249. 43 KJ much energy is required to heat 100.0 g H₂O from a liquid at 76°C to a gas at 132°C.
What is energy?The ability to work is defined as energy. The SI symbol for energy is Joule (J). The different types of energy include thermal energy, chemical energy, electrical energy, light energy, potential energy, kinetic energy, and heat energy. Chemical energy is defined as the power that is held inside chemical compound bonds (molecules and atoms). An exothermic reaction, in which it is released during a chemical reaction, primarily results in the production of heat as a byproduct.
Given that,
mass = 100 gm
c = 4.1843 J/C g
as we know,
q₁ = m × c × ΔT
here, q₁ = heat required to warm water
thus,
or, q₁ = 100 × 4.1843 × (132 - 76)
or, q₁ = 23430 J
Next, q₂ = m × ΔH (vap)
here, q₂ = heat required to vaporize the water
q₂ = 100 × 2260 [As ΔH (vap) = 2260 J/g]
q₂ = 22600 J
Now, energy required (Q) = q₁ + q₂
or, Q = (23430 + 22600) J
or, Q = 249430 J
or, Q = 249.43 KJ
249. 43 KJ much energy is required to heat 100.0 g H₂O from a liquid at 76°C to a gas at 132°C.
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while a substance is melting (such as water at 0 ^oc), which of the following statements is true? (select all that apply.)
a Energy is released by the system during the phase change b Energy has to be extracted from the system to induce the phase change c Temperature is increasing d Temperature is decreasing e Temperature remains constant f Energy is absorbed by the system during the phase change g Energy has to be given to the system to induce the phase change
Option d is the correct solution to this question here. The temperature is decreasing.
Standard sets of circumstances for experimental measurements are established at standard pressures and temperatures to enable comparisons between various sets of data.
Ice melting is a physical transformation rather than a chemical reaction. The addition of heat energy causes the phase of ice to transition from solid to liquid during melting. However, there is no chemical reaction taking place while it is melting, so the molecular composition remains unchanged.
NIST employs a 20 °C (293.15 K, 68 °F) temperature and a 1 atm absolute pressure (14.696 psi, 101.325 kPa). This standard is also known as normal pressure and temperature (abbreviated as NTP).
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select the answer that best completes the following statement: chemical reactions in living systems occur in an environment, within a narrow range of temperatures.
Chemical reactions in living systems occur in an environment, within a narrow range of temperatures causing chemical changes.
What are chemical changes?
Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.
There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.
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Determine whether the reactions with the following ΔH and ΔS values are spontaneous or non-spontaneous. State whether the reactions are exothermic or endothermic.(a) ΔH=−110kJ, ΔS=+40JK −1 at 400K(b) ΔH=+40kJ, ΔS=−120JK−1 at 250K
The equation ΔH=−110kJ, ΔS=+40JK −1 at 400K is spontaneous and exothermic.
And the equation ΔH=+40kJ, ΔS=−120JK−1 at 250K is non spontaneous and endothermic.
The spontaneity of a reaction is determined by the sign of the Gibbs free energy change (ΔG), which is related to enthalpy change (ΔH) and entropy change (ΔS) by the equation:
ΔG = ΔH - TΔS
where T is the temperature in Kelvin.
(a) ΔH=−110kJ, ΔS=+40JK−1 at 400K At 400K, ΔG = ΔH - TΔS = -110 kJ - (400 K)(+40 J/K) = -110 kJ - 16 kJ = -126 kJ. Since ΔG is negative, the reaction is spontaneous. The negative ΔH value indicates that the reaction is exothermic, meaning it releases heat.
(b) ΔH=+40kJ, ΔS=−120JK−1 at 250K At 250K, ΔG = ΔH - TΔS = +40 kJ - (250 K)(-120 J/K) = +40 kJ + 30 kJ = +70 kJ. Since ΔG is positive, the reaction is non-spontaneous. The positive ΔH value indicates that the reaction is endothermic, meaning it absorbs heat.
Therefore, the reaction in (a) is spontaneous and exothermic, while the reaction in (b) is non-spontaneous and endothermic
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Which of the following pure liquids is the best solvent for carbon disulfide ? HBr(l) CH_3OH(l) C_6H_6(l) H_20(l) NH_3(l)
The pure liquid that is the best solvent for carbon disulfide is C₆H₆, also known as benzene. Option C is correct.
The best solvent for carbon disulfide (CS₂) would be a liquid that has a similar polarity and intermolecular forces to CS₂. Carbon disulfide is a nonpolar molecule, which means that it is most likely to dissolve in a nonpolar solvent.
Benzene is a nonpolar molecule and has a similar structure and intermolecular forces to CS₂, which makes it a good solvent for nonpolar solutes such as carbon disulfide.
In contrast, the other options (HBr, CH₃OH, H₂O, and NH₃) are polar molecules, which means that they are unlikely to dissolve nonpolar solutes such as CS₂ as effectively as benzene.
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--The given question is incomplete, the complete question is
"Which of the following pure liquids is the best solvent for carbon disulfide ? A) HBr B) CH₃OH C) C₆H₆ D) H₂0 E) NH₃"--
Electrophilic nitration of benzoic acid gives almost exclusively 1,3-nitrobenzoic acid. By drawing the appropriate resonance forms of the intermediate cations resulting from attack of [NO2]+, explain this result.
The electrophilic nitration of benzoic acid involves the attack of the nitronium ion ([NO2]+) on the benzene ring of the benzoic acid. The intermediate formed is a positively charged arenium ion, which can resonate between different resonance forms.
One of the resonance forms of the intermediate cation is shown below:
O O
// //
/C+ <-----> /C
\ \
\ O-
In this resonance form, the positive charge is delocalized over the carbon atom and the adjacent oxygen atom. The resulting resonance hybrid is stabilized by the delocalization of the positive charge over the ring, which lowers the overall energy of the system.
The nitronium ion can attack at the meta position, which leads to the formation of 1,3-nitrobenzoic acid. The attack at the ortho or para positions would lead to the formation of other isomeric products.
The reason for the selective formation of 1,3-nitrobenzoic acid is due to the resonance stabilization of the intermediate cation. The meta position is less hindered than the ortho and para positions, and the resonance form shown above places the positive charge at the meta position. Therefore, the attack of the nitronium ion at the meta position is favored over the other positions. Additionally, the resonance form shown above places the negative charge of the carboxylate group at the para position, which makes it less favorable for the nitronium ion to attack at this position.
Overall, the resonance stabilization of the intermediate cation favors the selective formation of 1,3-nitrobenzoic acid in the electrophilic nitration of benzoic acid.
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A small amount of sodium chloride (NaCI) is dissolved in a large amount of water. Imagine separating this process into the four stages sketched below (These sketches show only a portion of the substances, so you can see the density and distribution of atoms and molecules in them.) NaCl H,O solution Use these sketches to answer the questions in the table below The enthalpy of solution AH NaCl dissolves in water. Use this information to list the stages in order of increasing enthalpy is positive when soln absorbed released neither absorbed nor released Would heat be absorbed or released if the system moved from Stage C to D? none ionic bonding force covalent bonding force metal bonding force hydrogen-bonding force dipole force ion-dipole force dispersion force What force would oppose or favor the system moving from Stage C to D?
Bond energy is described as a measure of the bond strength of a chemical bond, and it is the amount of energy required for breaking the atoms involved in a molecular bond into free atoms. The answers for multiple questions are given below.
i)
When bond is formed energy is released and to break a bond energy must be applied.
∆Hsoln is positive , so A < D
Comparing D to B and C , molecules are dissociated in B and molecules are more dissociated in C , so ∆H is order is D < B < C
Therefore, increasing order of enthalpy is A < D < B < C
ii)
Comparing to stage C to D molecules are associated , so heat is released.
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question at position 1 predict the most likely oxidation state for a titanium cation and which electrons will be lost to form that ion
The most likely oxidation state for a titanium cation is +4 and Titanium has four valence electrons in its outermost shell, and to form a cation it must lose those four electrons.
Since each electron has a negative charge, the loss of four electrons gives the titanium cation a net positive charge of 4+.
In its ground state, the electron configuration of neutral titanium is [Ar]3d2 4s2, with two electrons in the 4s orbital and two electrons in the 3d orbital.
When it loses its valence electrons, it becomes a Ti4+ cation with the electron configuration [Ar]3d0 4s0. Thus, the two electrons in the 4s orbital and two electrons in the 3d orbital are lost to form the Ti4+ cation.
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1st attempt d See Periodic Table The first eight ionization energies of a third row element are 1012, 1907,2914,4964,6274, 21267, 25431, and 29872 kJ/mol. Identify the element. Choose one: A. Na OB, P ○C.Cl ○ D. Mg O E. A OF.S OG. S O H. Ar
The element is aluminum (Al), which is in the third period of the periodic table and has the electron configuration [Ne]3s²3p¹.
The element can be identified by analyzing the ionization energies. The third row of the periodic table contains elements with the electron configuration [Ar] 3d¹⁰4s²4p¹. Based on the given ionization energies, we can identify the element by finding the period that contains the first eight ionization energies listed.
Looking at the given ionization energies, we can see that the first ionization energy is relatively low, indicating that the element is a metal. The next seven ionization energies increase gradually, indicating that they correspond to removing electrons from successive energy levels.
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Determine the empirical formula of a compound that contains 69.5% oxygen and 30.5% nitrogen, and then determine the molecular formula. The molar mass of the molecular formula is 138.06g/mol.
Answer:
NO2
Explanation:
the molar mass of oxygen = 138.06 x 0.695 = 95.9517
=> 95.9517/16 = 6
the molar mass of nitrogen = 138.06 x 0.305 = 42.1083
=> 42.1083 / 14 = 3
So there are 3 N and 6 O => N3O6 which is simplified empirically as NO2
After being given a solution containing a set of unknown ions, you perform the following sequence of tests to determine the identity of the cations in solution: 1. A flame test produces no color 2. You take a small portion of the solution and add NaOH. A vapor is produced that tums red litmus blue. 3. You add HCl to solution, which results in the formation of a white precipitate. The supernatant is separated and used for the remainder of the tests. 4. The addition of buffered NH,/NHg to a portion of the supernatant from step 3 does not produce a precipitato. 5. The addition of ammonium oxalate to a portion of the superntant from stop 3 does not produce a precipitato 6. The addition of a strong base to a portion of the supernatant from step 3 generates a precipitato, 7. The supernatant from step 6 is separated from the precipitate. The addition of HCI and KaFo(CN), to the solution produces no reaction. Based on these tests, Identity which of the following ions must be present, can be present, and cannot be present. Na+, NH4+, Ag+, Fo%+, Al3+, C.Cat, Mg2+, Ni2+, Zn2+
The cations can be identified by using various test, that are explained in the below section, Cations such as Na+, Ca2+, Cr3+, Fe3+, Al3+ and many other cation's test are explained below.
Na+ produces yellow flames in flame test. So it confirms the presence of Na+ ion.
If we add NaOH in sample and it produces vapor and turns red litmus, then NH4+ is present. But Here there is no change in litmus. So NH4+ ion CANNOT be present.
The addition of HCl does NOT produces white precipitate indicates that Ag+ CANNOT be present there. When we add HCl in Ag+ , a white precipitate of AgCl is formed.
The addition of buffered NH/NH3 to a portion of the solution produces a precipitate. Cr3+ , Fe3+ , Al3+ produces precipitate in addition of NH/NH3 buffer. So Cr3+ , Fe3+ , Al3+ CAN be present there but further test can confirm it.
Ammonium oxalate produces white precipitate in presence of calcium ion. So Ca2+ must be present there.
The addition of a strong base will precipitate Mg2+ and Ni2+ ion if they are present and Zn2+ forms soluble compound . But here they do not formed precipitate hence Mg2+ and Ni2+ CANNOT be Present there but Zn2+ CAN be present there , we need further tests to confirm Zn2+ ion.
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calculate the ph, poh, and percentage protonation of solute in each of the following aqueous solutions: (a) (b) (c) (d) 0.0073 m codeine, given that the of its conjugate acid is 8.21
A) pH = 9.44, pOH = 4.56, Percentage protonation = [codeine-H+] / [codeine-H+] + [codeine] = 0.72%
B)pH = pKa + log([codeine]/[codeine-H+]) = 10.09, pOH = 3.91, Percentage protonation = [codeine-H+] / [codeine-H+] + [codeine] = 0.17%
C)pH = 8.21, pOH = 5.79, Percentage protonation = [codeine-H+] / [codeine-H+] + [codeine] = 0.01%
The pKa value given is for the conjugate acid of codeine which is codeine-H+. We can use the pKa value to determine the Ka value for codeine-H+ and then use the Ka value to determine the concentrations of codeine-H+ and codeine in solution.
(a) For 0.0073 M codeine in water, we can assume that the concentration of codeine-H+ is very small and can be neglected. The Ka value can be determined from the pKa as follows:
pKa = -log(Ka)
8.21 = -log(Ka)
Ka = 7.94 x 10^-9
To determine the concentration of codeine-H+, we can use the equation for the dissociation of codeine-H+:
Ka = [H+][codeine-]/[codeine-H+]
Let x be the concentration of codeine-H+ that dissociates. Then:
7.94 x 10^-9 = x^2 / (0.0073 - x)
Solving for x gives x = 5.29 x 10^-5 M
Therefore, the concentration of codeine-H+ is 5.29 x 10^-5 M and the concentration of codeine is 0.0073 M. Since codeine is a weak base, we can assume that it does not significantly affect the pH of the solution.
(b) Since the concentration of codeine is greater than the concentration of codeine-H+, we can assume that the pH will be determined by the concentration of codeine-H+.
Let x be the concentration of codeine-H+. Then:
Ka = [H+][codeine-]/[codeine-H+]
7.94 x 10^-9 = x^2 / (0.1 - x)
Solving for x gives x = 2.08 x 10^-5 M
(c) Similar to (b), we can assume that the pH will be determined by the concentration of codeine-H+.
Let x be the concentration of codeine-H+. Then:
Ka = [H+][codeine-]/[codeine-H+]
7.94 x 10^-9 = x^2 / (0.01 - x)
Solving for x gives x = 2.21 x 10^-5 M
(d) Similar to (a), we can assume that the concentration of codeine-H+ is very small and can be neglected.
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The items listed below relate to the method used for solving problems in chemistry. Match each pair of items correctly.1. The known quantities in the problem...2. The unknown quantity...3. The units of conversion factors..4. Any units that do not cancel....1... often provide useful conversion factors.2... will be the solution to the problem.3... must cancel at each step, except for the units needed for the answer.4... are the units for the unknown quantity.
The correct matching is of pair of items is :
Known quantities in the problem --> Often provide useful conversion factors
Unknown quantity --> Will be the solution to the problem
Units of conversion factors --> Must cancel at each step, except for the units needed for the answer
Any units that do not cancel --> Are the units for the unknown quantity
1)The known quantities in the problem often provide useful conversion factors.
2)The unknown quantity will be the solution to the problem.
3)The units of conversion factors must cancel at each step, except for the units needed for the answer.
4)Any units that do not cancel are the units for the unknown quantity.
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What is the IUPAC name for CH3CH(OH)CH(CH3)CH(CH3)2 ?
The IUPAC name of the following structural formula CH3CH OH−CH3 is propan-2-ol.
What is the structural formula?
The molecular structure of a chemical compound is graphically represented by the structural formula of the complex, which demonstrates how the atoms may be arranged in three-dimensional space. Structural formulas offer a more comprehensive geometric depiction of the molecular structure than other forms ofof chemical formulas, which have fewer symbols and less descriptive power. For instance, many chemical compounds exist in many isomeric forms that have the same molecular formula but different enantiomeric structures..To know more about structural formula, click the link given below:
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in the bronsted Lowry definition of acids and the base, an acid and
A) is a proton donor B breaks stable hydrogen bonds C) corrodes metals D) form is stable hydrogen bonds
Answer: A (acid is a proton donor)
Explanation:
Bronsted-Lowry Definition
Acids are proton (H⁺) donors.Example: HCl is a strong acid and it donates a proton to water when both are in a solution. Water is the base because it accepted the proton from water to form H₃O⁺.Bases are proton (H+) acceptors.Example: NH₃(aq) + H₂O(l) <--> NH₄⁺(aq) + OH⁻(aq). In this reaction, NH₃ is the base because it accepted the proton from water, and water is the acid because it donated a proton to NH₃.· Amphoteric means some substances act as an acid or base, depending on the reaction. Water is a good example of an amphoteric substance because it can donate or accept a proton.How many moles of tetraphosphorous decaoxide will form when 3.30 moles of P4 react with oxygen gas in the given equation? P4 + 5 O2 → P4O10
3.30 moles of tetra phosphorous decaoxide (P4O10) will form when 3.30 moles of P₄ react with oxygen gas.
The balanced equation for the reaction between P4 and O2 is:
P₄ + 5O₂ → P₄O₁₀
According to the equation, 1 mole of P₄ reacts with 5 moles of O₂ to produce 1 mole of P₄O₁₀. Therefore, the number of moles of P₄O₁₀ formed is directly proportional to the number of moles of P₄ consumed in the reaction.
If 1 mole of P₄ produces 1 mole of P₄O₁₀, then 3.30 moles of P₄ will produce:
3.30 moles P₄ x (1 mole P₄O₁₀ / 1 mole P₄) = 3.30 moles P₄O₁₀
Therefore, 3.30 moles of tetra phosphorous decaoxide (P₄O₁₀) will form when 3.30 moles of P₄ react with oxygen gas.
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explain the relationship between no2 concentration and ozone concentration represented in this graph.
The relationship between NO2 and O3 concentrations is not straightforward and depends on a variety of factors. It is important to monitor both pollutants and take steps to reduce emissions of both NO2 and VOCs in order to improve air quality and reduce the risk of adverse health effects.
What is the relationship between NO2 concentration and ozone concentration?Since the question is incomplete and the graph that ought to be shown is missing, I will look at the relationship between NO2 concentration and ozone concentration in general.
The relationship between nitrogen dioxide (NO2) and ozone (O3) concentrations is complex and depends on a variety of factors such as sunlight, temperature, and the presence of other pollutants.
In general, high levels of NO2 can contribute to the formation of O3 through a complex series of chemical reactions involving other pollutants such as volatile organic compounds (VOCs). This process, known as photochemical smog formation, typically occurs during warm, sunny weather and can lead to elevated levels of O3 in urban and suburban areas.
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which of the following choices describe the steps required to determine the empirical formula of a compound from the mass percent? select all that apply. multiple select question. the ratio of atoms of each element must be a ratio of integer numbers. if the sample contains 52 g of carbon, it is assumed that this compound has 52 mass percent of carbon. if a compound contains 52% of mass of c, it can be assumed that there are 52 g of c in 100 g of the compound. the mass percentage of each element is converted to moles using the mass formula massmolarmass . the mass percentages are divided among the smallest number to calculate the ratio of atoms of each element present.
If carbon makes up 52 g of a compound's mass, then there are 100 g of the compound and 52 g of carbon.
What is meant by mass percentage?The percentage of a solution's mass that is made up of solutes is known as the mass percent. This percentage is measured in relation to the total mass of the solution.
A concentration or a component in a certain mixture can be expressed as mass percent. The mass percentage that indicates the mass of solute contained in a given mass of solution can be used to characterize the composition of the solution. In terms of mass or moles, the solute's concentration is expressed.
Each element must have an equal amount of atoms, which must be an integer ratio. If carbon makes up 52 g of a compound's mass, then there are 100 g of the compound and 52 g of carbon.
The mass percentage of each element is converted to moles using the mass formula mass / molar mass
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what charged group(s) are present in the amino acid gly at a ph of 12? select 1 correct answer(s) question 1 options: -nh3 -coo- -nh2 there are no ionized groups in gly at ph 12.
The side chain of the amino acid glycine is made up of hydrogen. The carboxyl group is deprotonated and negatively charged at basic pH (pH=12)
Organic substances known as amino acids have both amino and carboxylic acid functional groups. Alpha-amino acids, which make up proteins, are by far the most significant of the hundreds of amino acids found in nature. In the genetic code, only 22 alpha amino acids are present.
Body protein as well as other vital nitrogen-containing substances including creatine, peptide hormones, and some neurotransmitters cannot be produced without amino acids. Although allowances are expressed in terms of protein, amino acids are a biological necessity.
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consider a reaction in which 1.00 mole of so2(g) and 1.00 mole of o2(g) are added to a 1.00l container at 1000k and allowed to react until the equilibrium below is achieved. At equilibrium, the container has 0.919 moles of SO3 (g). What is the value of Kp?
2SO2 (g) + O2 (g) → 2SO3 (g)
The value of Kp for the given reaction is 100.4
The equilibrium constant expression (Kp) for the given reaction is:
Kp = (P_SO₃)² / (P_SO₂)² × (P_O₂)
where P is the partial pressure of the gas in atm.
We can use the given information to calculate the partial pressures of each gas at equilibrium.
The initial pressure of each gas is 1.00 atm (since each mole of gas occupies the same volume, and the total volume is 1.00 L).
At equilibrium, each mole of SO₂ that reacts produces 1 mole of SO₃. Therefore, the partial pressure of SO₃ at equilibrium is:
P_SO₃ = 0.919 mol / 1.00 L = 0.919 atm
Since two moles of SO₂ react for every one mole of O₂, the partial pressure of SO₂ at equilibrium is:
P_SO₂ = (1.00 mol - 0.919 mol) / 1.00 L = 0.081 atm
Partial pressure of O2 at equilibrium is:
P_O₂ = (1.00 mol - 0.919 mol) / 1.00 L = 0.081 atm
Now we can substitute these values into the equilibrium constant expression:
Kp = (0.919)² / (0.081)² × (0.081) = 100.4
Therefore, the value of Kp for the given reaction is 100.4 (with no units, since the units cancel out).
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g what is the iupac name for hoch2chohch2oh? selected answer: correctc. 1,2,3-propanetriol answers: a. 1-hydroxy-2,3-propanediol b. glycerol correctc. 1,2,3-propanetriol d. triisopropyl alcohol
The correct IUPAC name for HOCH2CHOHCH2OH is 1,2,3-propanetriol.
This compound is also commonly known as glycerol or glycerin. The IUPAC naming system uses a set of rules to assign unique and unambiguous names to chemical compounds.
The first step in naming this compound is to identify the longest continuous carbon chain, which in this case is three carbons long, giving us the base name "propane." Next, we need to identify any functional groups attached to the carbon chain.
In this case, there are three hydroxyl (-OH) groups attached to the carbon chain, giving us the suffix "triol." Putting these two parts together, we get the IUPAC name 1,2,3-propanetriol.
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Consider the reaction:
NaNO2(aq) + H2NSO3H(s) → NaHSO4(aq) + N2(g) + H2O(l)
If you start with 1.627 g of H2NSO3H and an excess amount of NaNO2,
what volume will the N2 occupy at 25°C and 0.978 atm?
212m^3 will be volume of N2
What do the ideal gas equations mean?
PV = nRT is the real gas equation or the ideal gas law. P stands for pressure, V for volume, n for moles of gas, R for the ideal gas constant, and T for Kelvin temperature.
The reaction is :
NaNO2(aq) + H2NSO3H(S) ⇒ NaHSO4(aq) +N2(g) +H2O(l)
1 mole of H2NSO3H produce 1 mole of N2
Molar mass of H2NSO3H is 97gm /mol
Molar mass of N2 will be 28gm/mol
No. of moles of N2 is 1
Using formula for number of moles, we get mass of N2 as
no. of moles * molar mass
i.e. 1*28 ⇒28g
Pressure is 0.978
Temperature is 25 degree
PV ⇒nRT
n is 1
R is 8.314
V ⇒ nRT/P
⇒ 1*8.314*25/0.978
⇒ 212m^3
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How many copper atoms are present in 3.271 x 10-19 grams of copper (II) sulfate?
The number of atoms of copper present in 3.271×10⁻¹⁹ grams of cupper (II) sulfate is 1234.1 atoms
How do I determine the number of atoms of copper present?First, we shall determine the mole of 3.271×10⁻¹⁹ grams of cupper (II) sulfate. Details below:
Mass of CuSO₄ = 3.271×10⁻¹⁹ grams Molar mass of CuSO₄ = 63.55 + 32 + (16 × 4) = 159.55 g/mol Mole of CuSO₄ =?Mole = mass / molar mass
Mole of CuSO₄ = 3.271×10⁻¹⁹ / 159.55
Mole of CuSO₄ = 2.05×10⁻²¹ mole
Next, we shall determine the mole of copper in the sample. Details below:
1 mole of CuSO₄ contains 1 mole of copper, Cu
Therefore,
2.05×10⁻²¹ mole of CuSO₄ will also contain 2.05×10⁻²¹ mole of copper.
Finally, we shall determine the number of atoms of copper present in the sample. Details below:
From Avogadro's hypothesis,
1 mole of copper = 6.02×10²³ atoms
Therefore,
2.05×10⁻²¹ mole of copper = 2.05×10⁻²¹ × 6.02×10²³
2.05×10⁻²¹ mole of copper = 1234.1 atoms
This, the number of atoms of copper is 1234.1 atoms
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devise a route to carry out the following conversion: (specify the reagents you would use to carry out the conversion by using letters from the table. the reaction may require more than one step, if so, write the letters in the order that they are used, e.g., hb. if two or more ways of conversion to the same product are possible, show only one of them.) reagents available a. , pyridine e. / i. 1. / 2. b. , heat f. j. c. , g. 1. / 2. k. / d. h. / lindlar catalyst l. 1. 2. , ,
Use reagent f. H2CrO4. Chromic acid is a strong acid used to oxidize alcohols into ketones and carboxylic acids. Chromium trioxide combines with water to produce chromic acid, which is deliquescent, light red or brown in color, and soluble in water.
What is a lindlar catalyst?Lindlar catalyst is a type of heterogeneous catalyst used in organic chemistry for hydrogenation reactions. It is composed of palladium metal supported on calcium carbonate, barium sulfate, or similar materials, and is commonly used in the partial hydrogenation of alkynes to alkenes. The catalyst is named after its inventor, Herbert Lindlar, who developed it in the 1950s.
Lindlar catalyst is a selective catalyst, which means that it allows for the hydrogenation of alkyne functional groups to alkenes while inhibiting further hydrogenation to alkanes. This is achieved by using poisoned or deactivated palladium that restricts the catalyst's activity and allows for partial hydrogenation to occur. This controlled hydrogenation is useful in organic synthesis because it provides a way to selectively reduce alkynes to alkenes without the formation of unwanted byproducts.
Reagent (f) can directly convert given alcohol to its carboxylic acid.
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Classify each of the following substances as an element, a compound, or a mixture.
Silver, Ag and fluorine, F, carbon monoxide, CO and calcium chloride, Ca Cl2, soft drink
Drag the appropriate items to their respective bins.
A mixture is a physical combination of two or more compounds that retain their identities.
What is mixture?A mixture in chemistry is a substance composed of two or more chemical components which are not chemically connected. A mixture is a physical combination of two or more compounds that retain their identities and therefore are mixed as solutions, suspensions, as well as colloids.
Silver element
Ag and fluorine mixture
F element
carbon monoxide compound
CO and calcium chloride mixture
CaCl[tex]_2[/tex] compound
soft drink mixture
Therefore, a mixture is a physical combination of two or more compounds that retain their identities.
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Which substance can dissolve only (is saturated at) 40 g in 60°C water?
a
KClO3
b
NaCl
c
KCl
(30 points)
KCl would be the substance can dissolve only (is saturated at) 40 g in 60°C water
Option C is correct.
What is a supersaturated solution?A supersaturated solution is described as a solution that contains more than the average solvent that can be dissolved at a given temperature
At 60ºC, the solubility of KCl is about 42g/100mL of water. Therefore, since we have 40g of KCl that were dissolved in less than 100mL of water, the solution would be supersaturated.
In conclusion, supersaturation occurs with a solution when the concentration of a solute exceeds the concentration specified by the value of solubility at equilibrium.
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using the bsa working solutions created above, how much and which concentration would you pipet into three different centrifuge tubes so you end up with 20 ug, 2 ug, and 0.2 ug of bsa? (note: remember to stay within the accurate range of your pipetman.)
The concentration on which we should pipette out are 4μl.
To recap, the working solutions are 20μg/μl, 2μg/μl, and 0.2μg/μl.
1.
It’s basically the same as the other question. Set it up algebraically:
50μg/μl x = 200 μg
Note the variable x in the equation. You can see easily that x = 4μl (remember to include the units, and recall that units cancel in equations).
So if you pipet 4μl of the 50μg/μl working solution into a centrifuge tube, you’ll have 20μg of BSA in there.
2.
Again, set it up algebraically:
5μg/μl x = 20 μg
Here, x = 4μl again. So again you are pipetting out 4μl, except this time it’s from the 5μg/μl working solution.
3.
I’m sure you can see that there’s a trend going on here (it’s not any special science trend or anything like that, though).
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Are there any mutations that can occur in lymphatic.
Answer:Lymphatic malformations (LM), also known as lymphangiomas, are characterized by the overgrowth of lymphatic vessels during pre- and postnatal development.
A/How many amide functional groups are present in vancomycin? B/ Which OH groups are bonded to SP³ hybridized carbon atoms and which are bonded to SP² hybridized carbons? C/. Would you expect vancomycin to be water soluble? Explain. D/. Label three different functional groups capable of hydrogen bonding.
A. There are two amide functional groups are present in vancomycin.
B. In phenol OH groups are bonded to SP³ hybridized carbon atoms and which are bonded to SP² hybridized carbons.
What is phenol ?Phenol is an aromatic organic compound. They have molecular formula C₆H₅OH. It is a white crystalline solid that is volatile. The molecule consists of a phenyl group bonded to a hydroxy group.
The OH group is attached to the sp2 hybridized carbon of an aromatic ring in phenols.
C. Vancomycin is a very water-soluble drug.
D. Alcohols, amines (primary and secondary), carboxylic acids, and amides are the four pure functional groups capable of hydrogen bonding (primary and secondary).
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as discussed in class, an important buffer for intracellular fluids (and thus a common buffer in the laboratory to mimic physiological conditions) is the phosphate system. despite being a polyprotic acid, the equilibrium of dihydrogen phosphate (h2po4-) and hydrogen phosphate (hpo42-) is the most useful due to its pka of 7.2. 6.2 - 8.2 in what ph range can this system be used as an effective buffer? 7.34 what is the ph of a mixture of 0.042m h2po4- and 0.058 m hpo42-? if 1.0 ml of 10.0 m naoh is added to 1.0 l of the buffer prepared above, what is the final ph? if 1.0 ml of 10.0 m naoh is added to 1.0l pure water at ph 7.0, what is the final ph? how does this compare to the change observed in the buffered
The phosphate system can be used as an effective buffer in the pH range of 6.2-8.2, which includes the pKa of 7.2 for the equilibrium of dihydrogen phosphate and hydrogen phosphate.
To calculate the pH of the given mixture of 0.042 M H2PO4- and 0.058 M HPO42-, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([HPO42-]/[H2PO4-])
pH = 7.2 + log(0.058/0.042)
pH = 7.34
Therefore, the pH of the mixture is 7.34.
When 1.0 mL of 10.0 M NaOH is added to 1.0 L of the buffer prepared above, we can use the equation for the reaction between NaOH and H2PO4-: NaOH + H2PO4- → NaHPO4 + H2O
The initial moles of H2PO4- in 1.0 L of the buffer are:
moles H2PO4- = 0.042 M × 1.0 L = 0.042 mol
Adding 1.0 mL of 10.0 M NaOH adds 0.01 mol of NaOH to the buffer solution. Assuming that all of the added NaOH reacts with H2PO4-, the final moles of H2PO4- and HPO42- can be calculated as follows:
moles H2PO4- = 0.042 mol - 0.01 mol = 0.032 mol
moles HPO42- = 0.058 mol + 0.01 mol = 0.068 mol
Using the Henderson-Hasselbalch equation with the new concentrations of H2PO4- and HPO42-, we can calculate the final pH: pH = pKa + log([HPO42-]/[H2PO4-])
pH = 7.2 + log(0.068/0.032)
pH = 7.76
Therefore, the final pH of the buffer after the addition of NaOH is 7.76.
When 1.0 mL of 10.0 M NaOH is added to 1.0 L of pure water at pH 7.0, we can use the equation for the reaction between NaOH and water:
NaOH + H2O → Na+ + OH- + H2O
The initial concentration of OH- can be calculated from the concentration of NaOH: [OH-] = 10.0 M × 1.0 mL / 1000 mL = 0.01 M
The initial concentration of H+ in the water is:
[H+] = 10^-7.0 = 1.0 × 10^-7 M
Assuming that all of the added NaOH dissociates completely, the final concentration of OH- can be calculated as:
[OH-] = 0.01 M + 10.0 M × 1.0 mL / 1000 mL = 0.100 M
The final concentration of H+ can be calculated from the equation for the ion product of water:
[H+][OH-] = 1.0 × 10^-14
[H+] = 1.0 × 10^-14 / 0.100 M = 1.0 × 10^-13 M
Therefore, the final pH of the water after adding NaOH is 13.0.
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Tell what you should do to prepare for equip calibration
-Check that all nozzles are the same and of the desired flow rate and pattern -clean all nozzle and screens -add clean water to the spray tank and adjust the pressure to the desired level while the pump is operating at normal speed. make sure water is flowing through the nozzles -Check the spray patterns from each nozzle -With the sprayer stationary and operating at the desired pressure, hold a measuring cup marked in fluid ounces directly under each nozzle for 1 min. -calculate the average output (flow rate) by dividing the total amount collected by the number of nozzles. -replace any nozzle whole flow rate is 5% greater or less than the average.
The equipment can be properly calibrated to ensure that the spray application rate is consistent and accurate, which is important for achieving the desired level of control while minimizing waste and avoiding off-target effects.
To prepare for equipment calibration, the following steps should be taken:
Check that all nozzles are the same and of the desired flow rate and pattern. Clean all nozzle and screens. Add clean water to the spray tank and adjust the pressure to the desired level while the pump is operating at normal speed. Make sure water is flowing through the nozzles.
Check the spray patterns from each nozzle. With the sprayer stationary and operating at the desired pressure, hold a measuring cup marked in fluid ounces directly under each nozzle for 1 min.
Calculate the average output (flow rate) by dividing the total amount collected by the number of nozzles. Replace any nozzle whose flow rate is 5% greater or less than the average.
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