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.
Learn more about Bullet identification here:https://brainly.com/question/32247848
#SPJ6
When light is absorbed by chlorophyll that has been extracted into a solvent solution, it _________ releases the energy in the form of ____________
Answer:
Photosystems PSI (P700) and PSII (P680) and photolysis of water releases energy
In the form of movement of electrons which is later extracted as ATP
Explanation:
On absorbing light energy electrons obtained from photolysis of water and those present in and those present in the reaction centres PSI and PSII move to the higher energy level and electron acceptors accept them. They then move down through the electron transport chain to the lower energy level during which ATP is produced which is a chemical form of energy.
When atmospheric carbon dioxide reacts with water in the atmosphere, it forms carbonic acid. This acid can then react with some forms of rock, weakening it and changing its chemical composition. What is this known as?
Answer:
The terms used to describe the given process in Chemical weathering.
Explanation:
Weathering of rock is defined as breaking down of the rock into small pieces.
There are two types pf weathering :
Mechanical weathering : This weathering is due to change in physical parameters : temperature change, pressure change etc.
For example : When water soaked up in cracks or crevices of rocks freezes it expands and physically breaks the rock.
Chemical weathering : This weathering is decomposition of rocks due to action of chemicals.This type of weathering also changes chemical composition of rocks.
For example : when gases like carbon dioxide, sulfur oxide get dissolves in water present in rock to form weak acid which ease up the dissolving of rock in that weak acid.
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%
Which results when an atom has such a strong attraction for electrons that it pulls one or more electrons completely away from another atom?
All people are tax people
-Turbo tax
Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 125 pm . Calculate the density of solid crystalline chromium in grams per cubic centimeter.
Answer:
[tex]\rho=7.15\ g/cm^3[/tex]
Explanation:
The expression for density is:
[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]
[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]
M is molar mass of Chromium = 51.9961 g/mol
For body-centered cubic unit cell , Z= 2
[tex]\rho[/tex] is the density
Radius = 125 pm = [tex]1.25\times 10^{-8}\ cm[/tex]
Also, for BCC, [tex]Edge\ length=\frac{4}{\sqrt{3}}\times radius=\frac{4}{\sqrt{3}}\times 1.25\times 10^{-8}\ cm=2.89\times 10^{-8}\ cm[/tex]
Thus,
[tex]\rho=\frac{2\times \:51.9961}{6.023\times \:10^{23}\times \left(2.89\times 10^{-8}\right)^3}\ g/cm^3[/tex]
[tex]\rho=7.15\ g/cm^3[/tex]
The density of solid crystalline chromium will be "7.15 g/cm³".
According to the question,
Radius,
125 pm or [tex]1.25\times 10^{-8} \ cm[/tex]Molar mass of Chromium,
M = 51.9961 g/molFor body-centered,
Z = 2For BCC,
The edge length will be:
= [tex]\frac{4}{\sqrt{3} }\times radius[/tex]
= [tex]\frac{4}{\sqrt{3} }\times 1.25\times 10^{-8}[/tex]
= [tex]2.89\times 10^{-8} \ cm[/tex]
hence
The expression for density will be:
→ [tex]\rho = \frac{Z\times M}{N_a(Edge \ length)^3}[/tex]
By substituting the values, we get
[tex]= \frac{2\times 51.9961}{6.023\times 10^{23}\times (2.89\times 10^{-8})}[/tex]
[tex]= 7.15 \ g/cm^3[/tex]
Thus the above approach is right.
Learn more:
https://brainly.com/question/23711097
A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.
If the gas was cooled from 20°C to -60°C, what was its new volume? ( Don't forget to convert Celsius to Kelvin)
The gas laws involve Charles' Law, showing a direct proportionality between volume and temperature for a gas. The new volume of the gas after being cooled from 20°C to -60°C (converted to Kelvin), which is under constant pressure, is calculated to be approximately 30.6 mL.
Explanation:This question involves the concept of the gas laws in Physics, specifically, Charles' Law which states that the volume of a given mass of an ideal gas is directly proportional to its temperature on an absolute scale if pressure and the amount of gas remain constant.
First, let's convert the Celsius temperatures to Kelvin by adding 273.15. This gives us initial temperature T1 = 20°C + 273.15 = 293.15 K and final temperature T2 = -60°C + 273.15 = 213.15 K. The initial volume V1 is 42 mL.
According to Charles' Law, V1/T1 = V2/T2. Substituting the given values, we find the new volume V2 = V1*(T2/T1) = 42 mL * (213.15 K / 293.15 K) = 30.6 mL.
Learn more about Charles' Law here:https://brainly.com/question/16927784
#SPJ12
Answer: that one
Explanation:
What must be true for diffusion of a solute to occur across a partition that separates two compartments?
Answer:
The following must be true for diffusion of a solute to occur across a partition that separates two compartments
The partition is semi permeable
There is concentration gradient across the partition
The process is known as Osmosis
Explanation:
Diffusion is the migration of particles of a substance, liquid or gas, from a region of high concentration to a region of lower concentration and it only occurs where there is a concentration gradient for example spraying a perfume at a corner of a room, the spray sent spreads across the room as there is a concentration gradient
Osmosis is the movement or diffusion of solvent molecules across a semi permeable membrane from a region of low solute concentration to a region of high solute concentration to create a balanced concentration of solute on both sides of the compartment.
Diffusion requires a concentration gradient and a membrane permeable to the solute; in osmosis, water diffuses to balance solute concentrations.
For the diffusion of a solute to occur across a partition that separates two compartments, there must be a concentration gradient present, where the solute molecules move from an area of higher concentration to an area of lower concentration. Additionally, permeability of the membrane to the solute is crucial, as only materials capable of passing through will diffuse.
In certain situations, such as osmosis, water will move across the membrane to balance the concentration gradient of solutes, when the solutes themselves cannot pass through the membrane. In living systems, osmosis is a dynamic and continuous process that maintains the balance of fluids.
Flammable materials, like alcohol, should never be
dispensed or used near
A. an open door.
B. an open flame.
C. another student.
D. a sink
Answer:
b.open flame because it is fundamental end of the alcohol mixes in with the flame then it will become a bigger fire
The safest practice is to never use flammable materials like alcohol near an open flame as it can easily ignite and cause a fire.
Explanation:Flammable materials such as alcohol should never be dispensed or used near an open flame (option B). These substances are highly reactive and can easily ignite, potentially causing a fire. It's crucial for safety purposes to avoid using flammable materials around direct sources of heat or open flames. This rule is applicable not only in chemistry laboratories, but also in any other scenario where flammable substances are present.
Learn more about flammable materials safety here:https://brainly.com/question/26230725
#SPJ6
We create an electron with wavefunction ψ = [ψ1s(r)+3ψ3s(r)]/ √ 10. Find the expected value hEi of the energy in that state. Use the energies of the hydrogen atom to extract your answer in electron volts.
Answer: r=132pm
Explanation:
r = ⟨r⟩=5a0/Z
r=(5(5.29177*〖10〗^(-11) m) x^2)/2
r=132pm
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.
At a certain temperature, the Kp for the decomposition of H2S is 0.842.
H2S(g) image from custom entry tool H2(g) + S(g)
Initially, only H2S is present at a pressure of 0.104 atm in a closed container. What is the total pressure in the container at equilibrium?
Answer:
0.1976 atm is the total pressure in the container at equilibrium.
Explanation:
The value of [tex]K_p[/tex] for the given reaction = 0.842
Partial pressure of the hydrogen sulfide = 0.104 atm
Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm
Partial pressure of the hydrogen gas at equilibrium = x
Partial pressure of the sulfur gas at equilibrium = x
[tex]H_2S(g)\rightleftharpoons 2H_2(g)+S(g)[/tex]
Initially
0.104 atm 0 0
At equilibrium
(0.104 -2x) 2x x
The expression of [tex]K_p[/tex] is given by ;
[tex]K_p=\frac{x\times x}{(0.104 -x) }[/tex]
[tex]0.842 =\frac{x^2}{(0.104-x)}[/tex]
Solving for x:
x = 0.0936 atm
Partial pressure of the hydrogen sulfide at equilibrium = (0.104-x) atm
= (0.104-0.0936 ) atm = 0.0104 atm
Partial pressure of the hydrogen gas at equilibrium = x = 0.0936 atm
Partial pressure of the sulfur gas at equilibrium = x = 0.0936 atm
Total pressure in the container will be sum of all the partial pressure of the hydrogen sulfide gas, hydrogen gas and sulfur gas at the equilibrium.
P = 0.0104 atm + 0.0936 atm + 0.0936 atm = 0.1683 atm
0.1976 atm is the total pressure in the container at equilibrium.
The total pressure in the system at equilibrium is calculated using the equilibrium constant Kp and the initial pressure of the system. By setting up a equilibrium constant expression and solving the quadratic equation for 'x', the change in pressure, we find that the total pressure at equilibrium in the system is ~0.101 atm.
Explanation:The decomposition reaction of H2S into H2 and S is a chemical equilibrium problem in which the equilibrium constant is given in terms of pressure, or Kp. We know that at the beginning, only H2S is present at a pressure of 0.104 atm, and at equilibrium, the pressure of H2S decreases by 'x', being 'x' the pressure of H2 and S at equilibrium. According to the expression of the equilibrium constant Kp, Kp = (pH2* pS) / p_H2S = x^2 / (0.104-x).
From here, if we solve for 'x' using the quadratic formula and the given Kp = 0.842, we find two solutions, ~0.0967 atm and ~0.0072 atm. We discard the result 0.0967 because it cannot be higher than the initial pressure of the system, so the correct 'x' is ~0.0072 atm. Therefore, the total pressure will be the sum of the individual pressures at equilibrium. Total Pressure = pH2 + pS + p_H2S = 2x + (0.104-x) = 2*0.0072 + (0.104-0.0072) = ~0.101 atm.
Learn more about Chemical Equilibrium here:https://brainly.com/question/3920294
#SPJ12
A 3.78-gram sample of iron metal is reacted with sulfur to produce 5.95 grams of iron sulfide. Determine the empirical formula of this compound.
Answer:FeS
Explanation:
Mass of sulphur=5.95-3.78=2.17g
Fe S
3.78/56. 2.17/32
0.0675/0.0675. 0.0678/0.0675
1:1
Empirical formula=Fe1S1=FeS
An empirical formula is a compound's chemical formula that only specifies the ratios of the elements it contains and not the precise number or arrangement of atoms. The empirical formula is FeS.
The formula of a material expressed with the smallest integer subscript is referred to as an empirical formula for a compound. The empirical formula provides details regarding the ratio of atom counts in the molecule. A compound's empirical formula is directly related to its % content.
Mass of sulphur = 5.95 - 3.78 = 2.17g
Moles of Fe and S are:
Fe = 3.78/56
n = 0.0675
S = 2.17/32
n = 0.0678
Dividing with the smallest number:
0.0675/0.0675 = 1
0.0678/0.0675 = 11
The ratio is:
1:1
Empirical formula = FeS
To know more about empirical formulas, visit;
https://brainly.com/question/29506551
#SPJ3
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.
Learn more about Volume change in Iron's allotropic transformation here:https://brainly.com/question/33518175
#SPJ12
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.
Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25∘C diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine △ Hrxn for C(diamond) → C(graphite) with equations from the following list: (1) C(dianond)+O2(g)⟶CO2(g)ΔH=−395.4kJ (2) 2CO2(g)⟶2CO(g)+O2(g)ΔH=566.0kJ (3) C(graphite)+O2(g)→CO2(g)ΔH=−393.5kJ (4) 2CO(g)⟶C(graphite)+CO2(g)ΔH=−172.5kJ
Answer:
-1.9 KJ/mol
Explanation:
In order to solve the problem, we have to rearrange the equations in a way in which all molecules of O₂ and CO₂ are eliminated:
2C(diamond) + 2O₂(g) → 2CO₂(g) ΔH₁= 2 x (-395.4 KJ) ------> we multiply by 2 both reactants and products
2 CO₂(g) → 2CO(g) + O₂(g) ΔH₂= 566.0 KJ
CO₂(g) → C(graphite) + O₂(g) ΔH₃= -1 x (-393.5 KJ) ------> we use reverse rxn
2CO(g) → C(graphite) + CO₂(g) ΔH₄= -172.5 KJ
When we cancel the molecules that appear both in reactants and products, the total reaction is the following:
2C(diamond) → 2C(graphite)
ΔHt= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = 2 x (-395.4 KJ) + 566.0 KJ + (-1 x (-393.5 KJ)) - 172.5 KJ
ΔHt= 347.2 KJ
This is for 2 mol of C(diamond) which are converted in 2 mol of C(graphite). To obtain ΔH for the reaction of 1 mol C(diamond) to 1 mol (graphite) we have to divide into 2:
ΔH= -3.8 KJ/2mol= -1.9 KJ/mol
To determine ΔHrxn for the reaction C(diamond) → C(graphite), we can use Hess's law and the given equations. By canceling out common compounds/products and summing the remaining equations, we can calculate ΔHrxn. The enthalpy change can be determined by summing the enthalpy changes of the canceled equations.
Explanation:This type of calculation usually involves the use of Hess's law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. For this specific reaction, we can use the given equations to determine the enthalpy change ΔHrxn for C(diamond) → C(graphite).
Step 1: C(diamond) + O2(g) → CO2(g) (ΔH = -395.4 kJ)Step 2: 2CO2(g) → 2CO(g) + O2(g) (ΔH = 566.0 kJ)Step 3: C(graphite) + O2(g) → CO2(g) (ΔH = -393.5 kJ)Step 4: 2CO(g) → C(graphite) + CO2(g) (ΔH = -172.5 kJ)To obtain the ΔHrxn for C(diamond) → C(graphite), we need to cancel out common compounds/products in these equations. From the given equations, we can see that CO2(g) is a common compound in steps 1, 2, and 3, and thus we can cancel it out. Likewise, O2(g) is present in steps 1, 2, and 3, so we can also cancel it out. By summing the canceled equations, we get the final equation:
C(diamond) → C(graphite)
The enthalpy change for this reaction is the sum of the enthalpy changes of the canceled equations:
ΔHrxn = ΔH1 + ΔH2 + ΔH3 + ΔH4 = -395.4 kJ + 566.0 kJ + (-393.5 kJ) + (-172.5 kJ) = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + (-393.5 kJ) + 566.0 kJ + (-172.5 kJ) = -395.4 kJ - 393.5 kJ + 566.0 kJ - 172.5 kJ = -395.4 kJ - 393.5 kJ + 566.0 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ
Learn more about Calculating Enthalpy Change here:https://brainly.com/question/28988531
#SPJ3
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
What potential difference is required to bring the proton to rest?
Answer:
0
Explanation:
Δk = (kf - Ki). Kf is 0 because your stopping it [m(v_i)^2 ]/2q = V
The potential difference required to bring a proton to rest can be calculated by knowing the initial kinetic energy of the proton. This energy, when applied in the opposite direction, would act to decelerate the proton and bring it to rest.
Explanation:The potential difference required to stop a moving proton can be calculated using the equation PEelec = qV, where q is the proton's charge and V is the potential difference (also known as voltage). Given that the proton's charge, q = qe = 1.6×10−¹⁹ Coulombs, we can rearrange the equation to solve for V: V = PEelec/q. If we know the initial kinetic energy of the proton (its energy due to motion, which can be thought of as the electrical energy it gained from accelerating through the potential difference), then the same voltage (potential difference) applied in the opposite direction will bring the proton to rest.
In other words, if the proton gained a certain amount of energy (in electron volts, eV) while accelerating through a potential difference, it will require the same amount of energy in the opposite direction to decelerate and come to rest.
For example, if the proton gained 1 electron Volt (1 eV) of energy from the potential difference, it would require a potential difference of -1 V to bring it to rest. This is because 1 eV = (1 V)(1.6×10−¹⁹ C).
Learn more about Potential Difference here:https://brainly.com/question/30893775
#SPJ12
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 compound contains carbon, hydrogen, and chlorine. It has a molar mass of 98.95 g/mol. Analysis of a sample shows that it contains 24.27% carbon and 4.07% hydrogen. What is its molecular formula?
Answer:
C2H4Cl2
Explanation:
Firstly, we know that the compound contains only three elements. These are carbon, hydrogen and oxygen. We have the percentage compositions of carbon and hydrogen, thus we need the one for chlorine. To get the one for chlorine, we simply subtract that of carbon and hydrogen from a total of 100%.
Hence percentage composition of chlorine = 100 - 24.27 - 4.07 = 71.66%
Now, we divide the percentage compositions by the atomic masses. The atomic masses of carbon, hydrogen and chlorine are 12, 35.5 and 1 respectively. We go on to the divisions as follows.
C = 24.27/12 = 2.0225
H = 4.07/1 = 4.07
Cl = 71.66/35.5 = 2.02
We then go on to divide each by the smallest which is 2.02
C = 2.0225/2.02 = 1
H = 4.07/2.02 = 2
Cl = 2.02/2.02 = 1
Hence the empirical formula is CH2Cl
Now, since the molecular mass is 98.95, we need to calculate the molecular formula
Hence, [CH2Cl]n = 98.95
[12 + 2(1) + 35.5]n = 98.95
[12 + 2 + 35.5]n = 98.95
49.5n = 98.95
n = 98.95/49.5 = 2
The molecular formula is thus [CH2Cl]2 = C2H4Cl2
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.
Learn more: https://brainly.com/question/25748313
You breathe in 3.0 L of pure oxygen at 298 K and 1,000 kPa to fill your lungs. How many moles of oxygen did you take in? Use the ideal gas law: PV = nRT where R=8.31 L−kPa/mol−K
A. 0.05 MOLE
B. 1.21 MOLES
C. 2.42 MOLES
D. 20.0 MOLES
Answer:
Option B.
Explanation:
Let's replace the data given in the formula of Ideal Gases Law.
P. V = n . R . T
3L . 1000kPa = n . 8.31 L.kPa/mol.K . 298K
(3L . 1000 kPa) / ( 8.31 L.kPa/mol.K . 298K) = n
1.21 moles
The number of moles of oxygen by using ideal gas equation is 1.21 moles. Hence, the correct option is B.
Ideal Gas equation[tex]PV=nRT[/tex]
Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the temperature.
We can calculate the number of moles by using the above equation as follows:-
[tex]PV = n R T\\ 1000kPa\times 3L = n \times 8.31 L.kPa/mol.K \times298K\\(3L . 1000 kPa) / ( 8.31 L.kPa/mol.K . 298K) = n\\n=1.21 moles[/tex]
So, the number of moles of oxygen is 1.21 moles.
Learn more about the ideal gas equation here
brainly.com/question/4147359
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
Pick the correct statement for the following isotope: a. 42Ca 42 is the mass number and 20 is the atomic number. b. 42 is the number of neutrons and 20 is the number of protons. c. 42 is the number of protons and 20 is the number of electrons. d. 42 is the atomic number and 20 is the number of neutrons.
Answer:
A
Explanation:
To label an element correctly using a combination of the symbol, mass number and atomic number furnishes some important information about the element.
We can obtain these information from the element provided that correct labeling of the element is presented. Firstly, after writing the symbol of the element, the atomic number is placed as a subscript on the left while the mass number of the atomic mass is placed as a superscript on the same left.
Looking at the question asked, we have the element symbol in the correct position as Ca, with 42 also in the correct position which is the mass number. The third number which is 20 is thus the atomic number of the element.
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
The average C - H bond energy in CH4 is 415 kJ / mol. Use the following data to calculate the average C - H bond energy in ethane ( C2H6;C - C bond ) , in ethane ( C2H4;C C bond ) , and in ethyne ( C2H2;C C bond ) .
C2H6 ( g ) + H2 ( g ) rightarrow 2CH4 ( g ) ΔHrxn = - 65.07 kj / mol
C2H4 ( g ) + H2 ( g ) rightarrow 2CH4 ( g ) ΔH = - 202.21kJ / mol
C2H2 ( g ) + 3H2 ( g ) rightarrow 2 CH4 ( g ) ΔH = - 376.74kj / mol
Average C - H bond
Average C - H bond energy in ethane = kJ / mol energy in ethane = kj / mol
Average C - H bond energy in ethyne = kJ / mol
Average C-H bond energy: Ethane = 764.93 kJ/mol, Ethene = 627.79 kJ/mol, Ethyne = 528.26 kJ/mol.
To find the average C-H bond energy in ethane (C₂H₆), ethene (C₂H₄), and ethyne (C₂H₂), we can use Hess's law and the given reactions along with the average C-H bond energy in methane (CH₄), which is 415 kJ/mol.
1. **Calculate the average C-H bond energy in ethane (C₂H₆):**
Given reaction:
[tex]\[C2H6 (g) + H2 (g) \rightarrow 2CH4 (g) \quad \Delta H_{\text{rxn}} = -65.07 \, \text{kJ/mol}\][/tex]
This reaction breaks one C-C bond and adds two C-H bonds.
Change in bond energy = Total energy of bonds broken - Total energy of bonds formed
[tex]\[= 1 \times (\text{C-C bond energy}) - 2 \times (\text{C-H bond energy})\][/tex]
From the given reaction, the change in bond energy is -65.07 kJ/mol.
Substituting the known values:
[tex]\[-65.07 \, \text{kJ/mol} = 1 \times (\text{C-C bond energy}) - 2 \times (415 \, \text{kJ/mol})\][/tex]
Solving for the C-C bond energy:
[tex]\[\text{C-C bond energy} = 2 \times 415 - 65.07 \, \text{kJ/mol} = 764.93 \, \text{kJ/mol}\][/tex]
2. **Calculate the average C-H bond energy in ethene (C₂H₄):**
Given reaction:
[tex]\[C2H4 (g) + H2 (g) \rightarrow 2CH4 (g) \quad \Delta H = -202.21 \, \text{kJ/mol}\][/tex]
This reaction breaks one C=C bond and adds two C-H bonds.
Change in bond energy = Total energy of bonds broken - Total energy of bonds formed
[tex]\[= 1 \times (\text{C=C bond energy}) - 2 \times (\text{C-H bond energy})\][/tex]
From the given reaction, the change in bond energy is -202.21 kJ/mol.
Substituting the known values:
[tex]\[-202.21 \, \text{kJ/mol} = 1 \times (\text{C=C bond energy}) - 2 \times (415 \, \text{kJ/mol})\][/tex]
Solving for the C=C bond energy:
[tex]\[\text{C=C bond energy} = 2 \times 415 - 202.21 \, \text{kJ/mol} = 627.79 \, \text{kJ/mol}\][/tex]
3. **Calculate the average C-H bond energy in ethyne (C₂H₂):**
Given reaction:
[tex]\[C2H2 (g) + 3H2 (g) \rightarrow 2CH4 (g) \quad \Delta H = -376.74 \, \text{kJ/mol}\][/tex]
This reaction breaks one C≡C bond and adds four C-H bonds.
Change in bond energy = Total energy of bonds broken - Total energy of bonds formed
[tex]\[= 1 \times (\text{C≡C bond energy}) - 4 \times (\text{C-H bond energy})\][/tex]
From the given reaction, the change in bond energy is -376.74 kJ/mol.
Substituting the known values:
[tex]\[-376.74 \, \text{kJ/mol} = 1 \times (\text{C≡C bond energy}) - 4 \times (415 \, \text{kJ/mol})\][/tex]
Solving for the C≡C bond energy:
[tex]\[\text{C = C bond energy} = 4 \times 415 - 376.74 \, \text{kJ/mol} = 528.26 \, \text{kJ/mol}\][/tex]
So, the average C-H bond energy in ethane [tex](C2H6) is \(764.93 \, \text{kJ/mol}\), in ethene (C2H4) is \(627.79 \, \text{kJ/mol}\), and in ethyne (C2H2) is \(528.26 \, \text{kJ/mol}\).[/tex]
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!
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!
What makes a nucleus unstable
Answer:
Unstable Nuclei. ... Too many neutrons or protons upset this balance disrupting the binding energy from the strong nuclear forces making the nucleus unstable. An unstable nucleus tries to achieve a balanced state by given off a neutron or proton and this is done via radioactive decay.
Explanation:
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.
Learn more about Anabolic and Catabolic reactions here:https://brainly.com/question/31913853
#SPJ3
A gas mixture of nitrogen and oxygen having a total pressure of 2.50 atm is above 2.0 L of water at 25 °C. The water has 51.3 mg of nitrogen dissolved in it. What is the molar composition of nitrogen and oxygen in the gas mixture? The Henry’s constants for N2 and O2 in water at 25 °C are 6.1×10–4 M/atm and 1.3×10–3 M/atm, respectively.
Answer: The molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
Given mass of nitrogen gas = 51.3 mg = 0.0513 g (Conversion factor: 1 g = 1000 mg)
Molar mass of nitrogen gas = 28 g/mol
Volume of solution = 2 L
Putting values in above equation, we get:
[tex]\text{Molarity of nitrogen gas}=\frac{0.0513g}{28g/mol\times 2L}\\\\\text{Molarity of nitrogen gas}=9.16\times 10^{-4}mol/L[/tex]
To calculate the partial pressure, we use the equation given by Henry's law, which is:
[tex]C_{N_2}=K_H\times p_{N_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]6.1\times 10^{-4}mol/L.atm[/tex]
[tex]C_{N_2}[/tex] = molar solubility of nitrogen gas = [tex]9.16\times 10^{-4}mol/L[/tex]
Putting values in above equation, we get:
[tex]9.16\times 10^{-4}mol/L=6.1\times 10^{-4}mol/L.atm\times p_{N_2}\\\\p_{N_2}=\frac{9.16\times 10^{-4}mol/L}{6.1\times 10^{-4}mol/L.atm}=1.50atm[/tex]
We are given:
Total pressure of the mixture = 2.50 atm
Partial pressure of oxygen gas = 2.50 - 1.50 = 1.00 atm
To calculate the mole fraction of a substance at 25°C, we use the equation given by Raoult's law, which is:
[tex]p_{A}=p_T\times \chi_{A}[/tex] ......(1)
where,
[tex]p_A[/tex] = partial pressure
[tex]p_T[/tex] = total pressure
[tex]\chi_A[/tex] = mole fraction
For nitrogen gas:We are given:
[tex]p_{N_2}=1.50atm\\p_T=2.50atm[/tex]
Putting values in equation 1, we get:
[tex]1.50atm=2.50\times \chi_{N_2}\\\\\chi_{N_2}=\frac{1.50}{2.50}=0.6[/tex]
For oxygen gas:We are given:
[tex]p_{O_2}=1.00atm\\p_T=2.50atm[/tex]
Putting values in equation 1, we get:
[tex]1.00atm=2.50\times \chi_{O_2}\\\\\chi_{O_2}=\frac{1.00}{2.50}=0.4[/tex]
Hence, the molar composition of nitrogen gas is 0.6 and that of oxygen gas is 0.4
The mole fraction of nitrogen and oxygen is 0.6 and 0.4 respectively.
Data Given;
Total pressure P = 2.50 atmvolume of the water = 2.0Lmass of water = 51.3mg = 0.0513gHenry constant for N2 = 6.1*10^-4 M/atmHenry constant for O2 = 1.3*10^-3 M/atmHenry's LawThe law states that the mass of a dissolved gas in a given volume of solvent at equilibrium will be proportional to the partial pressure of the gas.
Mathematically;
C = KP
c = concentration of the gasK = Henry's constantP = partial pressureThe number of moles Nitrogen dissolved is
[tex]n = mass/ molar mass\\ n = 0.0513/28\\ n = 0.00183127 moles[/tex]
The concentration of Nitrogen in water is
[tex]\frac{0.00183127}{2} *1 = 0.0009156M[/tex]
Applying Henry's law,
[tex]0.0009156 = 6.1*10^-^4 * P\\ P = 1.5atm[/tex]
The partial pressure of nitrogen in the mixture is 1.5atm
The total pressure of the gas is 2.50atm
Partial Pressure of oxygen = total pressure - partial pressure of nitrogen
partial pressure of oxygen = 2.50 - 1.50 = 1
Pressure FractionThe pressure fraction of the gas is the ratio between the partial pressure to the total pressure
pressure fraction of nitrogen = 1.5/2.5 = 0.6
pressure fraction of oxygen = 1/2.5 = 0.4
But partial pressure is equal to molar fraction.
This makes the mole fraction of nitrogen equals 0.6 and mole fraction of oxygen equals 0.4.
Learn more about Henry's law here;
https://brainly.com/question/12823901
A water bath in a physical chemistry lab is 1.85m long, 0.810m wide and 0.740m deep. If it is filled to within 2.57 in from the top, how many liters of water are in it
Final answer:
The water bath, filled to within 2.57 inches of the top, contains approximately 1014.48 liters of water. This calculation involves converting inches to meters, calculating the volume in cubic meters, and then converting that volume to liters.
Explanation:
To calculate the volume of water in the water bath, we need to take into account the dimensions given and the fact that it is filled to within 2.57 inches from the top. First, we need to convert the depth from which it's filled into meters, as the other dimensions are given in meters.
2.57 inches = 0.06533 meters (since 1 inch = 0.0254 meters)
The actual depth filled with water will be:
0.740 m (total depth) - 0.06533 m = 0.67467 m
Next, we multiply the length, the width, and the newly calculated depth to get the volume in cubic meters:
Volume = 1.85 m × 0.810 m × 0.67467 m = 1.01448 cubic meters
To convert the volume from cubic meters to liters, we use the conversion factor 1 cubic meter = 1000 liters:
Volume = 1.01448 m3 × 1000 L/m3 = 1014.48 liters
Therefore, there are approximately 1014.48 liters of water in the water bath.