Answer:
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
Explanation:
In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10
In the first rinse the concentration must be of 0.9M 10 = 0.09M
2nd = 0.009M
3rd = 0.0009M
4th = 0.00009M
5th = 0.000009M →
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
What is the Molarity of a 2 liter solution containing 43.55 grams of K2504?
Answer:
M = 0.125 M
Explanation:
Given data:
Molarity = ?
Volume of solution = 2 L
Mass of K₂SO₄ = 43.55 g
Solution;
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
Number of moles of solute:
Number of moles = mass/molar mass
Number of moles = 43.55 g / 174.26 g/mol
Number of moles = 0.25 mol
Molarity:
M = 0.25 mol / 2 L
M = 0.125 M
Three resonance structures of the given anion are possible. One is given, but it is incomplete. Complete the given structure by adding nonbonding electrons and formal charges. Draw the two remaining resonance structures (in any order), including nonbonding electrons and formal charges. Omit curved arrows.
Answer:
Explanation:
The missing incomplete resonance structure is attached in the image below. From there, we can see the addition of the nonbonding electrons and its' formal charge which makes the resonance structure a complete resonance structure. The others two resonance structure that can be derived from the complete structure is also shown in the image. Out of these three structures, the structure that contributes most to the hybrid is the structure with the negative charge on the oxygen.
(The stands for a number the student is going to calculate.)
The ___________ stands for a number the student is going to calculate. Fill in the missing part of this equation.
(87. 1/mmole. C). ______= _____ kJ/mol.C
Answer:
1000 kJ.mmole / 1000 J.mole
Explanation:
To solve this, we need to analyze the given data.
We have a number which is 87.1 J/mmole.C (I'm assuming it has the J at the beggining because if not, then you are missing some data) and the final result is kJ/mol.C
The only unit that has not changed in the process was the °C, while the mole and J change respectively. In this case, we need to know the conversion factor of mmole to mole and J to kJ.
In the case of a mole:
1 mole --------> 1000 mmole
In the case of Joule:
1 kJ ----------> 1000 J
So the first thing we will do is to change from J to kJ:
87.1 J * 1 kJ / 1000 J = 0.0871 kJ
Now let's convert mmol to mole:
0.0871 kJ/mmole.C * 1000 mmole / 1 mole = 87.1 kJ/mole.C
As you can see, there's is practicly no change at all with the units, so putting all together it would be:
87.1 J/mmole.C * 1000 kJ.mmole / 1000 J.mole = 87.1 kJ/mole.°CHope this helps
The 10x SDS gel electrophoresis buffer contains 250mM Tris HCl, 1.92M Glycine, and 1% (w/v) SDS. Buffers are always used at 1x concentration in the lab (unless specified otherwise in the protocol), so you will have to dilute the 10x buffer to 1x before use. What is the concentration of Tris and SDS in the 1x buffer
Answer:
25 mM Tris HCl and 0.1% w/v SDS
Explanation:
A 10X solution is ten times more concentrated than a 1X solution. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:
(250 mM Tris HCl)/10 = 25 mM Tris HCl
(1.92 M glycine)/10 = 0.192 M glycine
(1% w/v SDS)/10 = 0.1% w/v SDS
Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.
Photosynthesizing organisms use ____
to produce glucose.
Answer: Photosynthesizing organisms use carbon dioxide and water to produce glucose.
Explanation:
Photosynthesis is a phenomenon in which green plants containing chlorophyll use sunlight as a source of energy to convert carbon dioxide and water to form glucose and oxygen.
Photosynthesis is the process used by plants, algae and certain bacteria to convert energy from sunlight and turn it into chemical energy in the form of glucose which is used a s a source of energy by many organisms.
[tex]6CO_2+6H_2O\overset{sunlight}\rightarrow C_6H_{12}O_6+6O_2[/tex]
Where would a disease transmitted by person to person contact be most likely to spread quickly
Answer: inside
Explanation:
Answer:
direct and indirect contact
Explanation:
if you touch a doorknob right after an infected person than you make be exposed to the disease.
What produces the magnetic force of an electromagnet?
O magnetic fields passing through the device
O static charged particles on the wire
O movement of charged particles through the wire
O positive and negative charges repelling each other
Answer:
movement of charged particles through the wire .
Explanation:
When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .
For the reaction of ammonia (NH3) with oxygen (O2) to produce water and nitric oxide (NO), how many moles of water are produced when 2.2 moles of ammonia are reacted?
Answer:
3.3 moles of H₂O.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4NH₃ + 5O₂ —> 6H₂O + 4NO
From the balanced equation above,
4 moles of NH₃ reacted to produce 6 moles of H₂O.
Finally, we shall determine the number of mole of H₂O produced by the reaction of 2.2 moles of NH₃. This can be obtained as follow :
From the balanced equation above,
4 moles of NH₃ reacted to produce 6 moles of H₂O.
Therefore, 2.2 moles of NH₃ will react to produce = (2.2 × 6)/4 = 3.3 moles of H₂O.
Thus, 3.3 moles of H₂O were obtained from the reaction.
Calculate the percent composition (percent by mass of each element) of NH4Cl.
Round to the nearest ONES place ((example: 12.34% = 12%))
Answer:
[tex]\%N=26.2\%\\\\\%H=7.5\%\\\\\%Cl=66.3\%[/tex]
Explanation:
Hello!
In this case, since the calculation of the percent composition of an element in a chemical compound is computing considering its atomic mass, subscript in the formula and molecular mass of the compound it is; for nitrogen, hydrogen and chlorine we have that ammonium chloride has a molar mass of 53.49 g/mol so the percent compositions are:
[tex]\%N=\frac{14.01*1}{53.49}*100\% =26.2\%\\\\\%H=\frac{1.01*4}{53.49}*100\% =7.5\%\\\\\%Cl=\frac{35.45*1}{53.49}*100\% =66.3\%[/tex]
Best regards!
How does heat travel?
1. From cold things to hotter things
2. From hot things to colder things
3. Between things of the same temperature
Answer:
well heat travels by conduction, convection, and radiation but I think it's 2.
Explanation:
heat travels to colder things trying to make a balanced temperature for both of the objects.
An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using the periodic table, identify the element this is likely to be and explain your choice. (18 pts)
A) Isotope 1 – mass: 64 amu; percent abundance: 48.89%
B) Isotope 2 – mass: 66 amu; percent abundance: 27.81%
C) Isotope 3 – mass: 67 amu; percent abundance: 4.11%
D) Isotope 4 – mass: 68 amu; percent abundance: 18.57%
E) Isotope 5 – mass: 70 amu; percent abundance: 0.62%
Answer: Sol:-
Data provided in the question is :-
Atomic mass of isotope -1 = 64 amu
Atomic mass of isotope -2 = 66 amu
Atomic mass of isotope -3 = 67 amu
Atomic mass of isotope -4 = 68 amu
Atomic mass of isotope - 5 = 70 amu
Percentage abundace of isotope - 1 = 48.89 %
Percentage abundance of isotope -2 = 27.81 %
Percentage abundance of isotope - 3 = 4.11%
Percentage abundance of isotope-4 = 18.57%
Percentage abundance of isotope - 5 = 0.62 %
Formula used :-
Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]
Calculation :-
Put all the value in the formula :-
Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu
= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu
= {(6528.04) / 100} amu
= 65.2804 amu
Average atomic mass of an element is = 65.2804 amu
Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc
atomic mass of zinc = 65.38 \approx 65.2804 amu
The gas carbon dioxide is a pure substance. Which of the following is true about carbon dioxide? (5 points)
Select one:
a. Carbon and oxygen are chemically bonded in it.
b. Carbon and oxygen retain their original identity in it.
c. It can be separated into carbon and oxygen using physical methods.
d. The proportion of carbon and oxygen is different in different samples of the gas.
Answer:
Carbon and oxygen are chemically bonded in it.
Explanation:
The other answer choices do not apply for compounds, but rather for mixtures instead.
A sample of PCl5 weighting 2.69 gram was placed in 1.00 Litter container and completely vaporized at 250C. The pressure observed at that temperature was 1.00 atm. The possibility exists that some of the PCl5 dissociated according to PCl5 (g) ! PCl3 (g) Cl2 (g) . What must be the partial pressures of PCl5 PCl3 and Cl2 under these experimental conditions
Answer:
Partial pressures:
PCl₅ = 0.558 atm
PCl₃ = 0.22 atm
Cl₂ = 0.22 atm
Explanation:
From the given information:
The number of moles of PCl₅ associated with the evaporation is:
[tex]n_{PCl_5}= \dfrac {weight \ of \ PCl_5} {M.Wt. \ of \ PCl_5}[/tex]
[tex]n_{PCl_5}= \dfrac {2.69 \ g} {208.5 \ g/mol}[/tex]
[tex]n_{PCl_5}= 0.013 \ mol[/tex]
Temperature of the gas = 250° C = (250 + 273.15) K
= 523.15 K
Using the Ideal gas equation to determine the pressure exerted by the completely vaporized PCl₅
PV = nRT
[tex]P = \dfrac{nRT}{V}[/tex]
[tex]P = \dfrac{0.0013 \ mol \times 0.082 \ Latm^0 K^{-1} . mol ^{-1} \times 523.15 \ K}{1.0 \ L}[/tex]
P = 0.558 atm
Thus, at 250° C, decomposition of PCl₅ occurs.
In the container, PCl₅ decomposes to PCl₃ and Cl₂.
i.e.
[tex]PCl_{5(g)} \to PCl_{3(g)}+ Cl_{2(g)}[/tex]
Using Dalton's Law:
[tex]P_{total } =P_1 + P_2+P_3 +...[/tex]
[tex]P_1 = P_{Total} \times X_1[/tex]
where;
X = mole fraction
Then, the total no. of moles in the container is:
[tex]n = \dfrac{PV} {RT}[/tex]
[tex]n = \dfrac{1\ atm \times 1.0\ L}{0.0821 \ L \ atm \ K^{-1}.mol \times 523.15\ K}[/tex]
n = 0.023 mol
Now, the container contains a total amount of 0.023 mol where initially 0.013 mol are that of PCl₅ and remaining 0.005 mol of PCl₃ and 0.005 mol of Cl₂.
Thus, the partial pressure of PCl₃ is:
[tex]P__{PCL_3} }= P_{total} \times \dfrac{no. \ of \ moles \ of PCl_5}{total \ no. \ of \ moles}[/tex]
[tex]P__{PCL_3}} = 1 \ atm \times \dfrac{0.005}{0.023}[/tex]
[tex]P__{PCL_3}} = 0.22 \ atm[/tex]
Thus, since the no of moles of PCl₃ and Cl₂ are the same, then the partial pressure for Cl₂ is = 0.22 atm
A container holds 100.0 mL of nitrogen at 21° C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35° C?
Answer:
V₂ = 104.76 mL
Explanation:
Given data:
Initial volume = 100.0 mL
Initial temperature = 21°C (21 + 273.15 K = 294.15 K)
Final temperature = 35°C (35 + 273.15 K = 308.15 k)
Final volume = ?
Solution:
Charles Law:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ =100.0 mL × 308.15 K / 294.15 K
V₂ = 30815 mL.K /294.15 K
V₂ = 104.76 mL
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4.Calculate the Hydroxide, Hydrogen ion and POH of solution if the PH of solution is 7.b
5.Solution A Has PH =4 and solution B has PH = 7.How many times greater is the Hydroxide ion
concentration in solution A than the Hydronium ion concentration in solution B
the ph is gonna be your value and the 4 is gonna be your main subject
so as the ph is your value u gonna ad your ph and 7 and 4 toghter then multiple your answer 2 times because ph represent multiple and your value
Which water usage uses the least amount of water in a year in the United States? a industry
b livestock c irrigation d public water supply
plzzzzzzzz hurrrrry
A student dissolves of aniline in of a solvent with a density of . The student notices that the volume of the solvent does not change when the aniline dissolves in it. Calculate the molarity and molality of the student's solution. Be sure each of your answer entries has the correct number of significant digits.
Answer:
Molarity: 0.21M
Molality: 0.20m
Explanation:
...dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL...
To solve this question, we need to find the moles of aniline in 3.9g using its molar mass. Then, we need to find the kg and Liters of solution in order to find molarity (Moles/L solution) and molality (Moles/kg of solvent):
Moles aniline:
Molar mass:
6C: 6* 12.01g/mol = 72.06g/mol
7H: 7*1.008g/mol = 7.056g/mol
N: 1*14.007g/mol = 14.007g/mol
72.06g/mol+7.056g/mol+14.007g/mol = 93.123g/mol
Moles of 3.9g: 3.9g * (1mol / 93.123g) = 0.04188moles
Liters solution:
200mL * (1L / 1000mL) = 0.200L
kg solvent:
200mL * (1.05g/mL) * (1kg/1000g) = 0.210L
Molarity:
0.04188mol / 0.200L = 0.21M
Molality:
0.04188mol / 0.210L =0.20m
If the caffeine concentration in a particular brand of soda is 2.57 mg/oz, drinking how many cans of soda would be lethal
The lethal dose and how ounces of soda in a can of soda is not given, however, the standard lethal dose and volume of soda are given as below:
Lethal dose: 10 gm of caffeine
The volume of soda per can = 12oz/can
Answer:
The correct answer is - 324.254 cans or round up to 325 cans. Ans.
Explanation:
Given:
2.57 mg caffeine / 1oz
12oz / 1can
Lethal dose: 10.0g or 10,000mg of caffeine
Solution:
Caffeine per soda can = (2.57 mg caffeine / 1oz) * (12oz / 1can) = 30.84 mg caffeine / 1can.
lethal dose would be in =
(10,000mg caffeine) * (1can / 30.84 mg caffeine) = 324.254 cans or round up to 325 cans. Ans.
what state of matter travels in straight lines
Answer:
light
Explanation:
light is plasma, which is a state of matter
A clone has _________ chromosomes as its parent.
A. Half the number of
B. The same exact
C. double the number of
D. half of the same
Answer:
B. The same exact
Explanation:
I think B because in order to be a clone of your parent you have to have the exact same DNA and chromosomes.
Hope this helps :D
A clone has the same exact chromosomes as its parent.
CLONING:
Cloning is a genetic procedure in which identical copies of a cell or organism is made. Cloning can be done naturally or artificially, however, it follows the process of mitosis. In cloning, the genetic content of a parent cell is used as a template to replicate another cell or organism. Examples of cloning are biological twins, vegetative reproduction in plants etc. Therefore, a clone has the same exact chromosomes as its parent.Learn more at: https://brainly.com/question/12483409?referrer=searchResults
Use this equation for the next question:
2NaOH + H2SO4 ® Na2SO4 + 2H20
If a reaction produces 0.75 moles Na2SO4, how many moles of NaOH were used?
0.75 moles NaOH
2 moles NaOH
.375 moles NaOH
1.5 moles NaOH
Problem 3 A sample of 2.37 moles of an ideal diatomic gas experiences a temperature increase of 65.2 K at constant volume. (a) Find the increase in internal energy if only translational and rotational motions are possible. (b) Find the increase in internal energy if translational, rotational, and vibrational motions are possible. (c) How much of the energy calculated in (a) and (b) is translational kinetic energy?
Answer:
a) the increase in internal energy is 3211.78 J
b) dU = 3854.14 J
c) dU[tex]_{T}[/tex] = 1927.06 J
Explanation:
Given the data in question;
Foe a diatomic gas, the degree of freedom are as follow;
lets consider the positional degree of freedom
transitional df = 3
rotational df = 2
vibrational ff = 1
now, the internal energy given by;
U = Nf × 1/2NKT = Nf×1/2×nRT
where Nf is the number of degree of freedom
N is Number of atoms or molecules
n = number of molecules
L is Boltzmann constant
R is universal gas constant
so change in internal energy , change in T is given by
dU = Nf × 1/2 × nT dT
n = 2.37 moles
dT = 65.2 K
R = 8.314 J/mol.J
a)
Find the increase in internal energy if only translational and rotational motions are possible
since rotational and transitional motion are involved ;
Nf = 3(trasitional) + 2(rotational) = 5
so,
dU = 5 × 1/2 × nRdT
we substitute
dU = 5 × 0.5 × 2.37 × 8.314 × 65.2
dU = 3211.78 J
Therefore, the increase in internal energy is 3211.78 J
b)
Find the increase in internal energy if translational, rotational, and vibrational motions are possible.
Nf = 3 + 2 + 1 = 6
dU = 6 × 1/2 × nRdT
dU = 6 × 0.5 × 2.37 × 8.314× 65.2
dU = 3854.14 J
c)
How much of the energy calculated in (a) and (b) is translational kinetic energy?
dU[tex]_{T}[/tex] = 3 × 0.5 × 2.37 × 8.314 × 65.2
dU[tex]_{T}[/tex] = 1927.06 J
Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that liters per second of dinitrogen are consumed when the reaction is run at and the dinitrogen is supplied at . Calculate the rate at which ammonia is being produced. Give your answer in kilograms per second. Be sure your answer has the correct number of significant digits.
The question is incomplete. Here is the complete question.
In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 505. liters per second of dinitrogen are consumed when the reaction is run at 172.°C and 0.88 atm. Calculate the rate at which ammonia is being produced. Give your answer in kilogram per second. Be sure your answer has the correct number of significant digits.
Answer: Rate = 0.41 kg/s
Explanation: The balanced Haber reaction is
[tex]N_{2}+3H_{2}\rightarrow2NH_{3}[/tex]
As all the components are gases, we can use Ideal Gas Law, which relates Pressure (P), Volume (V), Temperature (T) and Moles (n) in the following formula:
PV = nRT
where
R is gas constant and, in this case, is R = 0.082 L.atm.K⁻¹mol⁻¹
T is in Kelvin
Converting Celsius in Kelvin:
T = 273 + 172
T = 445 K
Calculating moles
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{0.88(505)}{0.082(445)}[/tex]
n = 12.18 moles
According to the balanced equation, for 1 mol of dinitrogen gas consumed, 2 moles of ammonia is produced.
With 12.18 moles of dinitrogen, the reaction will result in
2(12.18) = 24.36 moles of ammonia
Molar mass of ammonia is M = 17.031 g/mol.
In 24.36 moles, there are
[tex]m=n.M[/tex]
m = 24.36.17.031
m = 414.87 grams
Since it's asking in kilograms: m = 0.41 kg.
In the beginning, it is said that dinitrogen gas is consumed at a rate of liters per second. So, the production rate of ammonia will be 0.41 kg/s.
The energy stored in an object is called potential energy
True or false
its true
Potential energy is the stored or latent energy in an object at rest. It’s fundamental to many physics-related concepts because its laws hold true on any level, from the planetary to the atomic level. The potential energy of an object is measurable.
A reaction between liquid reactants takes place at in a sealed, evacuated vessel with a measured volume of . Measurements show that the reaction produced of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to significant digits.
Answer:
0.41 atm
Explanation:
A reaction between liquid reactants takes place at 10.0 °C in a sealed, evacuated vessel with a measured volume of 5.0 L. Measurements show that the reaction produced 13. g of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.
Step 1: Given data
Temperature (T): 10.0 °CVolume of the vessel (V): 5.0 LMass of sulfur hexafluoride gas (m): 13. gStep 2: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 10.0 °C + 273.15 = 283.2 K
Step 3: Calculate the moles (n) of SF₆
The molar mass of SF₆ is 146.06 g/mol.
13. g × 1 mol/146.06 g = 0.089 mol
Step 4: Calculate the pressure (P) of SF₆
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T/V
P = 0.089 mol × (0.0821 atm.L/mol.K) × 283.2 K/5.0 L = 0.41 atm
Which of the following is the poorest conductor of electricity?
Calcium (Ca)
Silicon (Si)
Fluorine (F)
Sodium (Na)
Cobalt (Co)
Answer:
Fluorine (F)
Explanation:
The poorest conductor of electricity from the given choices is fluorine. This is because fluorine is a non - metal.
Like other non - metals, fluorine does not conduct electricity.
Only metals are known to conduct electricity and heat readily.
Semi - metals like silicon will conduct electricity under specific condition.
The free mobile electrons in metals makes it easy for them propagate electricity
Calculate the molarity (M) if 3.35g of H3PO4 is dissolved in water to give a total volume of 200mL
Answer:
0.171 M
Explanation:
Step 1: Given data
Mass of H₃PO₄ (solute): 3.35 gVolume of solution (V): 200 mLStep 2: Calculate the moles of solute
The molar mass of H₃PO₄ is 97.99 g/mol.
3.35 g × 1 mol/97.99 g = 0.0342 mol
Step 3: Convert "V" to liters
We will use the conversion factor 1 L = 1000 mL.
200 mL × 1 L/1000 mL = 0.200 L
Step 4: Calculate the molarity of the solution
We will use the definition of molarity.
M = moles of solute / liters of solution
M = 0.0342 mol/0.200 L = 0.171 M
An ionic compound has a generic formula of QR2.
Which elements could the Q and R represent?
Once you choose an answer, check it by plugging those elements into the QR2 formula to see if it looks right.
Q= Sodium R= Oxygen
Q= Magnesium R= Chlorine
Q= Oxygen R= Sodium
Q= Chlorine R= Magnesium
Answer:
Q= Magnesium R= Chlorine
Explanation:
The element Q should be magnesium and R is chlorine.
An ionic compound is a compound that is formed by the combination of a metal and non-metal. Such bonds forms when there is a transfer of electrons from the metals to the non-metals. This leaves a net positive charge on the metal and a negative charge on the non-metal.
The electrostatic attraction leads to the formation of the bond.
To solve this problem, the hypothetical compound is QR₂
Mg Cl
2 8 2 2 8 7
So, Mg transfers 2 electrons to two atoms of chlorine.
This leads to the formation of the compound MgCl₂
Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to approximately 20 wt.% V at room temperature. Determine the concentration in weight percent of V that must be added to iron to yield a unit cell edge length of 0.289 nm.
Answer:
Explanation:
To find the concentration; let's first compute the average density and the average atomic weight.
For the average density [tex]\rho_{avg}[/tex]; we have:
[tex]\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }[/tex]
The average atomic weight is:
[tex]A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }[/tex]
So; in terms of vanadium, the Concentration of iron is:
[tex]C_{Fe} = 100 - C_v[/tex]
From a unit cell volume [tex]V_c[/tex]
[tex]V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}[/tex]
where;
[tex]N_A[/tex] = number of Avogadro constant.
SO; replacing [tex]V_c[/tex] with [tex]a^3[/tex] ; [tex]\rho_{avg}[/tex] with [tex]\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }[/tex] ; [tex]A_{avg}[/tex] with [tex]\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }[/tex] and
[tex]C_{Fe}[/tex] with [tex]100-C_v[/tex]
Then:
[tex]a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }[/tex]
[tex]a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }[/tex]
[tex]a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }[/tex]
Replacing the values; we have:
[tex](0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }[/tex]
[tex]2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }[/tex]
[tex]2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})[/tex]
[tex]\mathbf{C_v = 9.1 \ wt\%}[/tex]