Cl(g)+2O_2(g)[/tex] [tex]\Delta H_1=-122.8kJ/mol[/tex]
(2) [tex]2O_3(g)\rightarrow 3O_2(g)[/tex] [tex]\Delta H_2=-285.3kJ/mol[/tex]
The expression for enthalpy of the reaction follows:
[tex]\Delta H^o_{rxn}=[1\times (-\Delta H_1)]+[1\times \Delta H_2][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1\times (-(-122.8))+(1\times (-285.3))=-162.5kJ/mol[/tex]
Hence, the [tex]\Delta H^o_{rxn}[/tex] for the reaction is -162.5 kJ/mol
245 g water sample initially at at 32 oC absorbs 17 kcal of heat. What is the final temperature of water?
Answer:62.66°C or 235.66K
Explanation:Q=McpT, the energy was given in calories so you first convert to Joules by multiplying the value in calories by 4.184J.
17*4.184=71.128kJ.
71.128kJ=mcpT
71.128kJ=245*4.187*(T-Tm)
Tm is the final temperature of the mixture. The T is the temperature given which should be converted to Kelvin by adding 273...T=32+273=305K.
71128J=245*4.187*(305-Tm)
71128=312873.575-1025.815Tm
1025.815Tm=312873.575-71128
1025.815Tm=241745.58
Tm=241745.58/1025.815
Tm=235.66K
Determine what type of functional group is present on formaldehyde (CH2O). What property is associated with this group?
Answer: carbonyl group C=O
Explanation:
Formaldehyde is an organic compound, it is the simplest form of Aldehydes. It formula is CH2O and has a carbonyl functional group, C=O. The general formula for adehydes is R-COH. The carbon atom is bonded to oxygen with a double bond and one of the two remaining bonds is occupied by hydrogen, and the other by an alkyl group.
One of the properties of adehydes is their solubility in water. The lower members (up to 4 carbons) of aldehydes are soluble in water due to H-bonding. Ofcourse the the higher members are not soluble in water because their hydrophobic long chains.
Aldehydes contain carbonyl group, therefore they undergo reactions like nucleophilic addition reactions, oxidation, reduction, halogenation.
A chemistry student needs 35.0 g of thiophene for an experiment. She has available 1.0 kg of a 14.0\%w/w solution of thiophene in ethanol Calculate the mass of solution the student should use. If there's not enough solution, press the "solution" button. Be sure your answer has the correct number of significant digits.
Answer:
The student should use 250g of the 14.0% w/w solution of thiophene in ethanolExplanation:
You must find how many grams of a 14.0% solution contains 35.0 g of thiophene (solute) and then evaluate if the amount available (1.0 kg) is enough.
Formula:
% w/w = (mass of solute / mass of solution) × 100Substitute and solve for the mass of solution:
[tex]14.0 =(35.0g/x)\times 100\\\\14.0/100=35.0g/x\\\\x=35.0g\times100/14\\\\x=250g[/tex]
Hence, the student should use 250g of the 14.0% w/w solution of thiophene in ethanol. Since, 1.0 kg is 1,000g there is enough available.
Final answer:
To acquire 35.0 grams of thiophene from a 14.0% w/w thiophene solution, the student should measure out 250 grams of the solution. The 1.0 kg of available solution is more than sufficient for the student's needs.
Explanation:
To calculate the mass of the thiophene solution that the student needs, we first need to understand the percentage concentration of the solution. A 14.0% w/w solution of thiophene in ethanol means that there are 14 grams of thiophene for every 100 grams of solution. To find out how many grams of solution contain the required 35.0 grams of thiophene, we use the following formula:
Mass of thiophene = (Percentage/100) × Total mass of solution
Therefore, we can rearrange this to solve for the total mass of solution:
Total mass of solution = Mass of thiophene / (Percentage/100)
Total mass of solution = 35.0 g / (14.0/100) = 250 g
The student should use 250 grams of the solution to obtain 35.0 grams of thiophene. Since the available solution is 1.0 kg, which is 1000 grams, there is enough solution for the experiment.
Based upon the mass of baking soda (NaHCO3) and using an excess of HCl in this experiment, you will 1) determine the mass of CO2 produced (actual yield); 2) calculate a theoretical yield for CO2; and 3) calculate a percent yield for CO2.
Equation of reaction
NaHCO3 + HCl ---------> NaCl + H2O + CO2(g)
1)The mass(actual yield) of CO2 can be gotten by isolating it from other products and getting its mass.
Assume 1.0g of CO2 was gotten as a product of the experiment
2) For the theoretical yield
The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow
Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3
Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles
1 mole of NaHCO3 yielded 1 mole of CO2
0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2
Mass of CO2 = Number of moles * Molar Mass
Mass of CO2 = 0.0238 * 44 = 1.0472g
Theoretical yield of CO2 = 1.0472 grams
3) Percentage yield of CO2 = Actual yield/Theoretical yield *100%
Percentage yield of CO2 = 1.0/1.0472 *100
Percentage yield of CO2 = 95.49%
Answer:
1) The mass of [tex]CO_{2}[/tex] produced is 1.29g
2) The theoretical yield for [tex]CO_{2}[/tex] is 1.311g
3) The percent yield is 98.4%
Explanation:
Step 1: in an experiment let's allow sodium bicarbonate (baking soda) to react with hydrochloric acid HCl to obtain high yield [tex]CO_{2}[/tex].
[tex]NaHCO_{3}+HCl[/tex] ⇒ [tex]NaCl + H_{2} O + CO_{2}[/tex]
from the balanced equation it can be seen that 1 mole of [tex]NaHCO_{3[/tex] produces 1 mole of [tex]CO_{2}[/tex].
Step 2: Let assume 2.5g of [tex]NaHCO_{3[/tex] were used
mole of [tex]NaHCO_{3[/tex] = 2.5g/molecular mass of [tex]NaHCO_{3[/tex]
molecular mass of[tex]NaHCO_{3[/tex] = 23+1+12+(16×2) = 84g/mol
converting mass of [tex]NaHCO_{3[/tex] into mole we have:
[tex]\frac{2.5g}{84gmol^{-1} }[/tex]= 0.0298 moles
Step 3: since [tex]NaHCO_{3[/tex] : [tex]CO_{2}[/tex] mole is ratio 1:1, then 0.0298 mole of [tex]NaHCO_{3[/tex] will produce 0.0298 mole of [tex]CO_{2}[/tex]
The mass of this amount of [tex]CO_{2}[/tex] = 0.0298 × molecular mass of [tex]CO_{2}[/tex]
= 0.0298 mol × ( 12+(16×2))g/mol = 1.3111g of [tex]CO_{2}[/tex]
therefore; the theoretical yield expected is 1.311g
Step 4: Let assume for the experiment that we obtained 1.29g of [tex]CO_{2}[/tex] as the actual yield
then, percent yield for [tex]CO_{2}[/tex] = [tex]\frac{Actual yield obtained}{theoretical yield that should have been obtained}[/tex] × 100%
= 1.2g/1.311g × 100% =98.4%
Identify a chemical that is used to counteract the effects of acid precipitation on aquatic ecosystems.
Calcium carbonate
Explanation:
The chemical most commonly used to counteract the effects of acid precipitation on aquatic ecosystems is calcium carbonate.
Acid rain or acid deposition or acid precipitation is any form of precipitation with an elevated level of hydrogen ion concentration in them.
To nullify this acidic precipitation in aquatic ecosystem, we need to use an environmentally friendly alkaline agent.
The most desired is calcium carbonate. The carbonate neutralizes the acid by producing carbon dioxide, water and calcium salts.
A student is given a stock solution of 1.00 M NaCl in water. They are asked to make 5 mL of 0.0500 M NaCl. How much of the stock solution should they dilute to 5 mL to make the correct concentration?
a) 10.0 mL
b) 55.5 mL
c) 1.00 mL
d) 35.3 mL
e) 27.8 mL
Answer:
The answer to your question is None of your answers is correct, maybe the data are wrong.
Explanation:
Data
Concentration 1 = C1 = 1 M
Volume 2 = 5 ml
Concentration 2 = 0.05 M
Volume 1 = x
To solve this problem use the dilution formula
Concentration 1 x Volume 1 = Concentration 2 x Volume 2
Solve for Volume 1
Volume 1 = (Concentration 2 x Volume 2)/ Concentration 1
Substitution
Volume 1 = (0.05 x 5) / 1
Simplification
Volume 1 = 0.25/1
Result
Volume 1 = 0.25 ml
Anabolic reactions _______ bonds, whereas catabolic reactions __________ bonds. A. decrease; increase. B. break; make C. weaken; strengthen D. loosen; tighten. E. make; break
Answer:
The corrext answer is E. make; break
Explanation:
In living organisms, the metabolism is either anabolic or catabolic where anabolic metabolism is energy consuming and catabolic metabolism is eneegy releasesing. It should however be noted that anabolic reaction builds or biosynthesize new mollecular structures while catabolic reaction breaks down complex structure bonds into simple structures
The braking down of bonds in catabolic reations realeses energy to sustain the anabolic rection process for the formation of new bonds
Anabolic reactions create or 'make' new bonds while forming complex structures, requiring energy. Catabolic reactions involve 'break' down bonds to create simpler structures, often releasing energy.
Explanation:The correct answer is E. Anabolic reactions are those that 'make' or form new bonds, they involve the joining of smaller molecules to form larger, more complex ones. This process often requires energy. Conversely, catabolic reactions are those that 'break' or degrade bonds, they involve the breaking down of larger molecules into smaller, simpler ones. This process often releases energy.
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Which of the following solutes has the greatest effect on the colligative properties for a given mass of pure water? Explain.
a. 0.01 mol of CaCl2 (an electrolyte)
b. 0.01 mol of KNO3 (an electrolyte)
c. 0.01 mol of CO(NH2)2 (a nonelectrolyte)
Answer:
Option a.
0.01 mol of CaCl₂ will have the greatest effect on the colligative properties, because it has the biggest i
Explanation:
To determine which of the solute is going to have a greatest effect on colligative properties we have to consider the Van't Hoff factor (i)
These are the colligative properties:
ΔP = P° . Xm . i → Lowering vapor pressure
ΔT = Kb . m . i → Boiling point elevation
ΔT = Kf . m . i → Freezing point depression
π = M . R . T → Osmotic pressure
Van't Hoff factor are the numbers of ions dissolved in the solution. For nonelectrolytes, the i values 1.
CaCl₂ and KNO₃ are two ionic solutes. They dissociate as this:
CaCl₂ → Ca²⁺ + 2Cl⁻
We have 1 mol of Ca²⁺ and 2 chlorides, so 3 moles of ions → i = 3
KNO₃ → K⁺ + NO₃⁻
We have 1 mol of K⁺ and 1 mol of nitrate, so 2 moles of ions → i = 2
Option a, is the best.
0.01 mol of CaCl2 has the greatest effect on colligative properties because it dissociates into three ions per formula unit, resulting in a higher number of dissolved particles compared to the same amount of KNO3, which dissociates into two ions, and CO(NH2)2, which does not dissociate.
Explanation:The solute that has the greatest effect on the colligative properties for a given mass of pure water from the options provided would be 0.01 mol of CaCl2 (an electrolyte). This is because the colligative properties such as freezing point depression, boiling point elevation, and osmotic pressure depend on the total number of dissolved particles in solution.
When CaCl2 dissolves in water, it dissociates into three ions: one Ca2+ ion and two Cl- ions. This means for every mole of CaCl2 we get three moles of particles. In contrast, 0.01 mol of KNO3, another electrolyte, produces two ions per formula unit, and 0.01 mol of CO(NH2)2 (urea, a nonelectrolyte) does not dissociate into ions when dissolved, hence producing only one mole of particles per mole of solute.
Therefore, for the given mass of 0.01 mol, CaCl2 produces the greatest number of particles and therefore has the greatest effect on colligative properties when compared to KNO3 and urea.
A cylinder contains oxygen at a pressure of 10.0 atm and a temperature of 300.0 K. The volume of the cylinder is 10,000 mL. What is the mass of oxygen gas?
Answer:
130 g of O₂
Explanation:
Let's apply the Ideal Gases Law, to solve this.
Pressure . volume = moles . R . T° (K)
First of all, we need to convert the volume in mL to L, because the units of R
10000 mL . 1L/1000mL = 10L
Let's replace → 10 atm . 10L = n . 0.082L.atm/mol.K . 300K
(100 atm.L) / (0.082L.atm/mol.K . 300K) = n
4.06 moles = n
Let's convert the moles to mass → 4.06 mol . 32 g/1mol = 130 g
What term best describes when cryptography is applied to entire disks instead of individual files or groups of files?
Answer:
Full disk encryption
Explanation:
Disk encryption is a protection technique used in securing information by converting it into unreadable codes that cannot decrypt in other to prevent unauthorized persons from accessing the information.
When cryptography is applied to entire disks, it is termed Full disk encryption.
how many moles of H2 are produced from 5.8 moles of NH3
2NH3 -> N2 + 3H2
How many moles of O2 are needed to produced 1.8 moles of H2O?
C3H8 + 5O2 -> 3CO2 + 4H2O
Answer:
1. 8.7moles of H2
2. 2.25moles of O2
Explanation:
1. 2NH3 —> N2 + 3H2
From the equation,
2moles of NH3 produce 3 moles of H2.
Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e
Xmol of H2 = (5.8x3)/2 = 8.7moles
2. C3H8 + 5O2 —> 3CO2 + 4H2O
From the equation,
5moles of O2 produced 4moles of H2O.
Therefore, Xmol of O2 will produce 1.8mol of H2O i.e
Xmol of O2 = (5x1.8)/4 = 2.25moles
Describe the Sun’s interior. Include references to the main physical processes that occur at various depths within the sun.
Answer:
The sun is the large astronomical body that is located at the center of the solar system. The interior structure of the sun are as follows-
Core- It represents the extreme interior portion of the sun. This is the hottest region of the sun and there occurs the process of constant nuclear fusion reactions that fuel the sun and helps in burning. Here the nucleus of two hydrogen atoms is fused together to form a heavy nucleus of a helium atom. It has a thickness of about 500 km. Radiative zone- This is the layer that surrounds the core of the sun. From this layer, the energy is radiated outward where the protons carry these energies in the form of thermal radiation. Here, the process of radiation takes place. Convective zone- This is the outer layer of the sun's interior. It is about 200,000 km thick, and in this layer, the energy constantly flows, allowing the heat to move upward by undergoing the process of convection.How did oxygen (O2) get into Earth's atmosphere? How did oxygen (O2) get into Earth's atmosphere? It was captured from the solar nebula. It came from chemical reactions with surface rocks. It was outgassed from volcanoes. It was released by life through the process of photosynthesis.
Answer:
Option (4)
Explanation:
The primitive atmosphere was comprised of lesser or no amount of oxygen. With the progressive time, primitive organisms appeared such as the primitive aquatic plants, algae, and cyanobacteria, that carried out the process of photosynthesis in the presence of sunlight, water, and CO₂, and in return produced the food and oxygen. This is how the rapid photosynthesis process led to the increasing amount of oxygen in the atmosphere that facilitates the growth and evolution of different life forms on earth.
Thus, the correct answer is option (4).
Would you expect the water solubility of the resulting molecule to be higher than, lower than, or about the same as the solubility of glucose?
Answer:
It'll be lower.
Explanation:
Water is a universal solvent, which means it can dissolve virtually anything except oils and non-polar substance because of its polarity and ability to form hydrogen bond .Water molecules are also attracted to other polar molecules and to ions. A charged or polar substance that interacts with and dissolves in water is said to be hydrophilic: hydro means "water," and philic means "loving." In contrast, nonpolar molecules like oils and fats do not interact well with water
Two small metal spheres are 26.50 cm apart. The spheres have equal amounts of negative charge and repel each other with a force of 0.03500 N. What is the charge on each sphere?
Answer:
[tex]-5.226\times 10^{-7} C[/tex] is the charge on each sphere.
Explanation:
Coulomb's law is given as ;
[tex]F=K\times \frac{q_1\times q_2}{r^2}[/tex]
[tex]q_1,q_2[/tex] = Charges on both charges
r = distance between the charges
K = Coulomb constant =[tex]9\times 10^{9} N m^2/C^2[/tex]
We have ;
Charge of ion =[tex]q_1=-q[/tex]
Charge of electron =[tex]q_2=-q[/tex]
[tex]r=26.55 cm =0.2655 m[/tex]
Force between the charges at r distance will be : F
F = 0.03500 N
[tex]0.03500 N=9\times 10^{9} N m^2/C^2\times \frac{(-q)\times (-q)}{(0.2655 m)^2}[/tex]
[tex]q=5.226\times 10^{-7} C[/tex]
[tex]-5.226\times 10^{-7} C[/tex] is the charge on each sphere.
The melting point of sodium metal is 97.8 ∘C and the melting point of sodium chloride is 801 ∘C. What can you infer about the relative strength of metallic and ionic bonding from these melting points?
Answer:
The metallic bond of the sodium metal is weaker than the ionic bond of the sodium chloride
Explanation:
The metallic bond of the sodium metal is weaker than the ionic bond of the sodium chloride. This is seen in the fact that the melting point of the sodium metal which is 97.8°C is lower than the melting point of the sodium chloride which is 801°C. This shows that the inter-atomic bonds of the sodium metal are weaker than that of the inter-ionic bonds of the sodium chloride which will require higher energy to break which is shown by its high melting point of 801°C. The melting point of the sodium metal of 97.8 °C which is lower requires less energy to break its bonds. This is shown by its lower melting point.
Organism in kingdoms archaebacteria live in environments that are very hot or very cold Which other characteristics is common to organisms in this kingdom
Answer:
Archaebacteria belong to the kingdom 'Archaea'. They resemble bacteria when observed under a microscope.They are single-cell organisms. However,they differ from prokaryotes in many ways.
•They have a completely different cell membrane so that they can survive and thrive in harsh environments
•Unlike bacteria,their cell wall and membrane can be stiff and gives a specific shape such as flat,rod shaped or cubic.
•the absence of peptidoglycan in cell wall,instead contain lipid and protein to give them strength and resistance against chemical.
•The cell membrane of archae are made up of phospholipid bilayer but unlike bacteria and eukaryotes,the bilayers have ether bonds.These ether bonds have the ability to resist chemical and helps them to survive in very extreme environments that might otherwise kill them.
•They have multiple RNA polymerase that contain multiple polypeptide
•They have initiator tRNA with unmodified methionine(unlike bacteria).
•They have the ability to survive in very high temperatures,in acidic environment and alkaline environments and even have tolerance for high salt content.
A rocket is launched with a thrust of 5 x 106 N at an angle of 37 degrees above the horizontal. The rocket has a total mass of 200,000 kg. What direction is the rocket's acceleration?
Answer:
[tex]\beta=63.85^{\circ}[/tex] from the vertical OR [tex]16.15^{\circ}[/tex] from horizontal.
Explanation:
Given:
Thrust of launching the rocket, [tex]F=5\times 10^6\ N[/tex]
angle of launch from the horizontal, [tex]\theta=37^{\circ}[/tex]
mass of the rocket, [tex]m=200000\ kg[/tex]
Now the direction of acceleration due to the thrust force is in the direction of force:
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{5000000}{200000}[/tex]
[tex]a=25\ m.s^{-2}[/tex]
And the acceleration due to gravity is always directed towards the center of the earth i.e. vertically downwards.
Since acceleration is a vector quantity we, approach accordingly:
[tex]\tan\beta=\frac{a\cos\theta}{g}[/tex]
[tex]\tan\beta=\frac{25\cos37^{\circ}}{9.8}[/tex]
[tex]\beta=63.85^{\circ}[/tex] from the vertical OR [tex]16.15^{\circ}[/tex] from horizontal.
One mole of aspartame (C14H18N2O5) reacts with two moles of water to produce one mole of
aspartic acid (C4H7NO4), one mole of methanol (CH3OH) and one mole of phenylalanine.
a. What is the molecular formula of phenylalanine?
b. What mass of phenylalanine is produced from 378 g of aspartame?
Answer:
a. Molecular formula of phenylalanine → C₉H₁₁NO₂
b. 234.2 g of C₉H₁₁NO₂
Explanation:
Let's think the reaction:
C₁₄H₁₈N₂O₅ + 2H₂O → CH₃OH + C₉H₁₁NO₂ + C₄H₇NO₄
In order to work with this reaction, we assume that water is on excess.
Let's determine the moles of aspartame → molar mass = 294 g/mol
Moles = mass / molar mass → 378 g / 294 g/mol = 1.28 moles
As ratio is 1:1, 1 mol of aspartame can produce 1 mol of phenylalanine; therefore 1.28 moles of aspartame must produce 1.28 moles of phenylalanine.
Let's convert the moles to mass → molar mass C₉H₁₁NO₂ = 183 g/mol
Moles . molar mass = Mass → 1.28 mol . 183 g/mol = 234.2 g
A. The molecular formula of phenylalanine is C₉H₁₁NO₂
B. The mass of phenylalanine, C₉H₁₁NO₂ produced from 378 g of aspartame is 212.14 g
A. Determination of the molecular formula of phenylalanine.
To obtain the molecular formula of phenylalanine, we shall write the balanced equation. This is given below
C₁₄H₁₈N₂O₅ + 2H₂O → C₄H₇NO₄ + CH₃OH + C₉H₁₁NO₂
Thus, the molecular formula of phenylalanine is C₉H₁₁NO₂
B. Determination of the mass of phenylalanine, C₉H₁₁NO₂ produced from 378 g of aspartame
C₁₄H₁₈N₂O₅ + 2H₂O → C₄H₇NO₄ + CH₃OH + C₉H₁₁NO₂
Molar mass of C₁₄H₁₈N₂O₅ = (14×12) + (18×1) + (14×2) + (16×5) = 294 g/mol
Mass of C₁₄H₁₈N₂O₅ from the balanced equation = 1 × 294 = 294 g
Molar mass of C₉H₁₁NO₂ = (9×12) + (11×1) + 14 + (16×2) = 165 g/mol
Mass of C₉H₁₁NO₂ from the balanced equation = 1 × 165 = 165 g
From the balanced equation above,
294 g of C₁₄H₁₈N₂O₅ reacted to produce 165 g of C₉H₁₁NO₂
Therefore,
378 g of C₁₄H₁₈N₂O₅ will react to produce = (378 × 165) / 294 = 212.14 g of C₉H₁₁NO₂
Thus, 212.14 g of C₉H₁₁NO₂ were obtained from the reaction.
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Some velomobile seats have been designed to hold ice packs inside their cushions. If you started a ride with ice packs that held 1100 gg of frozen water at 0 ∘C∘C , and the temperature of the water at the end of the ride was 32 ∘C∘C , how many calories of heat energy were absorbed?
Final answer:
Approximately 34.97 kilocalories of heat energy were absorbed by the water in the velomobile seats.
Explanation:
In this scenario, we need to calculate the amount of heat energy absorbed by the water in the velomobile seats. To do this, we can use the formula:
q = mcΔTWhere:
q is the heat energy absorbedm is the mass of the waterc is the specific heat capacity of water, which is 4.18 J/g°CΔT is the change in temperature, which is 32°C - 0°C = 32°CLet's calculate the heat energy absorbed:
q = m * c * ΔTq = 1100 g * 4.18 J/g°C * 32°Cq = 146,176 JConverting this to kilocalories:
1 Joule = 0.000239006 kJ (approximately)q = 146,176 J * 0.000239006 kJ/Jq = 34.97 kJTherefore, approximately 34.97 kilocalories of heat energy were absorbed by the water in the velomobile seats.
things required for life, including heat, water, nutrients, salts and oxygen are called "___________ substances."
Answer:
Vital
Explanation:
Hello! The factors mentioned previously are essential substances for the life and health of our bodies. Not only in humans, but also for animals. They are factors that are linked and contribute to physical and mental energy, growth and life. Are essentials because humans can't live without that.
Thanks for your question! Feel free to ask more!
The carbon-carbon double bond in ethene is ________ and ________ than the carbon-carbon triple bond in ethyne.
Answer:
weaker and longer
Explanation:
Since there are 3 bonds in ethyne in comparision with the 2 bonds of ethyne between carbon atoms, they are attracted more to each other → the bond gets shorter . And since there are one more bond that supports the union → the bond gets stronger
thus the carbon-carbon double bond in ethene is weaker and longer than the carbon-carbon triple bond in ethyne
Iron (Fe) undergoes an allotropic transformation at 912C: upon heating from a BCC (phase) to an FCC ( phase). Accompanying this transformation is a change in the atomic radius of Fe—from rBCC = 0.12584 nm to rFCC = 0.12894 nm—and, in addition a change in density (and volume). Compute the percent volume change associated with this reaction. Does the volume increase or decrease?
The volume change during Iron's BCC to FCC transformation is computed via cubing the atomic radii and applying them to a percent change formula. The increasing radius hints at a likely volume increase, however, the exact percent change requires numerical calculation.
Explanation:To answer your question on whether the volume increases or decreases during the allotropic transformation of Iron (Fe) from a BCC phase (with atomic radius rBCC = 0.12584 nm) to an FCC phase (rFCC = 0.12894 nm), we need to recognize that the volume of an atom in a crystal structure can be determined by cubing its atomic radius, and the volume change can be obtained by taking the difference between the initial and final volumes.
The initial volume (of the BCC phase) is (0.12584 nm)³, while the final volume (of the FCC phase) is (0.12894 nm)³. The percent volume change can then be computed by [(final volume - initial volume)/initial volume] x 100%.
If this calculation yields a positive value, this would mean the volume increases; if the result is a negative value, the volume decreases. While it's evident that the radius increases during this transformation, due to the cubic relationship between radius and volume, it's likely that the volume also increases, although the actual percent volume change would require numerical computation.
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Final answer:
The percent volume change during the allotropic transformation of iron from BCC to FCC can be calculated using the given atomic radii and results in an increase in volume.
Explanation:
To compute the percent volume change associated with the allotropic transformation of iron (Fe) from a BCC (body-centered cubic) structure to an FCC (face-centered cubic) structure at 912°C, we can use their respective atomic radii and the equation for the volume of a sphere, V = (4/3)πr3. Given the atomic radii for BCC as rBCC = 0.12584 nm and for FCC as rFCC = 0.12894 nm, the respective volumes can be calculated.
After obtaining the volumes, we calculate the percent volume change with the formula: Volume Change (%) = (VFCC - VBCC)/VBCC x 100%. Using the given radii, we find that the volume of FCC iron is larger than BCC iron, indicating that the volume increases during the transformation. The actual numerical percent volume change can be computed using the given radii values and the volume equation stated above.
A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field produced by the disk at a point on its central axis at distance z = 12 cm from the disk?
Answer:
the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C
Explanation:
Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.
The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀([tex]1 - \frac{z}{\sqrt{z^{2} + R^{2} } }[/tex])
Substituting the values into the equation, it becomes
E = σ/ε₀([tex]1 - \frac{z}{\sqrt{z^{2} + R^{2} } }[/tex]) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²([tex]1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }[/tex]) = 7.12 × 10⁵([tex]1 - \frac{0.12}{0.1216}[/tex]) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C
Therefore, the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C
What amount of ammonia, NH3(g), can be produced from 15 mol of hydrogen reacting with excess nitrogen?
3 H2(g) + N2(g) → 2NH3 (g)
Answer:
10mol
Explanation:
3H2 + N2 -> 2NH3
Stoichiometry is a tool that chemists can use to find the amount of substance present in any part of a reaction. The arrow (->) suggests that the reaction goes to completion (100%), so assume that left side = right side.
3H2
15 mol
You can divide the amount of moles by the coefficient to find the number of moles when you have a coefficient of 1. This number can then be used to find the value of moles for the rest of the products/reactants:
15/3=5mol
NH3 has a coefficient of 2, so we have to multiply the value we got (5mol) by 2. This results in having 10mol of ammonia as the end result.
Final answer:
The production of ammonia from hydrogen and excess nitrogen follows a 3:2 mole ratio, according to the balanced equation N2(g) + 3H2(g) → 2NH3(g). From 15 moles of hydrogen, 10 moles of ammonia are produced.
Explanation:
The question asks about the production of ammonia (NH3) from hydrogen (H2) in the presence of excess nitrogen (N2) based on the chemical reaction provided. According to the stoichiometry of the balanced equation, N2(g) + 3H2(g) → 2NH3(g), there is a 3:2 mole ratio between hydrogen and ammonia. Therefore, for every 3 moles of hydrogen, 2 moles of ammonia are produced.
To calculate the amount of ammonia produced from 15 mol of hydrogen, we use the mole ratio from the balanced equation. Since 3 moles of hydrogen produce 2 moles of ammonia, the amount of ammonia produced from 15 moles of hydrogen can be found using cross-multiplication:
(2 mol NH3) / (3 mol H2) = (x mol NH3) / (15 mol H2)
x = (15 mol H2 × 2 mol NH3) / 3 mol H2
x = 10 mol NH3
The answer is that 10 moles of ammonia can be produced from 15 moles of hydrogen reacting with excess nitrogen.
At high trns phosphine (pH2) dissociates into phosphorus and hydrogen by the following reaction: At SOW C the rate at which phosphine dissociates is for I in seconds. The reaction occurs in a constant volume. 3-L vessel, and the initial concentration of phosphine is 5 kmol/rn3. If 3 rnol of the phosphine reacts. how much phosphorus and hydrogen is produced?
Answer: 0.75 moles of Phosphorus and 4.5 moles of Hydrogen are produced respectively from 3 moles of Phosporus.
Explanation:
4PH3 --------> P4 + 6H2
From the stochiometry of the reaction,
4 moles of Phosphine gives 1 mole of Phosphorus
3 moles of Phosphine will give (3×1)/4 moles of Phosphorus.
Therefore, 0.75 moles of Phosphorus is produced.
Similarly, 4 moles of Phosphine gives 6 moles of Hydrogen
3 moles of Phosphine will give (3×6)/4 moles of Hydrogen.
Therefore, 4.5 moles of Hydrogen is produced.
QED!
Which of the following structural bonding patterns for bricks features a single wythe? Select one: a. English bond b. Flemish bond c. Running bond d. Common bond
Answer:
C. Running bond
Explanation:
The common bond is a structural bonding pattern for bricks with a single wythe.
Explanation:The structural bonding pattern for bricks that features a single wythe is the common bond.
When a bullet is retrieved, how is it marked for identification purposes? What should be avoided?
Answer:
The retrieved bullet is normally marked at the tip or base with the initials of the investigator while ensuring that no markings are placed on the sides of the retrieved bullet. It is required to ensure that the markings made on the bullets does not go over or obscure striations or markings already present on the bullet.
Where the retrieved bullet is mutilated whereby it is impossible to engrave the required markings on it, it should be placed in a marked envelope, container or pill box
Explanation:
It is possible to trace retrieved casings and bullets from crime scene back to the gun from which it was fired or to the suspect's gun. The retrieved casings and bullet, when scrutinized at the crime lab, can reveal the gun model and make from which the casing or bullet was fired. The retrieved bullet or casing could also be traced back to the lot or batch of ammunition in possession of the suspect.
When a bullet is retrieved for identification purposes, it is marked with a unique identification number and other relevant information. Care must be taken to avoid damaging the bullet's markings and altering its shape or surface. Proper documentation and careful handling are crucial for accurate identification of the bullet.
Explanation:When a bullet is retrieved for identification purposes, it is marked using a variety of methods. One common method is to assign a unique identification number to the bullet, typically by inscribing or engraving it on the base or side of the bullet. This number can be used to match the bullet to a specific firearm. Additionally, the bullet may be photographed and cataloged, and any unique characteristics such as rifling marks or imperfections can be documented and compared to the firearm that fired it.
When marking a bullet for identification, it is important to avoid damaging the bullet's crucial markings or altering its shape in any way that could affect its comparison to a firearm. The use of permanent markers or corrosive substances should be avoided, as they can damage the bullet surface. Careful handling and proper documentation are critical to preserving the integrity of the bullet and ensuring accurate identification.
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In a neutralization reaction, 24.6 mL of 0.300 M H2SO4(aq) reacts completely with 20.0 mL of NaOH(aq). The products are Na2SO4(aq) and water. What is the concentration of the NaOH solution?
Answer : The concentration of the NaOH solution is, 0.738 M
Explanation :
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=0.300M\\V_1=24.6mL\\n_2=1\\M_2=?\\V_2=20.0mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.300M\times 24.6mL=1\times M_2\times 20.0mL[/tex]
[tex]M_2=0.738M[/tex]
Thus, the concentration of the NaOH solution is, 0.738 M
Isooctane, C8H18, is the component of gasoline from which the term octane rating derives.
A. Write a balanced equation for the combustion of isooctane to yield CO2 and H2O.
B. Assuming that gasoline is 100% isooctane, the isooctane burns to produce only CO2 and H2O, and that the density of isooctane is 0.792 g/mL, what mass of CO2 in kilograms is produced each year by the annual US gasoline consumption of 4.6 x 10^10 L?
C. What is the volume in L of this CO2 at STP?
D. How many moles of air are necessary for the combustion of 1 mol of isooctane, assuming that air is 21.0% O2 by volume.? What is volume in L of this air at STP?
Answer:
a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O
b) Mass of CO₂ produced annually from this combustion of isooctane gasoline = 1.12 × 10⁵ Kg
c) CO₂ produced from the combustion of the gasoline in a year will occupy 5.632 × 10⁷ L
d) There needs to be a minimum of 1.752 × 10⁷ moles of air and 3.92 × 10⁸ L of air for the oxygen to be in excess all through the year of gasoline combustion.
Explanation:
a) C₈H₁₈ + (23/2)O₂ -----> 8CO₂ + 9H₂O
b) C₈H₁₈ has a density of 0.792 mg/L.
Since density = mass/volume;
mass = density × volume
Mass of C₈H₁₈ with 4.6 x 10^10 L volume = 0.792 × 4.6 x 10^10 = 3.643 × 10^10 mg = 3.643 × 10⁷ g.
To obtain the mass of CO₂ produced, we need the number of moles of C₈H₁₈ that burned.
Number of moles = mass/molar mass
Molar mass of C₈H₁₈ = (8×12) + 18 = 114g/mol
Number of moles of C₈H₁₈ = (3.643 × 10⁷)/114 = (3.2 × 10⁵) moles.
From the chemical reaction,
1 mole of C₈H₁₈ burns to give 8 moles of CO₂
(3.2 × 10⁵) moles will give 8 × 3.2 × 10⁵ = (2.56 × 10⁶) moles of CO₂
Mass of CO₂ produced = number of moles × Molar mass
Molar mass of CO₂ = 44 g/mol
Mass of CO₂ produced = 2.56 × 10⁶ × 44 = 1.12 × 10⁸ g = 1.12 × 10⁵ kg
c) 1 mole of any gas at stp occupies 22.4L
2.56 × 10⁶ moles of CO₂ will occupy 2.56 × 10⁶ × 22.4 = 5.632 × 10⁷ L
d) 1 mole of C₈H₁₈ requires 23/2 moles of O₂ for complete combustion yearly.
3.2 × 10⁵ moles would require 3.2 × 10⁵ × 23/2 = 3.68 × 10⁶ moles of O₂
O₂ makes up 21% of the air
That is,
0.21 moles of O₂ would be contained in 1 mole of air
3.68 × 10⁶ moles of O₂ would be contained in (3.68 × 10⁶ × 1)/0.21 = 1.752 × 10⁷ moles of air.
1 mole of any gas at stp occupies 22.4L
1.752 × 10⁷ of air will occupy
1.752 × 10⁷ × 22.4/1 = 3.92 × 10⁸ L of air!
Final answer:
The balanced equation for the combustion of isooctane is C8H18(l) + 11O2(g) → 8CO2(g) + 9H2O(g). Assuming that gasoline is 100% isooctane, the mass of CO2 produced each year by the annual US gasoline consumption of 4.6 x 10^10 L is 9.04 x 10^6 kg. The volume of this CO2 at STP is 2.03 x 10^11 L. We use the molar ratios from the balanced equation to determine the number of moles of air necessary for the combustion of 1 mol of isooctane. The volume in L of this air at STP is 1173 L.
Explanation:
The balanced chemical equation for the combustion of isooctane (C8H18) is:
C8H18(l) + 11O2(g) → 8CO2(g) + 9H2O(g)
Assuming that gasoline is 100% isooctane, we can use the balanced equation to calculate the mass of CO2 produced each year by the annual US gasoline consumption of 4.6 x 1010 L. We start by calculating the mass of isooctane consumed, then use the molar ratios from the balanced equation to calculate the mass of CO2 produced. Given that the density of isooctane is 0.792 g/mL, the mass consumed is 4.6 x 1010 L x 0.792 g/mL = 3.64 x 1010 g. The molar mass of CO2 is 44.01 g/mol, so the number of moles of CO2 produced is 28 x 3.64 x 1010 g / 114.22 g/mol = 9.04 x 109 mol. Finally, we convert to kilograms by dividing by 1000, so the CO2 produced is 9.04 x 106 kg.
To calculate the volume of CO2 at STP, we use the ideal gas law. At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. So, the volume of CO2 at STP is 9.04 x 109 mol x 22.4 L/mol = 2.03 x 1011 L.
We use the molar ratios from the balanced equation to determine the number of moles of air necessary for the combustion of 1 mol of isooctane. From the equation, we can see that 11 moles of O2 are required for the combustion of 1 mol of isooctane. Since air is 21.0% O2 by volume, the volume of air required is 11 / 0.21 = 52.4 mol. At STP, 1 mole of gas occupies 22.4 L, so the volume of air required is 52.4 mol x 22.4 L/mol = 1173 L.