Answer:
The correct answer to a) is sulfuric acid
b) 1/6 moles of aluminum sulfate is produced
Explanation:
To solve this we need to write out the balaned chemical equation as follows
Al(OH)3(s) + 3 H2SO4(aq) -----> Al2 (SO4)3(aq) + 6 H2O(l)
Here we see that one mole of aluminium hydroxide reacts with three moles of sulphuric acid to form one mole of aluminium sulphate and six ,oles of water
Hence the limiting reactant in this question is the sulphuric acid as 0.5 moles of alumininium hydroxide requires 1.5 moles of sulphuric acid to completely use up the aluminium hydroxide present
Dividing the number of moles of the reactants present by the amount of moles of sulphuric acid required we have
Hence 3 mole of sulphric acid combines with 1 mole of Aluminium hydroxide
0.5 mole of sulfuric acid combines with 0.5÷3 or 1/6 mole of Aluminium hydroxide
Hence the correct answer is sulfuric acid
b) to solve this, since three moles of sulfuric acid produces one mole of aluminium sulfate then 0.5 moles of sulfuric acid produces 0.5/3 or 0.166 mole of aluminum sulfate
hence 1/6 moles of aluminum sulfate is produced
How many moles of aluminum sulafte is produced when 125 moles of aluminum hydroxide and 136 moles of sulfuric acid react?
Answer:
The answer to this question is 45.33 moles of aluminum sulfate is produced when 125 moles of aluminum hydroxide and 136 moles of sulfuric acid react
Explanation:
To solve this, we write out the chemical equation of he reaction thus
Al(OH)3(s) + 3 H2SO4(aq) -----> Al2 (SO4)3(aq) + 6 H2O(l)
here it is seen that one moles of aluminum hydroxide reacts with three moles of sulfuric to produce one mole of aluminum sulfate and six moles of water
hence
136 moles of sulfuric acid reacts with 136/3 or 45.33 moles of aluminum hydroxide to produce 136/3 or 45.33 moles of aluminum sulfate and 2× 136 moles of water
Hence the amount in moles of aluminum sulfate produced is 45.33 moles
Final answer:
Using the balanced chemical equation 3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O, and knowing that aluminum hydroxide is the limiting reactant, 125 moles of aluminum hydroxide will produce 41.67 moles of aluminum sulfate.
Explanation:
The question asks how many moles of aluminum sulfate will be produced when reacting 125 moles of aluminum hydroxide with 136 moles of sulfuric acid. To answer this, we need the balanced chemical equation:
3 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O
The stoichiometry of the reaction shows that 3 moles of aluminum hydroxide react with 3 moles of sulfuric acid to produce 1 mole of aluminum sulfate. Since there are more moles of sulfuric acid present, aluminum hydroxide is the limiting reactant. Therefore, we can calculate the moles of aluminum sulfate produced by dividing the moles of aluminum hydroxide by 3, which gives us:
125 moles Al(OH)3 ÷ 3 = 41.67 moles Al2(SO4)3 (rounded to two decimal places as per the significant figures in the provided moles of reactants).
A scientist is preparing an experiment. She needs to collect 10 moles of helium gas in a 7.5-liter container before he can begin. The gas temperature inside the helium container will be a constant 20 degrees C. The scientist wants make sure that the pressure exerted by the helium will not burst the gas container. What would be the pressure of the helium gas inside the container?
Answer : The pressure of the helium gas inside the container would be, 32.1 atm
Explanation :
To calculate the pressure of the gas we are using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P = Pressure of [tex]He[/tex] gas = ?
V = Volume of [tex]He[/tex] gas = 7.5 L
n = number of moles [tex]He[/tex] = 10 mole
R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]
T = Temperature of [tex]He[/tex] gas = [tex]20^oC=273+20=293K[/tex]
Putting values in above equation, we get:
[tex]P\times 7.5L=10mole\times (0.0821L.atm/mol.K)\times 293K[/tex]
[tex]P=32.1atm[/tex]
Thus, the pressure of the helium gas inside the container would be, 32.1 atm
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
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.
When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, a precipitation reaction occurs. Write the balanced net ionic equation of the reaction. Include charges on the ions, where applicable. Include coefficients only when they are different than ?
The net ionic equation for the reaction of strontium chloride and potassium sulfate, forming strontium sulfate solid, is Sr2+(aq) + SO42-(aq) → SrSO4(s)
Explanation:When an aqueous solution of strontium chloride is mixed with an aqueous solution of potassium sulfate, strontium sulfate precipitates out and potassium and chloride ions remain in the solution. The balanced net ionic equation for this reaction is as follows:
Sr2+(aq) + SO42-(aq) → SrSO4(s)
This equation represents the change where strontium ions from strontium chloride and sulfate ions from potassium sulfate are combined to form strontium sulfate solid. The charges are balanced with the positive 2 charge of the strontium ion balancing the negative 2 charge of the sulfate ion.
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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 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.
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.
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.
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
A client comes to the emergency department with status asthmaticus. The client's respiratory rate is 48 breaths/minute, and the client is wheezing. An arterial blood gas analysis reveals a ph of 7.52, a partial pressure of arterial carbon dioxide (paco2) of 30 mm hg, pao2 of 70 mm hg, and bicarbonate (hco3--) of 26 meq/l. What disorder is indicated by these findings?
Answer:
The complete question is:
Question: What disorder is indicated by these findings? A client comes to the emergency department with status asthmaticus. His respiratory rate is 48 breaths/minute, and he is wheezing. An arterial blood gas analysis reveals a pH of 7.52, a partial pressure of arterial carbon dioxide (PaCO2) of 30 mm Hg, PaO2 of 70 mm Hg, and bicarbonate (HCO3−) of 26 mEq/L.
A. Metabolic acidosis
B. Respiratory acidosis
C. Metabolic alkalosis
D. Respiratory alkalosis
Answer: The correct answer is:
D. Respiratory alkalosis
Explanation:
In Respiratory alkalosis the Partial Pressure of Arterial Carbondioxide (PaCO2) become decreased (i.e. less than 35 mm Hg) and the pH of blood become increased (i.e. more than 7.45). Alveolar hyperventilation causes respiratory alkalosis.
Alveolar hyperventilation occurs when alveolar ventilation is increased than the arterial carbondioxide tension and carbondioxide production.
Alveolar ventilation is the gaseous exchange between alveoli and the external environment.
Whereas, in metabolic acidosis, bicarbonate (HCO3) become decreased (i.e. less than 22 mEq/l and the pH of blood become decreased (i.e. less than 7.35); in respiratory acidosis, the pH of blood also become decreased (i.e. less than 7.35) and the PaCO2 become increased (i.e. more than 45 mm Hg); and in metabolic alkalosis, the bicarbonate (HCO3) become increased (i.e. more than 26 mEq/l and the pH become increased (i.e. more than 7.45).
Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same?
a. Because in the gas phase molecules do not interact with each other.
b. Because molecules of a gas have very low kinetic energy.
c. Because these factors compensate each other.
d. Because most of the volume occupied by the substance is empty space.
Answer:
option d
Explanation:
Molecular sizes of gaseous molecules are very less. Volume occupied by the all the molecules of the gases are very less or negligible as compared to the container in which it is kept. Therefore, most of the volume occupied by gaseous molecules are negligible.
Volume occupied by the gaseous molecules are actually the volume of the container and its does not depend upon the amount, molecular mass or dipole moment of the gaseous molecules.
Therefore, the correct option is d ‘Because most of the volume occupied by the substance is empty space.’
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.
Calculate the mass of 1.9 • 10^24 atoms of Pb
Answer:
65.4 is the mass for 1.9×10²⁴ atomsof Pb
Explanation:
1mol of atoms of Pb has → NA (6.02×10²³ atoms) and weighs → 207.2 g
Therefore 1.9×10²⁴ atomsof Pb may weigh (1.9×10²⁴ . 207.2) / NA = 65.4 g
What amount of sucrose (C12H22O11) should be added to 5.83 mol water to lower the vapor pressure of water at 50 °C to 72.0 torr? The vapor pressure of pure water at 50 °C is 92.6 torr. 1
Answer:
The amount of sucrose that must be added is 1.66 moles
Explanation:
Colligative property of lowering vapor pressure has this formula:
Vapor pressure of pure solvent (P°) - Vapor pressure of solution = P° . Xm
We have both vapor pressure (pure solvent and solution9, so let's determine the ΔP
ΔP = 92.6 Torr - 72 Torr = 20.6 Torr
Let's add the data in the formula
20.6 Torr = 92.6 Torr . Xm
Xm = Mole fraction of solute → (mol of solute/ mol of solute + mol of solvent)
Mol of solvent = 5.83 mol (data from the problem)
Therefore Xm = 20.6 Torr / 92.6 Torr → 0.222
Let's find out the moles of solute (our unknown value)
0.22 = moles of solute / moles of solute + 5.83 moles of solvent
0.222 (moles of solute + 5.83 moles of solvent) = moles of solute
0.222 moles of solute + 1.29 moles of solvent = moles of solute
1.29 moles of solvent = moles of solute - 0.222 moles of solute
1.29 moles = 0.778 moles of solute
1.29 / 0.778 = moles of solute → 1.66 moles
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
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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Which is a characteristic of a Lewis base? It behaves as the electron donor. It behaves as the electron acceptor.
Answer: The correct statement is, It behaves as the electron donor.
Explanation:
According to the Lewis concept:
A Lewis-acid is defined as a substance that accepts electron pairs.
A Lewis-base is defined as a substance which donates electron pairs.
For example : Acid + Base ⇄ Acid-base adduct
[tex]H^++NH_3\rightleftharpoons NH_4^+[/tex]
As per question, the characteristic of a Lewis base is that it behaves as the electron donor.
Hence, the correct statement is, It behaves as the electron donor.
Answer:
A
Explanation:
It behaves as the electron donor
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
Why is the expectation value of the energy associated with the 1-D "particle-in-a-box" the same as the eigen value of the Hamiltonian associated with the 1-D "particle-in-a-box" wave function?
Answer: The average potential energy of the PIB is 0 irrespective of the wave function.
Explanation:
⟨H⟩=⟨KE⟩+⟨V⟩
the nn quantum number
⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )
the average kinetic energy of the wavefunction is dependent on
⟨V⟩=∫sin(kx)0sin(kx)dx=0
The average potential energy of the PIB is 0 irrespective of the wave function.
⟨H⟩=⟨KE⟩=(π^2 ℏ^2)/(2mL^2 )
While performing a recrystallization, a chemist notices that a small amount of the sample will not dissolve, even after the recrystallization solution has been boiling for some time. Select the correct course of action.
A. Perform a gravity filtration with a stemless funnel.
B. Perform a gravity filtration with a long-stemmed funnel.
C. Isolate the insoluble material via filtration, grind it into a powder, and add it back to the recrystallization solution
D. Cool the recrystallization solution to room temperature and then heat it again to dissolve the insoluble material.
Answer: A. Perform a gravity filtration with a stemless funnel.
Explanation: Gravity filtration with a stemless funnel prevent blocking up the funnel. Undissolved sample may come out in the stem when the solution cools down thus will be blocking the funnel. Using stemless funnel prevent this problem.
The graph represents a moderately weak acid. How would the graph change to represent a relatively strong acid?
A) The HA bar on the left must be much taller.
B) H3O+ would be converted into H+.
C) The right side of the bar graph would have only one bar: H3O+.
D) The HA bar on the right must be converted completely to H3O+ and A-
Answer:
the correct answer is d
Explanation:
The HA bar on the right must be converted completely to H3O+ and A-. Strong acids completely dissociate in solution. Complete dissociation would mean that there is no HA bar left on the right of the arrow.
To represent a strong acid in the graph, the 'HA' bar has to be almost non-existent while the H3O+ and A- bars significantly increase reflecting the fact that strong acids completely disassociate in water. The correct answer is 'The HA bar on the right must be converted completely to H3O+ and A-'.
The correct answer to the given question is option D).
In the context of this question, the graph represents a moderately weak acid and we're asked to determine what changes would occur in the graph for a relatively strong acid.
Here, the 'HA' would represent the weak acid that partially disassociates into H3O+ (hydronium ions) and A- (the conjugate base). One characteristic of a strong acid is that it completely disassociates in water.
Therefore, to represent a strong acid, the 'HA' bar on the right would need to be much lower or even non-existent (to indicate complete disassociation). In turn, the H3O+ and A- bars on the right would need to increase significantly/acquire all the 'HA' bar's original height to represent the products of the strong acid's complete disassociation.
So, the correct answer would be (D) 'The HA bar on the right must be converted completely to H3O+ and A-'.
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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.
How much energy is required to vaporize 48.7 g of dichloromethane (CH2Cl2) at its boiling point, if its ΔHvap is 31.6 kJ/mol?
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
Consider the reaction 4PH3(g) → P4(g) + 6H2(g) At a particular point during the reaction, molecular hydrogen is being formed at the rate of 0.137 M/s.__(a) At what rate is P4 being produced? M/s (b) At what rate is PH3 being consumed? M/s
Answer:
The rate at which [tex]P_4[/tex] is being produced is 0.0228 M/s.
The rate at which [tex]PH_3[/tex] is being consumed is 0.0912 M/s.
Explanation:
[tex]4PH_3\rightarrow P_4(g)+6H_2(g)[/tex]
Rate of the reaction : R
[tex]R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]
The rate at which hydrogen is being formed = [tex]\frac{d[H_2]}{dt}=0.137 M/s[/tex]
[tex]R=\frac{1}{6}\frac{d[H_2]}{dt}[/tex]
[tex]R=\frac{1}{6}\times 0.137 M/s=0.0228 M/s[/tex]
The rate at which [tex]P_4[/tex] is being produced:
[tex]R=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]
[tex]0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}[/tex]
The rate at which [tex]PH_3[/tex] is being consumed :
[tex]R=\frac{-1}{4}\frac{d[PH_3]}{dt}[/tex]
[tex]0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}[/tex]
[tex]\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s[/tex]
The rate at which P4 is being produced is 0.034 M/s and the rate at which PH3 is being consumed is 0.2055 M/s.
Explanation:The given reaction is 4PH3(g) → P4(g) + 6H2(g). We are given the rate at which molecular hydrogen is being formed, which is 0.137 M/s. To find the rate at which P4 is being produced, we need to consider the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of PH3 consumed, 1 mole of P4 is produced. Therefore, the rate at which P4 is being produced is 0.137/4 or 0.034 M/s.
Similarly, to find the rate at which PH3 is being consumed, we can use the stoichiometry of the reaction. From the balanced equation, we can see that for every 4 moles of PH3 consumed, 6 moles of H2 is produced. Therefore, the rate at which PH3 is being consumed is (6/4) * 0.137 or 0.2055 M/s.
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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.
If the entire solar system were about the size of a quarter (roughly 1" in diameter), approximately how far away would the nearest star be?
Answer:
Explanation:
If our entire Solar System were the size of a quarter, the planets and the sun is now a tiny speck of dust. The flat disc of the coin can represent the orbits of the planets.
Using this scale, the diameter of our Milky Way galaxy will be about the size of North America.
The nearest star, other than our own Sun, is about four light years away. That means it takes four years for its light to reach us. Since light travels at a speed of 3.0 x 10^8 meters per second, each light year is such a great distance. Proxima Centauri Milky way would be another quarter, two soccer fields away.
A much further star, Deneb is actually 1,800 light years away, the nearest star would be about 24,000 miles away.
A solution of sugar contains 35 gramsof sucrose, C12H22O11in 100 mL of water. What is the percent composition of the solution?
Answer:
Percent composition of the solution is 26 % of sucrose and 74 % of water
Explanation:
Percent composition is the mass of solute, either of solvent in 100 g of solution.
Mass of solution = Mass of solvent + Mass of solute
Mass of solute = 35 g
Mass of solvent = 100 g
As we know, water density = 1g/mL
So 1g/mL . 100 mL = 100 g
35 g + 100 g = 135 g → Mass of solution
(Mass of solute / Mass of solution) . 100 =
(35 g / 135 g) . 100 = 26 %
(Mass of solvent / Mass of solution) . 100 =
(100 g / 135 g) . 100 = 74 %
To calculate the percent composition of sucrose in the solution, divide the mass of sucrose (35 grams) by the total mass of the solution (sucrose plus water, which is 135 grams) and multiply by 100%. The solution has a percent composition of approximately 25.93% sucrose.
Explanation:The question involves calculating the percent composition of a solution by mass. If a solution contains 35 grams of sucrose (C12H22O11) in 100 mL of water (noting that the density of water is roughly 1 g/mL, so we have 100 grams of water), the total mass of the solution is the sum of the mass of the solute (sucrose) and the solvent (water), which is 35 g + 100 g = 135 g. To find the percent by mass of sucrose in the solution, we use the formula:
Percent by mass of sucrose = (Mass of sucrose / Total mass of solution) × 100%
Inserting the values we have:
Percent by mass of sucrose = (35 g / 135 g) × 100% ≈ 25.93%
Therefore, the percent composition of sucrose in the solution is approximately 25.93%.
The amount of heat absorbed by the alcohol was determined to be 1.17kJ. Given that the specific heat of the alcohol is 2.42J/gC , calculate the change in temperature
Answer:
ΔT = 20.06 °C
Explanation:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 1.17 kJ = 1170 J
m = mass = 24.1 g
Cp = Specific Heat Capacity = 2.42 J.g⁻¹.°C⁻¹
ΔT = Change in Temperature = ??
Solving eq. 1 for ΔT,
ΔT = Q / m Cp
Putting values,
ΔT = 1170 J / 24.1 g × 2.42 J.g⁻¹.°C⁻¹
ΔT = 20.06 °C
High self-monitors prefer situations in which clear expectations exist regarding how they're supposed to communicate. True False
Answer:
True
Explanation:
This are acts or actions that concur with situational expectations.