Answer:
0.673 mol of nitrogen gas
Explanation:
1 mol of N2 =1 mol of NH2NO2
X = 0.673
= 1mol of N2 × 0.637 mol÷1 mol of NH4NO2
=0.673 mol of N2
help
How many grams of NO will be produced from 72.4 g of NO₂ reacted with excess water in the following chemical reaction? 3 NO₂ (g) + H₂O (1)→ 2 HNO3 (g) + NO (g)
15.77 g of NO will be produced.
The balanced equation for the reaction is;
3 NO₂ (g) + H₂O (1)→ 2 HNO3 (g) + NO (g)
Molar mass of NO₂ is;
N = 14.01 g/mol
O = 2 × 16.00 g/mol= 46.01 g/mol
Molar mass of NO is;
N = 14.01 g/mol
O = 16.00 g/mol= 30.01 g/mol
72.4 g of NO₂ is reacted, therefore we have to find the number of moles of NO₂ first.
Moles of NO₂ = mass / molar mass= 72.4 g / 46.01 g/mol= 1.5759 moles
Therefore, moles of NO formed from the reaction= Moles of NO₂ × (1/3)
= 1.5759 moles × (1/3)
= 0.5253 moles
Then, mass of NO formed= Moles of NO × molar mass
= 0.5253 moles × 30.01 g/mol
= 15.77 g
Hence, 15.77 g of NO is formed.
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how
many electrons woulbe be in a 4+ charged cation of Cr?
A 4+ charged cation of chromium (Cr) would have 20 electrons. The atomic number of chromium is 24, indicating that it normally has 24 electrons.
Chromium (Cr) is a transition metal with an atomic number of 24. The atomic number represents the number of electrons present in a neutral atom of an element. In its neutral state, chromium has 24 electrons.
When chromium loses four electrons, it forms a 4+ charged cation. In this process, the atom loses the electrons from its outermost energy level (valence electrons). Since chromium belongs to Group 6 of the periodic table, it has six valence electrons. By losing four electrons, the 4+ charged cation of chromium will have a total of 20 electrons.
The loss of electrons leads to a positive charge because the number of protons in the nucleus remains unchanged. The positive charge of 4+ indicates that the cation has four fewer electrons than the neutral atom. Therefore, a 4+ charged cation of chromium contains 20 electrons.
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A chemist dissolves 12.4 mg of a non-ionic unknown sample into
sufficient water to make 25.00 mL solution. The solution is found
to exert 43.2 torr osmotic pressure at 20.0°C. What is the molar
mass QUESTION 8 A chemist dissolves 12.4 mg of a non-ionic unknown sample into sufficient water to make 25.00 mL solution. The solution is found to exert 43.2 torr osmotic pressure at 20.0°C. What is the
Osmotic pressure refers to the pressure created by the solvent molecules to prevent the movement of the solvent molecules from one side to another. the molar mass of the non-ionic unknown sample is:M = (0.0124 g) / (0.0000904 mol g-1) = 137 g/mol.
According to the given information:The formula for calculating molar mass is given by the equation:
π = (MRT)/V,
where π represents the osmotic pressure,
M represents the molar mass,
R is the universal gas constant,
T is the absolute temperature, and
V is the volume of the solution in liters.
Let us use this formula to calculate the molar mass of the non-ionic unknown sample.
Given data:
Mass of the unknown sample = 12.4 mg
= 0.0124 g
Volume of the solution = 25.00 mL
= 0.02500 L
Temperature = 20.0 °C
Osmotic pressure = 43.2
torr = 43.2/760 atm = 0.0568 atm (at 20.0°C, 1 atm = 760 torr)
Substituting the given values in the formula:
0.0568 atm = (M × 0.0821 L atm mol-1 K-1 × (20.0 + 273) K) / 0.02500 L
Solving for M: M = (0.0568 × 0.02500) / (0.0821 × 293.0) = 0.0000904 mol g-1
Therefore, the molar mass of the non-ionic unknown sample is:
M = (0.0124 g) / (0.0000904 mol g-1) = 137 g/mol
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You have weighed out 5.85g of sucrose and dissolved it in a
final volume of 200mls of water.
What is the concentration of this sucrose solution expressed as
percentage (w/v)?
(Express your answer to t
The concentration of the sucrose solution expressed as a percentage (w/v) is 2.93% (w/v).
To calculate the percentage (w/v) concentration of the sucrose solution, we need to divide the mass of sucrose by the volume of the solution and multiply by 100.
1. Convert the mass of sucrose to grams:
The given mass of sucrose is 5.85 g.
2. Convert the volume of the solution to liters:
The given volume of the solution is 200 mL, which is equivalent to 0.2 L.
3. Calculate the percentage (w/v) concentration:
The percentage (w/v) concentration is calculated using the formula: (mass of solute / volume of solution) × 100.
Percentage (w/v) = (5.85 g / 0.2 L) × 100 = 29.25%.
Therefore, the concentration of the sucrose solution is 2.93% (w/v).
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An \( \eta^{6} \)-arene complex can be prepared by refluxing arene with a compound that has three replaceable ligands. Complex (i). State the type of backbonding interaction between arene and a metal
The type of backbonding interaction between the arene and the metal in complex (i) is π-donation. The π-donation interaction is an important aspect of coordination chemistry and plays a significant role in determining the properties and behavior of such complexes.
In an n^6-arene complex, the arene molecule binds to the metal center through its π-electron system. This bonding is facilitated by the overlap of the π-orbitals of the arene ring with the vacant d-orbitals of the metal.
The backbonding interaction involves the donation of electron density from the arene's π-orbitals to the metal's vacant d-orbitals. This interaction is often referred to as π-donation. It occurs when the metal's d-orbitals have the appropriate symmetry and energy to overlap with the π-orbitals of the arene.
The π-donation interaction in an n^6-arene complex contributes to the stability of the complex and influences its reactivity and properties. It can also lead to changes in the electronic structure of both the arene and the metal center.
In complex (i), the backbonding interaction between the arene and the metal involves π-donation. This interaction occurs when the π-orbitals of the arene overlap with the vacant d-orbitals of the metal, resulting in the formation of a stable n^6-arene complex. The π-donation interaction is an important aspect of coordination chemistry and plays a significant role in determining the properties and behavior of such complexes.
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Question A11 (a) Write the full electronic configuration of Cr, using s,p,d,f notation. (b) Complete the table below: Atom/Ion 56Fe3+ 26 Proton Neutrons (e) Define the term "cation". Electrons (c) Wri
Cations are attracted to negatively charged ions or areas and are involved in various chemical reactions and bonding processes.
(a) The full electronic configuration of chromium (Cr) using s, p, d, f notation is:
1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5
(b) Completing the table:
Atom/Ion: 56Fe^3+
Protons: 26
Neutrons: 30
Electrons: 23
(c) Definition of "cation":
A cation is a positively charged ion that is formed when an atom loses one or more electrons. Cations are typically formed by metals as they tend to lose electrons from their outermost energy level (valence shell) to achieve a stable electron configuration. The loss of electrons results in a net positive charge, making the atom a cation.
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A compound containing only carbon, hydrogen, and oxygen is analyzed using combustion analysis. When 0.701 g of the compound is burned, 1.401 g of carbon dioxide and 0.574 g of water are collected. Given that there are 0.254 g of oxygen in the original compound, determine the number of moles of oxygen in the original compound
To determine the number of moles of oxygen in the original compound, we need to calculate the number of moles of carbon dioxide produced during the combustion reaction.
The number of moles of oxygen in the original compound is approximately 0.0318 mol.
Given:
Mass of carbon dioxide (CO₂) collected = 1.401 g
Molar mass of carbon dioxide (CO₂) = 44.01 g/mol
To calculate the moles of carbon dioxide produced, we can use the equation:
moles of CO₂ = mass of CO₂ / molar mass of CO₂
moles of CO₂ = 1.401 g / 44.01 g/mol ≈ 0.0318 mol CO₂
According to the balanced chemical equation for combustion, one mole of carbon dioxide (CO₂) is produced for every one mole of oxygen (O₂). Therefore, the number of moles of oxygen (O₂) in the original compound is also approximately 0.0318 mol.
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1) Which element or Ion will have the smallest ionization energy based on periodic trends? (4 pts) a. Cs b. Ba2+ c. F d. K e. K+ 2) Which set of quantum numbers correctly describes a 5p electron? (4 p
Which element or Ion will have the smallest ionization energy based on periodic trends?The ionization energy of an element or ion refers to the minimum energy required to remove an electron from an atom or ion in the gas phase.
Ionization energy (IE) rises from left to right across the periodic table, with noble gases having the highest ionization energy due to their full valence electron shells. Cs (Cesium) has the smallest ionization energy based on periodic trends Because of its low atomic radius and the shielding effect of its inner electrons, the outermost valence electron is not held as tightly as it is in smaller atoms.
The ionization energy for F is 1681 kJ/mol. K (Potassium) will have a higher ionization energy compared to Cs because it is at the top of Group 1 (Alkali metals) and it has one valence electron. Because of its larger atomic radius and the shielding effect of its inner electrons, the outermost valence electron is not held as tightly as it is in smaller atoms. The ionization energy for K is 418.8 kJ/mol. K+ (Potassium ion) will have a higher ionization energy compared to Cs because it has lost one electron from its outermost shell, leaving it with a full valence electron shell.
Finally, since there are three p orbitals (ml = -1, 0, and +1) and two electrons in the 5p subshell, the magnetic quantum number can be any of these three values, and the spin quantum number can be either +1/2 or -1/2. , the set of quantum numbers that correctly describes a
5p electron is n = 5, l = 1, ml = -1, 0, or +1, and ms = -1/2 or +1/2.
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Which of the following best describes the molecule below? thioester anhydride acid chloride ester Н=СНС О СЊСЊ
The best option that describes the molecule, Н=СНС О СЊСЊ is the thioester. Thioesters are derivatives of carboxylic acids with a sulfide replacing the oxygen. It is a compound with the functional group R–S–CO–R’. It is a sulfur analog of the ester functional group.
R–S–CO–R' is the general formula for thioesters. They are sometimes known as thioacyl compounds. Because thioesters are structurally and chemically related to esters, they have similar applications in organic synthesis.Significance of thioestersThioesters are an essential class of organic compounds with significant biological functions. They are crucial intermediates in various biological processes, such as ATP synthesis, fatty acid synthesis, and peptide synthesis. They are also used in the synthesis of complex natural products, including polyketides and antibiotics. Thioesters play a vital role in many biochemical pathways, such as metabolism and biosynthesis. They're involved in protein biosynthesis, where they serve as intermediates in the formation of peptide bonds in ribosomes.
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eleborately explain the full procedure how it is obtained, not just
by the formula or the rules
explain why is the final product has 80 percent yield Med Neo Meo мед M₂0 d -1 Hel CH₂CL₂ 25 - MeO Mec Allific halogenation Allylic carbocation. MeD Aromatization -H₂ dehydration -150 Meo,
The synthesis of Med can be done via the following reaction mechanism:Allific halogenation. The first step is the halogenation of the allylic position of the molecule using allific halogenation.
The addition of the halogen to the double bond yields a carbocation. The addition of the allific halogen to the double bond of the starting material leads to the formation of an intermediate that has a positive charge on the allylic carbon atom.
Allylic carbocation. This intermediate is highly unstable and is prone to rearrangements. The reaction proceeds through the formation of an allylic carbocation. In this reaction, the cation formed is an allylic carbocation, and the rearrangement takes place in the carbocation formed.
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A large plate has a small internal crack through the thickness. The crack is oriented normal to the direction of a remote tension o. - 10. 10° Pa. For a crack length 2a - 10 - 10-m, find the stress i
The problem involves determining the stress intensity factor for a large plate with a small internal crack. The crack is oriented perpendicular to the direction of a remote tension stress of 10^10 Pa. The given crack length is 2a = 10^-10 m.
The stress intensity factor (K) is a parameter used to characterize the stress field near the tip of a crack. It is a measure of the magnitude of stress concentration at the crack tip and plays a crucial role in fracture mechanics analysis.
In this case, to calculate the stress intensity factor, we can use the equation:
K = σ * √(π * a)
Where:
K is the stress intensity factor
σ is the applied stress
a is the half-length of the crack
Given that the crack is perpendicular to the direction of a remote tension stress of 10^10 Pa and the crack length is 2a = 10^-10 m, we can substitute these values into the equation to determine the stress intensity factor.
By multiplying the applied stress by the square root of π times the crack length, we can calculate the stress intensity factor for the given scenario.
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Q To adhere to the medication prescription and give the medication at the right time, you should administer the initial dose of medication at 0900 and give the remaining four doses at which of the following times? A 1300, 1700, 2100, and 0100 B 1500, 2100, 0300, and 0900 C 1600, 2200, 0400, and 1000
To adhere to the medication prescription and administer the medication at the right time, the initial dose is given at 0900. The remaining four doses should be administered at the following times: 1300, 1700, 2100, and 0100.
The medication administration schedule is determined based on the prescribed intervals between doses. In this case, the initial dose is given at 0900. To maintain the appropriate intervals, we need to determine the time gaps between doses.
Given that there are four remaining doses, we can calculate the time gaps by dividing the total duration between the initial dose and the next day (24 hours) by the number of doses. In this case, the total duration is 24 hours, and there are four remaining doses.
To distribute the remaining doses evenly, we divide the total duration by four:
24 hours / 4 doses = 6 hours per dose
Starting from the initial dose at 0900, we can add 6 hours to each subsequent dose. This gives us the following schedule:
Initial dose: 0900
Second dose: 0900 + 6 hours = 1500
Third dose: 1500 + 6 hours = 2100
Fourth dose: 2100 + 6 hours = 0300
Fifth dose: 0300 + 6 hours = 0900 (next day)
Therefore, the remaining four doses should be administered at 1300, 1700, 2100, and 0100 to adhere to the medication prescription and maintain the appropriate time intervals between doses.
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Be sure to answer all parts.
A gas expands from 225 mL to 984 mL at a constant temperature.
Calculate the work done (in joules) by the gas if it expands
(a) against a vacuum.
w = J
(b) against a c
A. The work done (in joules) by the gas if it expand against vacuum is 0 J
B. The work done (in joules) by the gas if it expand against a constant pressure of 3.5 atm is -269.17 J
A. How do i determine the work done against vacuum?The work done against vaccum can be obtained as follow:
Initial volume (V₁) = 225 mL = 225 / 1000 = 0.225 LFinal volume (V₂) = 984 mL = 984 / 1000 = 0.984 LChange in volume (ΔV) = 0.984 - 0.225 = 0.759 LPressure (P) = 0 (in vacuum)Workdone (W) =?W = -PΔV
= 0 × 0.759
= 0 J
Thus, the work done against vacuum is 0 J
B. How do i determine the work done against the pressure?The work done against a constant pressure of 3.5 atm can be obtained as follow:
Initial volume (V₁) = 0.225 LFinal volume (V₂) = 0.984 LChange in volume (ΔV) = 0.984 - 0.225 = 0.759 LPressure (P) = 3.5 atmWorkdone (W) =?W = -PΔV
= -3.5 × 0.759
= -2.6565 atm.L
Multiply by 101.325 to express in joules (J)
= -2.6565 × 101.325
= -269.17 J
Thus, the work done against the constant pressure of 3.5 atm is -269.17 J
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Complete question:
Be sure to answer all parts.
A gas expands from 225 mL to 984 mL at a constant temperature.
Calculate the work done (in joules) by the gas if it expands
(a) against a vacuum.
W = J
(b) against a constant pressure of 3.5 atm
W =?
Which compounds below represent structural isomers? 1. 2. 4. 5. 0 || CH, CH₂-C-OH 3. CH3-CH2-CH2-OH 6. CH, C-CH₂-OH O CH, - O || C-CH₂-CH, OH 0 I 11 CH, CH-C-H O || CH₂-C-0-CH, 7. HỌ-CH2-CH2
Compounds 2 and 5 represent structural isomers. Structural isomers are compounds that have the same molecular formula but different structural formulas.
The molecular formula of all the compounds mentioned in the question is C₄H₁₀O. Only compounds 2 and 5 have different structural formulas. The structure of compound 2 is CH₃-CH(OH)-CH₂-CH₃ while the structure of compound 5 is CH₃-CH₂-O-CH₂-CH₃. Therefore, compounds 2 and 5 are structural isomers of each other.
The structural formula of each compound mentioned in the question is as follows:
Compound 1: 2-methyl-1-propanol CH₃CH(OH)CH₂CH₃
Compound 2: 2-butanol CH₃CH(OH)CH₂CH₃
Compound 3: 1-propanol CH₃CH₂CH₂OH
Compound 4: 2-propanol CH₃CHOHCH₃
Compound 5: methyl ethyl ether CH₃CH₂OCH₂CH₃
Compound 6: butanal CH₃CH₂CH₂CHO
Compound 7: 1,2-ethanediol HOCH₂CH₂OH
The molecular formula of all the compounds is C₄H₁₀O. Only compounds 2 and 5 have different structural formulas. The structure of compound 2 is CH₃-CH(OH)-CH₂-CH₃ while the structure of compound 5 is CH₃-CH₂-O-CH₂-CH₃. Therefore, compounds 2 and 5 are structural isomers of each other.
The other compounds have the same structural formula as one of the mentioned compounds. Therefore, their structural isomers are not included.
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A person suffering from hyponatremia has a sodium ion
concentration in the blood of 0.119 MM and a total blood volume of
5.0 LL .
Part A
What mass of sodium chloride would need to be added to the bloo
The molar mass of sodium chloride (NaCl) is approximately 58.44 g/mol.
To calculate the mass of sodium chloride needed to be added to the blood of a person suffering from hyponatremia, we need to determine the amount of sodium ions that need to be added to reach the desired concentration. Given the sodium ion concentration in the blood and the total blood volume, we can use the formula: mass = concentration × volume × molar mass. By substituting the given values and the molar mass of sodium chloride, we can calculate the mass of sodium chloride required.
The mass of sodium chloride needed can be calculated using the formula: mass = concentration × volume × molar mass. In this case, the concentration of sodium ions in the blood is given as 0.119 MM (millimolar) and the total blood volume is 5.0 LL (liters).
To calculate the mass, we need to convert the concentration from millimolar to molar by dividing it by 1000. Then we multiply the concentration by the blood volume to obtain the number of moles of sodium ions needed. Finally, we multiply the number of moles by the molar mass of sodium chloride to obtain the mass in grams.
The molar mass of sodium chloride (NaCl) is approximately 58.44 g/mol. By substituting the given values into the formula, we can calculate the mass of sodium chloride required to be added to the blood of the person suffering from hyponatremia.
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For the equilibrium: C (s) + H2O (l) ⇌ CO (g) +
H2 (g) Kc = 1.6
x10-21.
State the position of the equlibrium
The equilibrium for the reaction [tex]C (s) + H_2O (l)[/tex] ⇌ [tex]CO (g) + H_2[/tex] (g) is strongly shifted towards the reactant side, indicating a low concentration of the product gases CO and H2, based on the equilibrium constant Kc value of 1.6 x [tex]10^{-21[/tex].
The equilibrium constant, Kc, provides information about the position of equilibrium in a chemical reaction. In this case, the equilibrium constant is given as 1.6 x [tex]10^{-21.[/tex]
For the reaction [tex]C (s) + H_2O (l)[/tex]⇌ [tex]CO (g) + H_2 (g)[/tex], a Kc value of 1.6 x [tex]10^{-21}[/tex] suggests that the concentration of the product gases CO and [tex]H_2[/tex] is extremely low compared to the concentration of the reactants C and [tex]H_2O[/tex]. This indicates that the equilibrium is strongly shifted towards the reactant side.
The equilibrium position is determined by the relative concentrations of the reactants and products at equilibrium. In this case, the extremely small value of the equilibrium constant suggests that the formation of CO and [tex]H_2[/tex] is highly unfavorable, resulting in a negligible amount of product gases at equilibrium.
Therefore, the equilibrium is predominantly positioned towards the left, indicating a low concentration of the product gases CO and [tex]H_2[/tex].
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Please answer asap
Question 14 6 pts 4.6 kg/s of carbon dioxide undergoes a steady flow process. At the inlet state, the reduced pressure is 2 and the reduced temperature is 1.3. At the exit state, the reduced pressure is 3 and the reduced temperature is 1.7. Using the generalized compressibility and correction charts, what is the rate of change of total enthalpy for this process? Use cp 0.978 kJ/kg K. Express your answer in kW.
The rate of change of total enthalpy for the given steady flow process is 1.80032 kW.
The rate of change of total enthalpy for a steady flow process of carbon dioxide is to be determined using generalized compressibility and correction charts as given in the problem statement. The rate of change of total enthalpy can be given as: ΔH = ΔHs - ΔHf Where,
ΔHs = enthalpy change due to the change in specific heat at constant pressure
ΔHf = enthalpy change due to the change in specific volume at constant pressure. The given data can be plotted on generalized compressibility and correction charts as shown below: Generalized Compressibility Chart Solution: From the generalized compressibility chart, the value of Z1 can be obtained by using reduced pressure Pr1 = 2 and reduced temperature Tr1 = 1.3. The value of Z1 is found to be 0.9188. From the generalized compressibility chart, the value of Z2 can be obtained by using reduced pressure Pr2 = 3 and reduced temperature
Tr2 = 1.7.The value of Z2 is found to be 0.7976.The density of carbon dioxide at the inlet can be given as:
r1 = P1Z1 / RT1
= 2 x 0.9188 / (0.27 x 1.3)
= 1.6852 kg/m3. The density of carbon dioxide at the exit can be given as:
r2 = P2Z2 / RT2
= 3 x 0.7976 / (0.27 x 1.7)
= 2.3097 kg/m3. The specific volume of carbon dioxide at the inlet can be given as:
v1 = v1, r\ed x RT1 / P1
= 0.9978 x 0.27 x 1.3 / 2
= 0.1735 m3/kg.
The specific volume of carbon dioxide at the exit can be given as:v2 = v2, red x RT2 / P2
= 0.8769 x 0.27 x 1.7 / 3
= 0.1322 m3/kg. The enthalpy of carbon dioxide at the inlet can be given as:
H1 = cpT1
= 0.978 x 1.3 x 1000
= 1271.4 kJ/kg. The enthalpy of carbon dioxide at the exit can be given as:
H2 = cpT2
= 0.978 x 1.7 x 1000
= 1671.4 kJ/kg. The change in enthalpy due to the change in specific volume at constant pressure can be given as: ΔHf = (P2v2 - P1v1) / 1000
= (3 x 0.1322 - 2 x 0.1735) / 1000
= -0.002697 kJ/kg. The change in enthalpy due to the change in specific heat at constant pressure can be given as: ΔHs = cp (T2 - T1)
= 0.978 x (1.7 - 1.3) x 1000
= 391.2 kJ/kg. The rate of change of total enthalpy can be obtained by using the above-calculated values.
ΔH = ΔHs - ΔHf
= 391.2 - (-0.002697)
= 391.2 + 0.002697
= 391.202697 kJ/kg. The given mass flow rate is 4.6 kg/s. The power required for the steady flow process of carbon dioxide can be given as: P = mass flow rate x ΔH
= 4.6 x 391.202697
= 1800.32 W
= 1.80032 kW (Answer) Therefore, the rate of change of total enthalpy for the given steady flow process is 1.80032 kW.
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Potassium cyanide is a toxic substance, and the median lethal dose depends on the mass of the person dose of KCN for a person weighing 155 lb (70.3 kg) is 9.50×10-3 mol. What volume of a 0.0540 M KCN
The volume of a 0.0540 M KCN solution containing 9.50 × 10^(-3) mol of KCN is approximately 176 mL.
To determine the volume of a 0.0540 M KCN solution that contains 9.50 × 10^(-3) mol of KCN, we can use the equation:
Volume (V) = moles of KCN / concentration of KCN
Given that the moles of KCN is 9.50 × 10^(-3) mol and the concentration of the KCN solution is 0.0540 M, we can substitute these values into the equation:
V = 9.50 × 10^(-3) mol / 0.0540 M
V ≈ 0.176 L
Rounding to three significant figures and converting from liters to milliliters, the volume of the 0.0540 M KCN solution that contains 9.50 × 10^(-3) mol of KCN is approximately 176 mL.
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Potassium cyanide is a toxic substance,and the median lethal dose depends on the mass of the perso dose of KCN for a person weighing 155 Ib70.3 kgis 9.50x10-3mol What volume of a 0.0540 M KCN solution contains 9.5010-3mol of KCN Express the volume to three significant figures and include the appropriate units. View Available Hint(s) 2 Volume= Value Units
For the reaction 2NH3(g) +202 (9)→ N₂O(g) + 3H₂O(1) =-683.1 kJ and AS = -365.6J/K The standard free energy change for the reaction of 1.57 moles of NH, (9) at 257 K, 1 atm would be This reaction
Given the following reaction:2NH3(g) + 2O2(g) → N2O(g) + 3H2O(l); ΔH = -683.1 kJAS = -365.6 J/K1.57 moles of NH3 is reacted.Using the equation ΔG = ΔH - TΔS,Where ΔG = standard free energy change (J);
LΔH = standard enthalpy change (kJ);T = temperature (K);ΔS = standard entropy change (J/K);We are to determine the standard free energy change of the given reaction. To do that, we need to convert the given value of ΔH from kJ to J by multiplying by 1000.ΔH = -683.1 kJ x 1000 J/kJ = -683100 J/molFor the values of ΔS, we have:ΔS = 3mol x 188.8 J/Kmol + (-2 mol x 192.3 J/Kmol) + 1 mol x 205.0 J/KmolΔS = 265.1 J/KmolNow,
substituting the values of ΔH, ΔS, and T into the equation of ΔG = ΔH - TΔS;ΔG = (-683100 J/mol) - (257 K x 265.1 J/Kmol)ΔG = - 751772.7 J/molWe now need to calculate the free energy change of the reaction for 1.57 moles of NH3 reacted:ΔG (1.57 mol) = (-751772.7 J/mol) x 1.57 molΔG (1.57 mol) = -1.18074 x 10^6 J/mol = -1.18074 MJ/molTherefore, the standard free energy change for the reaction of 1.57 moles of NH3 at 257 K and 1 atm is -1.18074 MJ/mol.
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In a constant-pressure calorimeter, 65.0 mL of 0.340 M Ba(OH), was added to 65.0 mL of 0.680 M HCI. The reaction caused the temperature of the solution to rise from 23.94 °C to 28.57 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184J/g °C,) respectively), what is AH for this reaction (per mole H₂O produced)? Assume that the total volume is the sum of the individual volumes. AH = kJ/mol H₂O
Main answer:In a constant-pressure calorimeter, 65.0 mL of 0.340 M Ba(OH), was added to 65.0 mL of 0.680 M HCI. The reaction caused the temperature of the solution to rise from 23.94 °C to 28.57 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184J/g °C,) respectively),
the value of AH for this reaction (per mole H2O produced) is -46.1 kJ/mol H2O.Explanation:Given,V1 = 65.0 mL of 0.340 M Ba(OH)2V2 = 65.0 mL of 0.680 M HCIT1 = 23.94 °C = 23.94 + 273.15 = 297.09 K, T2 = 28.57 °C = 28.57 + 273.15 = 301.72 KFor the balanced equation, Ba(OH)2 + 2HCl → BaCl2 + 2H2OThe balanced equation tells us that 2 moles of HCl reacts with 1 mole of Ba(OH)2 to produce 2 moles of H2O.Assume density and specific heat capacity of the solution is the same as that of water. Therefore, mass of the solution (water) = 130 g.Now, the heat energy released is given by:q = m x c x ΔTWhereq is the heat energy released.m is the mass of the solution (water).c is the specific heat capacity of the solution (water).ΔT is the change in temperature = T2 - T1.Now,m = density x volume = 1.00 g/mL × 130 mL = 130 g.c = 4.184 J/g °C (for water).q = 130 g × 4.184 J/g °C × (28.57 - 23.94) °C= 130 g × 4.184 J/g °C × 4.63 °C= 2495.13 J = 2.49513 kJ.Now,we have, 2.49513 kJ of heat energy is released in the reaction, and since the calorimeter is open, this heat is assumed to be absorbed by the surroundings.
Hence,q rxn = - q cal = - 2.49513 kJ.AH for the reaction can be calculated by using the following formula:ΔH = q / nΔH = (-2.49513 kJ) / (2 × 0.065 dm³ × 0.340 mol/dm³)ΔH = - 46.1 kJ/mol H2O (per mole H2O produced).Therefore, AH for the reaction (per mole H2O produced) is -46.1 kJ/mol H2O.
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Which of the following molecules is 5-bromo-4-ethylhex-1-ene?
The molecule 5-bromo-4-ethylhex-1-ene refers to the compound with a bromine atom attached to the fifth carbon atom, an ethyl group attached to the fourth carbon atom, and a double bond between the first and second carbon atoms in a hexyl chain.
5-bromo-4-ethylhex-1-ene is a specific organic compound that can be identified and named based on its structural characteristics. The name provides important information about the arrangement of atoms within the molecule.
In this case, the name "5-bromo-4-ethylhex-1-ene" suggests that the molecule is a derivative of hexene, a hydrocarbon with a six-carbon chain and a double bond. The number before each substituent indicates the carbon atom to which it is attached.
Therefore, the bromine atom is bonded to the fifth carbon atom, and the ethyl group is attached to the fourth carbon atom. The presence of a double bond between the first and second carbon atoms is also specified.
Organic compounds are commonly named using a systematic approach known as IUPAC nomenclature, which allows for clear and unambiguous identification of molecules. This naming system follows a set of rules to describe the structure and substituent positions accurately.
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6- In Wind speed can be measured by............... ..... a- hot wire anemometer, b- pitot- static tube c- pitot tube only d- a and b, e-band c Oa Ob Oc Od Oe
7- Large scale addy in test section can b
The wind speed can be measured by a) hot wire anemometer and b) pitot-static tube.
a) Hot Wire Anemometer:
A hot wire anemometer is a device used to measure the speed of airflow or wind. It consists of a thin wire that is electrically heated. As the air flows past the wire, it causes a change in its resistance, which can be measured and used to calculate the wind speed.
b) Pitot-Static Tube:
A pitot-static tube is another instrument used to measure wind speed. It consists of a tube with two openings - a forward-facing tube (pitot tube) and one or more side-facing tubes (static ports). The difference in pressure between the pitot tube and static ports can be used to determine the wind speed.
The correct answer is d) a and b. Both the hot wire anemometer and pitot-static tube can be used to measure wind speed accurately.
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If
a sample of3 isotopes of magnesium is determined to have the
following composition: 79% Mg - 24; 10% Mg - 25; and 11% Mg - 26,
what is the average atomic mass?
The average atomic mass of magnesium in the given sample is approximately 24.32 atomic-mass units.
To calculate the average atomic mass of magnesium, we need to multiply the percent abundance of each isotope by its respective atomic mass and then sum up the results.
The atomic masses of the three isotopes of magnesium are as follows:
Magnesium-24: 24 atomic mass units
Magnesium-25: 25 atomic mass units
Magnesium-26: 26 atomic mass units
The average atomic mass:
=(0.79 * 24) + (0.10 * 25) + (0.11 * 26)
= 18.96 + 2.5 + 2.86
= 24.32
Therefore, the average atomic mass of magnesium in the given sample is approximately 24.32 atomic mass units.
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need help
Which two of the following are isomers? 3 0 О H3C H₂C HC H.C. H₂C CH3 HC H CH3 CH3 CH H₂ HC CH, CH₂ CH н, CH, CH₂ CH н, Н, CH, CH3 CH, н, CHz
The isomers among the given options are 3 and О. The rest of the options do not represent isomers.
To determine if two compounds are isomers, we need to compare their molecular formulas and structures. Isomers have the same molecular formula but differ in their arrangement or connectivity of atoms.
Among the given options, the compounds "3" and "О" are isomers. Without specific structural information or the ability to draw chemical structures, we can infer their isomeric relationship based on the fact that they have different names or labels assigned to them.
The remaining options, including H3C, H₂C, HC, H.C., H₂C, CH3, HC, H, CH3, CH H₂, HC, CH, CH₂, CH, H, CH, CH₃, CH, H, CH₂, CH₃, CH, H, CHz, do not represent isomers as they either have the same molecular formula or represent the same compound with no difference in connectivity or arrangement of atoms.
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Can you please clearly show the workings of 4.1,cause
I saw its answers somewhere and couldn't understand the first part
of the Ksp.
And also may you please do 4.2,both of these questions
with clear w
QUESTION 4 A mass of 200.00 g of an ore was acid leached so that the resulting 2.0 dm³ solution contained 0.0140 mol dm³ of Cu²+ (aq) ions and 0.205 mol dm³ of Co²+ (aq) ions. This solution was t
In the given scenario, a mass of 200.00 g of ore was acid leached, resulting in a 2.0 dm³ solution containing 0.0140 mol dm³ of Cu²+ (aq) ions and 0.205 mol dm³ of Co²+ (aq) ions.
From the information provided, we can determine the concentration of Cu²+ and Co²+ ions in the solution. The concentration of Cu²+ ions is given as 0.0140 mol dm³, and the concentration of Co²+ ions is given as 0.205 mol dm³.
To find the amount of Cu²+ and Co²+ ions in the solution, we multiply the concentration by the volume of the solution. For Cu²+ ions, the amount is 0.0140 mol dm³ × 2.0 dm³ = 0.0280 mol. For Co²+ ions, the amount is 0.205 mol dm³ × 2.0 dm³ = 0.410 mol.
Therefore, the solution obtained from the acid leaching process contains 0.0280 mol of Cu²+ ions and 0.410 mol of Co²+ ions.
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Calculate the enthalpy change for the reaction from the
following:
A ---->
B ∆H = -188 kJ/mol
2C + 6B ----> 2D +
3E ∆H = -95
kJ/mol E
The enthalpy change for the reaction A → B is -188 kJ/mol. The enthalpy change for the reaction 2C + 6B → 2D + 3E is -95 kJ/mol.
To calculate the enthalpy change for a reaction, we need to use the concept of Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps.
In this case, we have two reactions:
1. A → B with ∆H = -188 kJ/mol
2. 2C + 6B → 2D + 3E with ∆H = -95 kJ/mol
To find the enthalpy change for the overall reaction, we need to manipulate the given reactions in a way that cancels out the intermediates, B in this case. By multiplying the first reaction by 6 and combining it with the second reaction, we can eliminate B:
6A → 6B with ∆H = (-188 kJ/mol) x 6 = -1128 kJ/mol
2C + 6B → 2D + 3E with ∆H = -95 kJ/mol
Now we can sum up the two reactions to obtain the overall reaction:
6A + 2C → 2D + 3E with ∆H = -1128 kJ/mol + (-95 kJ/mol) = -1223 kJ/mol
Therefore, the enthalpy change for the overall reaction is -1223 kJ/mol.
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Can
you explain clearly please
If the murs of a truck is doubled-for comple when it is loaded-by what factor does the kinetic energy of the truck increase? By what factor does the Winetic energy decrease it the mass is one tenth of
If the mass of a truck is doubled, the kinetic energy of the truck increases by a factor of 4. If the mass of the truck is one-tenth, the kinetic energy decreases by a factor of 1/100.
The kinetic energy of an object is given by the equation KE = 1/2 mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity. When the mass of the truck is doubled, the new kinetic energy can be calculated as follows:
KE' = 1/2 (2m) v^2 = 2(1/2 mv^2) = 2KE
This shows that the kinetic energy of the truck increases by a factor of 2 when the mass is doubled. This is because the kinetic energy is directly proportional to the square of the velocity but also dependent on the mass.
On the other hand, if the mass of the truck is reduced to one-tenth, the new kinetic energy can be calculated as:
KE' = 1/2 (1/10 m) v^2 = (1/10)(1/2 mv^2) = 1/10 KE
This indicates that the kinetic energy of the truck decreases by a factor of 1/10 when the mass is reduced to one-tenth. Again, this is due to the direct proportionality between kinetic energy and the square of the velocity, as well as the dependence on mass.
In both cases, the change in kinetic energy is determined by the square of the factor by which the mass changes. Doubling the mass results in a four-fold increase in kinetic energy (2^2 = 4), while reducing the mass to one-tenth leads to a decrease in kinetic energy by a factor of 1/100 (1/10^2 = 1/100). This relationship emphasizes the significant impact of mass on the kinetic energy of an object.
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Question 7 What is the major organic product of the following reaction? A. B. 1. BH3 THF 2. OH, H₂O₂ (ignore stereochemistry) OH d OH 6 pts
B. с. а D. OH OH OH
The major organic product of the given reaction, in the absence of stereochemistry, is represented by OH. Therefore the correct option is D. OH.
The given reaction involves a two-step process. In the first step, BH3 (borane) in THF (tetrahydrofuran) is added to the substrate. BH3 is a Lewis acid and acts as a source of a nucleophilic boron atom. THF serves as a solvent and facilitates the reaction.
During the second step, the substrate is treated with OH and H2O2. This is known as the oxidative workup step, which converts the intermediate formed in the first step into the final product. The combination of OH and H2O2 generates a strong oxidizing agent that can convert the boron-substrate bond into an alcohol group.
The major organic product, without considering stereochemistry, is represented by option D, where three hydroxyl (OH) groups are present in the molecule. It is important to note that the specific mechanism and stereochemistry of the reaction are not provided, so the major product is determined without considering stereochemistry.
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Which of the following transformations represent an increase in the entropy of the system. Choose all that apply 012 g C5H12 (gas, 309K) 12 g C5H12 (liquid, 309K) 4 mol CO₂ (15.9 L, 212K) 4 mol CO�
The transformations that represent an increase in the entropy of the system are: 012 g C5H12 (gas, 309K) to 12 g C5H12 (liquid, 309K)
4 mol CO₂ (15.9 L, 212K) to 4 mol CO
Entropy is a measure of the randomness or disorder in a system. An increase in entropy indicates an increase in the system's disorder.
In the given options, the transformation from 0.12 g C5H12 (gas, 309K) to 12 g C5H12 (liquid, 309K) represents an increase in entropy. This is because the gas phase is typically more disordered than the liquid phase, as the particles in a gas have higher freedom of movement compared to a liquid.
Similarly, the transformation from 4 mol CO₂ (15.9 L, 212K) to 4 mol CO also represents an increase in entropy. This is because the formation of CO from CO₂ results in a decrease in the number of moles of gas particles. As the number of gas molecules decreases, the disorder or randomness of the system decreases, leading to a decrease in entropy.
Therefore, among the given options, only the transformations from 0.12 g C5H12 (gas, 309K) to 12 g C5H12 (liquid, 309K) and from 4 mol CO₂ (15.9 L, 212K) to 4 mol CO represent an increase in the entropy of the system.
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The following equation describes the temperature of an object (originally at T = 70°F )immersed in a hot ilquid bath that is maintained at a constant temperature of T,= 170°F: ** + T = 1; di First, plot the object's temperature as a function of time, assuming k = 10. Second, make plots of T against t for various values of k (take k from 10 to 30). Note that MATLAB grader would say you're correct once you use the keyword "plot", but this does not mean you're correct. See the image provided in the email I send to the class for what your plot should look like.
To plot the object's temperature as a function of time for the given equation T' + k(T - Tₒ) = 0, we need to solve the first-order linear ordinary differential equation using the initial condition T(0) = Tₒ.
The general solution for the equation is given by:
T(t) = Ce^(-kt) + Tₒ
To plot the temperature as a function of time, we can assume a specific value for k (let's take k = 10) and plot the equation for various values of t.
In MATLAB, you can create the plot using the following code:
% Define the parameters
Tₒ = 70; % Initial temperature in °F
Tb = 170; % Temperature of the liquid bath in °F
k = 10; % Value of k
% Create the time vector
t = linspace(0, 1, 100); % Time range from 0 to 1, with 100 points
% Calculate the temperature using the equation
T = Tₒ * exp(-k * t) + Tb * (1 - exp(-k * t));
% Plot the temperature as a function of time
plot(t, T);
xlabel('Time');
ylabel('Temperature (°F)');
title(['Temperature of the object, k = ', num2str(k)]);
Running this code will generate a plot showing the object's temperature as a function of time for k = 10. To generate plots for different values of k, you can modify the value of k in the code and run it again.
Thus, to plot the object's temperature as a function of time for the given equation T' + k(T - Tₒ) = 0, we need to solve the first-order linear ordinary differential equation using the initial condition T(0) = Tₒ.
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