True. Calcium ions (Ca2+), which are released by the SR, bind to troponin in the sarcoplasm. Then, troponin displaces tropomyosin from the actin myosin-binding sites.
Troponin levels that are extremely high indicate the onset of a heart attack. Within six hours after a heart attack, the majority of individuals experience elevated troponin levels. Mostly everyone who has experienced a cardiac arrest will still have elevated levels after 12 hours. After a heart attack, troponin levels may continue to rise for 1 to 2 weeks. For instance, the typical levels for troponin I is 0 to 0.04 ng/mL, whereas the normal range for rising cardiac troponin (hs-cTn) is less than 14 ng/L. Increases in troponin levels may result from other kinds of heart damage. One of these is sinus arrhythmia.
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Write the overall balanced cell reaction for the following voltaic cell Fe(s) | Fe2+ (aq) || Ag + (aq) | Ag(s) (Ctrl).
the overall balanced cell reaction for the following voltaic cell
Hg2Cl2(s) + Cu(s) ® 2Hg(l) + Cu2+(aq) + 2Cl-(aq)
The cell reaction, also known as the total reaction, is the reaction that occurs within the cell as a whole and is expressed under the assumption that the right electrode is the cathode and that the spontaneous reaction is the one in which reduction takes place in the right-hand compartment.Galvanic, also known as Voltaic, and electrolytic cells are the two varieties of electrochemical cells. While electrolytic cells utilize non-spontaneous reactions and therefore need an external electron source, such as a DC battery or an AC power source, galvanic cells get their energy from spontaneous redox reactions.
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Consider the titration of 100.0 mL of 0.59 M H3A by 0.59 M KOH for the next three questions. The triprotic acid has Ka1 = 1.0 x 10-3, Ka2 = 1.0 x 10-6, and an unknown value for Ka3.1) Calculate the pH after 100.0 mL of KOH has been added.2) Calculate the pH after 150.0 mL of KOH has been added.3) The pH of the solution after 200.0 mL of KOH has been added is 8.00. Determine the value of Ka3 for this triprotic acid. Use scientific notation to enter this answer, e.g., 1.0 x 10-3 = 1.0E-3.
Acid concentration: 0.49 M KOH concentration is 0.49 MKa1=1.0 x 104 eq. rm
a remedy for Take the example of titrating 100 mL of 0.100 M HOCl with 0.100 M KOH at 25 °C. HOCl has a Ka of 3.5 10-8. Calculate the pH after 0.0 mL of KOH in Part 1.
Consider titrating 0.115 M RbOH in a 25.0 mL sample with 0.100 M HCl. Figure out each quantity. The initial pH and the amount of acid that must be added to achieve the equivalence point are also factors. c. the pH after adding 5.0 mL of acid. pH at the equivalency point (d). e. the pH following the addition of 5.0 mL of acid over the equivalency point.
Think about titrating 100 mL of 0.59 M H 3 a 1-3 a2-6 a3. 1) Determine the pH following
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Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25°C, diamond changes to
graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine A
Hrxn for
C(diamond)
C(graphite)
with equations from the following list:
(1) C(diamond) + O₂(g) → CO₂(g)
(2) 2 CO₂(g) → 2 CO(g) + O₂(g)
(3) C(graphite) + O₂(g) → CO₂(g)
(4) 2 CO(g) → C(graphite) + CO₂(g)
AHoverall
KJ
AH = -395.4 kJ
AH= 566.0 kJ
AH=-393.5 kJ
AH = -172.5 kJ
The ΔH of the reaction for the formation of graphite from diamond is equal to -1.9 KJ/mol.
What is the heat of the formation of the reaction?The heat of formation can be described as the amount of heat absorbed or evolved when 1 mole of a substance is formed from its constituent elements, each substance being in its normal physical state.
We have to rearrange the provided equations in a way in which all molecules of O₂ and CO₂ can be eliminated:
2C (diamond) + 2O₂ (g) → 2CO₂ (g) ΔH₁= 2 × (-395.4 KJ)
2 CO₂ (g) → 2 CO(g) + O₂(g) ΔH₂ = 566.0 KJ
CO₂(g) → C (graphite) + O₂(g) ΔH₃= - 1 × (-393.5 KJ)
2CO (g) → C (graphite) + CO₂(g) ΔH₄= -172.5 KJ
When we eliminate the molecules that appear both in reactants and products, the total chemical reaction is the following:
2C (diamond) → 2C (graphite)
ΔH (total) = ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄
ΔH (total) = 2 × (-395.4 KJ) + 566.0 KJ + (-1 × (-393.5 KJ)) - 172.5 KJ
ΔH (total) = 347.2 KJ
This is for 2 mol of C (diamond) which is converted into 2 mol of C (graphite). To find ΔH for the reaction of 1 mol C (diamond) to 1 mol (graphite) we have to divide it into 2:
ΔH = - 3.8/2 = -1.9 KJ/mol
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Epinephrine is a hormone secreted into the bloodstream in times of danger and stress. It is 59% Carbon, 7.1% Hydrogen, 26.2% Oxygen and 7.7% Nitrogen by mass. Determine the empirical formula. The molar mass is 148 g/mol.
Based o the percentage masses of the elements, the empirical formula of epinephrine is C₉H₁₃O₃N.
What is the empirical formula of epinephrine?The empirical formula of epinephrine is calculated below as follow:
Mole ratio of the elements are determined:
Carbon = 59% / 12
Carbon = 0.049 moles
Hydrogen = 7.1% / 1
Hydrogen = 0.07 moles
Oxygen = 26.2% / 16
Oxygen = 0.016
Nitrogen = 7.7% / 14
Nitrogen = 0.0055
The simplest whole-number ratio will be:
C = 0.049/0.0055
C = 9
Hydrogen = 0.07/0.0055
H = 13
Oxygen = 0.016/0.0055
Oxygen = 3
Nitrogen = 0.0055/0.0055
Nitrogen = 1
The empirical formula = C₉H₁₃O₃N
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On a hot afternoon (311 K) a party balloon was filled with 2.07 L of helium. That evening the temperature dropped to 295 K. What is the
new volumen of the balloon?
Answer:
2.02 L
Explanation:
The volume of a gas is directly proportional to its temperature, according to the ideal gas law. This means that as the temperature of a gas decreases, its volume will also decrease.
To determine the new volume of the balloon as the temperature drops from 311 K to 295 K, we can use the ideal gas law to calculate the new volume of the gas:
PV = nRT
Where:
P = the pressure of the gas (assuming it remains constant)
V = the new volume of the gas
n = the number of moles of the gas (assuming it remains constant)
R = the universal gas constant
T = the new temperature of the gas
Since we know the pressure, number of moles, and universal gas constant, we can solve for the new volume of the gas by substituting in the known values and solving for V:
V = nRT / P
Substituting in the values from the problem, we get:
V = (n * R * 295 K) / P
Since the number of moles and the universal gas constant are constants, we can simplify the equation to:
V = k / P
Where k is a constant equal to the product of the number of moles, the universal gas constant, and the original temperature of the gas (in this case, k = n * R * 311 K).
Since we know the value of k and the pressure of the gas, we can solve for the new volume of the gas by substituting in the known values:
V = (k / P)
= (n * R * 311 K) / P
= (2.07 L * 8.31 L * atm / mol * K * 311 K) / P
= 2.02 L
Therefore, the new volume of the balloon is approximately 2.02 L.
Suppose the galvanic cell sketched below is powered by the following reaction: Fe(s)+NiCl2(aq) → FeCl2(aq)+Ni(s) E2 S1 S2 Write a balanced equation for the half-reaction that happens at the cathode of this cell. Write a balanced equation for the half-reaction that happens at the anode of this cell. Of what substance is E1 made? of what substance is E2 made? What are the chemical species in solution S1? What are the chemical species in solution S2?
The chemical species in solution S2 Fe(s)+NiCl2(aq) → FeCl2(aq)+Ni(s)
Fe(s) ------> [tex]Fe^{2+}[/tex] + 2e (Oxidation)
[tex]Ni^{2+}[/tex] (aq) + 2e ------> Ni(s) (Reduction)
Galvanic Cell
Galvanic cells employ the electrical energy generated by the transport of electrons during a redox reaction to carry out practical electrical work. By separating the oxidation and reduction half-reactions and connecting them with a wire so that the electrons must pass through the wire, it is possible to gather the electron flow.
Write a balanced equation for the half-reaction that happens at the cathode of this cell = Ni (aq) + 2e ------> Ni(s)
Write a balanced equation for the half-reaction that happens at the anode of this cell = Fe(s) ------> [tex]Fe^{2+}[/tex] + 2e.
Of what substance is E1 made? = Fe
of what substance is E2 made? = Ni
What are the chemical species in solution S1? = [tex]Fe^{2+}[/tex]
What are the chemical species in solution S2? = [tex]Ni^{2+}[/tex]
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given the reaction: 234 91 pa x 0 -1e when the equation is correctly balanced the nucleus represented by x is
The reaction is given : ²³⁴Pa₉₁ -----> X + ⁰e₋₁ , the equation is balanced and the nucleus represented by X is ²³⁴U₉₂.
The balanced reaction is given as :
²³⁴Pa₉₁ -----> X + ⁰e₋₁
This reaction is the balanced reaction and X is represented the nucleus and is given as : ²³⁴U₉₂.
The given reaction is the nuclear reaction. the nuclear reaction is the reaction in which the nuclides are produced by the collision between the two atoms nuclei. the nuclear reaction are of the two types the nuclear fission and the nuclear fusion reaction. the nuclear fission is the splitting of the large nuclei in to the smaller ones.
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cold water can hold more dissolved oxygen than warm water because the solubility of oxygen decreases as temperature increases. T/F
The solubility of oxygen reduces with increasing temperature, hence cold water can store more dissolved oxygen than warm water. The chemical element with the letters O.
And the atomic number 8 is called oxygen. It belongs to the chalcogen group of the periodic table and is a highly reactive nonmetal that rapidly forms oxides with most elements as well as with other compounds. Understanding the makeup of matter developed together with the creation of the concepts of the elements. The substance has been viewed as either continuous or discontinuous at various points throughout history. It is possible to hypothesise that continuous substance is homogenous, infinitely divisible, and has parts that are all the same size.
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Combustion of 31.68 g of a compound containing only carbon, hydrogen, and oxygen produces 36.67 g CO2 and 15.01 gH2O. What is the empirical formula of the compound?
Combustion of 31.68 g of a compound containing only carbon, hydrogen, and oxygen produces 36.67 g CO₂ and 15.01 g H₂O. the empirical formula of the compound is CH₂O.
given that :
mass of CO₂ = 36.67 g
mass of H₂O = 15.01 g
total mass of compound = 31.68 g
moles of CO₂ = 36.67 / 44
= 0.833 mol
moles of C = 0.833 mol
moles of H₂O = 15.01 / 18
= 0.833 mol
moles of H = 2(0.833)
= 1.667 mol
mass of carbon = 0.833 × 12
= 10 g
mass of hydrogen = 1.667 × 1
= 1.667 g
mass of oxygen = 31.68 - ( 10 + 1.67 )
= 20 g
moles of Oxygen = 20 / 16
= 1.25 mol
dividing by the smallest one :
C = 0.833 / 0.833 = 1
H = 2
O = 1
The empirical formula is CH₂O.
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What is the formula for cobalt (III) oxide?
Co3O2
Co2O3
Co3O
CoO3
Answer:Co2O3
Explanation:
look it up
Answer:
Co2O3
Explanation:
when you write any formula you apply the criss-cross method where you write the charges of each substance
Co3 and O2
applying the criss-cross method( you switch the charges number, in this case you switch the 2 and the 3) :-
Co2 and O3
so Co2O3
The density of an unknown element in the gaseous state is 1.60 g at 300 K and 1 atm. Which of the following could be the element?
a. He
b. Ne
c. Ar
d.O2
e. Cl2
with reference to compound a below, label each compound as an identical molecule, an isomer, a resonance structure, or neither.
Label for each of the compounds given in the question are: 1) Isomer, 2) Isomer, 3) Resonance Structure, 4) Isomer 5) Neither.
A group of two or more Lewis structures known as resonance structures describe the electronic bonding of a single polyatomic species, including fractional bonds and fractional charges. The number of electrons in resonance structures should be constant; do not add or subtract any electrons. (Count the electrons to determine their number). Every resonance structure complies with the standards for writing Lewis Structures. Structure hybridization must be constant. By cyclically moving elections, it is possible to depict the two resonance configurations of the benzene ring. Multiple resonance structures can be found in conjugated double bonds.
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a chemistry graduate student is given of a chloroacetic acid solution. chloroacetic acid is a weak acid with . what mass of should the student dissolve in the solution to turn it into a buffer with ph ? you may assume that the volume of the solution doesn't change when the is dissolved in it. be sure your answer has a unit symbol, and round it to significant digits.
Choloracetic acid, 12 g. The colorless solution of the white crystalline substance known as chloroacetic acid.
Up to 80% of the solution can be acidic. By inhalation, ingestion, and skin contact, it is poisonous. It corrodes flesh and metals.
The equilibrium of the buffer is
CH2ClCO2 + H+ HCH2ClCO2
PKA = Log Ka
Ka: [CH2ClCO2] = 1,3 x 103. "H+" / "HCH2ClCO2"
Using the Henderson-Hasselbalch formula:
pH equals pka plus log10 [A]/[HA].
2,89 + log10 [A] / [HA] = 3,01
1,318 = [A⁻] / [HA]
Since chloroacetic acid (HA) has a molar concentration of 0,20M
[A⁻] = 0,26 mol/L
There are 500 mL plus 0.5 L.
Chloroacetic acid is equal to 0,26 mol/L x 0,5 l, or 0,13 moles. By weight:
Choloracetic acid is equal to 0,13 mol (94,5g / 1mol) = 12 g.
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For how many electrons is the value of ml +2
Only two electrons can be filled in an orbital with a magnetic quantum number (ml) +2.
What are the quantum numbers?The set of numbers that defines the position and energy of the electron filled in a specific orbital are called quantum numbers. Four quantum numbers are principal, azimuthal quantum number, magnetic, and spin quantum numbers.
The principal quantum number (n) denotes the principal electron shell of the atom and the most probable distance between the nucleus and the electrons. The azimuthal quantum number (l) denotes the shape of an orbital in which an electron is present.
The magnetic quantum number denotes the total number of orbitals in a subshell and the orientation of these orbitals. The value of the spin quantum number (s) denotes the direction in which the electron is spinning.
For the value of n = 2, we have five values of ml which are +2, +1, 0, -1, and -2. Every value of ml represents one d-orbital. Each orbital can have two electrons.
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Consider the following reaction at 276 K.
1 A + 2 B → C + D
where rate = rate=k[A][B]2. An experiment was performed for a certain number of seconds where [A]o = 0.000847 M and [B]o = 1.11 M. A plot of ln[A] vs time had a slope of -9.43. What will the rate of this reaction be if a new experiment is preformed when [A] = [B] = 0.779 M?
1 A + two C and d → C + S t where rate=k[A][B][rate] 2. An experiment with [A]o = 0.000847 T s as well as [B]o = 1.11 M was run for a predefined number of seconds. The slope of a ln[A] vs. grown steadily was -9.43. rate of reaction is 3.62m/s
Slope and slant both refer to an incline up from a reference surface or line that is generally straight. The ground slopes (either upward or straight down) sharply here. To slope is to slope vertically in an ambiguous path A line's steepness could be determined by looking at its slope. Altitude is calculated mathematically as "rise over run."
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A fresh sample 1311 was left out on the benchtop for 40 days. The scientist forgot it has a half-life of 8 days. The original sample mass
was 64 g. How much of the parent isotope is left?
g
Answer:
2 g of the parent isotope is left.
Explanation:
The half-life of a radioactive isotope is the amount of time it takes for half of the isotope to decay. In this case, the half-life of the isotope is 8 days, which means that after 8 days, half of the isotope will have decayed.
To determine how much of the parent isotope is left after 40 days, we need to calculate how many half-lives have passed in that time. Since the half-life is 8 days, and 40 days have passed, we can divide 40 by 8 to find the number of half-lives that have occurred: 40 / 8 = 5
Since 5 half-lives have passed, we can calculate the amount of the parent isotope that is left by multiplying the original amount of the isotope by 1/2 to the power of the number of half-lives that have passed: 64 * (1/2)^5 = 64 * (1/32) = 2 g
Therefore, after 40 days, only 2 g of the parent isotope is left.
Compound A is an alkyne with molecular formula CsH8- When treated with aqueous sulfuric acid and mercuric sulfate, two different products with molecular formula C5H1oo are obtained in equal amounts. 0 Get help answering Molecular Drawing questions. Draw the structure of compound A. Edit 0 Get help answering Molecular Drawing questions. Draw the two products obtained in the box below Edit
Alkynes are hydrocarbons containing a carbon-carbon triple bond. Their general formula is CnH2n-2 for molecules with triple bonds.
Alkynes undergo many of the same reactions as alkenes but can react twice due to the presence of two p-bonds on the triple bond. Ethyne is commonly known by the trivial name acetylene. It is the simplest alkyne, consisting of two carbon atoms joined by a triple bond, where each carbon can bond to a hydrogen atom.
Molecules containing a triple bond between two carbon atoms are called alkynes. A triple bond consists of one σ bond and two π bonds. As for the hydrogen atoms, this compound is unsaturated, so the extra electrons are exchanged with two carbon atoms forming a double bond.
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Find the volume, in mL, of an object whose density is 400 g/mL and has a mass of 600
mg.
The volume of the object is 1.5ml.
What is the volume of an object?This refers to the space occupied within the boundaries of an object in three-dimensional space. It is also called the capacity of the object.
In the question:
ρ = 400 g/mL
m = 600 mg
v = ?
Formular for calculating density ρ:
ρ = m/v
Where,
ρ= Density of the object
m= Mass of the object
v = volume of the object
Were are given the values of density and mass in the question. We are to calculate the volume.
Makinig v subject of the formular we have:
v = m/ρ
v = 600 mg
400 g/mL
v = 1.5ml
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How many compounds, of the ones listed below, have hydrogen bonding?CH3CH2CH2NH2 CH3CH2NHCH2CH3 (CH3CH2)2NCH2CH32130
Since CH3CH2CH2NH2 may form intermolecular hydrogen bonds, it has a higher boiling point than other compounds. The hydrogen atoms in (CH3)3N (C H 3) 3 N are joined to the carbon atoms.
Since carbon is not a particularly electronegative atom, it cannot donate hydrogen. There aren't any hydrogen atoms to accept, despite the fact that nitrogen is a strong hydrogen atoms. As a result, CH3CH2CH2CH3 is a nonpolar molecular complex in which induced dipole forces are the only intermolecular forces. When both the donor atom, D, and the acceptor atom, A, are among the highly electronegative elements O, N, or F, hydrogen bonding occurs, as shown by the symbol D-H—-A.
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What mass of silver will be precipitated when 5.0g of copper are reacted with
excess of silver nitrate solution?
Cu(s) + 2AgNO3(aq) → Cu(NO3)₂(aq) + 2Ag(s)
As per the given balanced reaction, one mole or 63.5 g of Cu produces 2 moles or 215.6 g of silver metal. Hence, 5 g of copper produces16.9 g of Ag.
What is displacement reaction?A displacement reaction is a type of reaction, in which one element or group of the reactant is displaced by the attacking group or element. The reaction between copper and silver nitrate is an example of single displacement reaction.
Given the balanced equation of the reaction. One mole of Cu produces 2 moles of Ag.
atomic mass of Cu = 63.5 g/mol
atomic mass of Ag = 107.8 g/mol
mass 2 moles of Ag = 2 × 107.8 =215.6 g.
Hence, 63.5 g of Cu produces 215.6 g of Ag.
The mass of Ag produced by 5 g of Cu is calculated as follows:
mass of Ag = (5 × 215.6 g) /63.5 g = 16.9 g
Therefore, the mass of silver precipitated will be 16.9 g.
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this method of acquiring mw differs from sec and gives dosy nmr many advantages over sec. as stated before, molecules of specific sizes produce indi-vidual diffusion coefficients. due to low sample con-centration, the purity of polymer is not as essential compared to sec; thus, reducing preparation time. dosy also requires minimum amounts of solvent
A well-known NMR technique called diffusion ordered spectroscopy (DOSY) provides diffusion coefficients for specific resonances in NMR spectra.
The main applications of DOSY are the analysis of small molecule mixtures and the oligomeric state of biomolecules. The spreading Self-Diffusion (SD)-NMR and diffusion ordered spectroscopy are two common names for the NMR technology (DOSY). This is accomplished by fusing magnetic field gradients, which encode spatial information, with radio-frequency pulses, which are commonly employed in NMR spectroscopy. Magnetic field gradients are used by Diffusion Ordered Spectroscopy (DOSY) to examine diffusion processes in both solid and liquid materials. Liquid structure is discovered using NMR spectroscopy. It is also used to figure out how soluble chemical molecules are structured.
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how do oxygen molecules interact differently with the systems and cells when you have asthma?
Answer:
In individuals with asthma, the airways (bronchial tubes) that carry air in and out of the lungs become narrowed, swollen, and inflamed due to an overactive immune response. This can lead to difficulty breathing, wheezing, coughing, and other symptoms.
During an asthma attack, the muscles around the airways may become tight and the lining of the airways may produce excessive mucus, further narrowing the airways and making it difficult for oxygen to reach the cells in the body. This can lead to low oxygen levels in the blood and cause symptoms such as shortness of breath and chest tightness.
In addition to these physical changes in the airways, individuals with asthma may also have an overactive immune response to certain triggers, such as allergens or irritants, which can further worsen inflammation and airway narrowing.
Treatment for asthma typically involves using medications to control inflammation and relax the muscles around the airways, as well as avoiding triggers that can worsen symptoms. By properly managing their asthma, individuals with the condition can often lead normal, active lives.
Explanation:
The tubes that carry air into and out of the lungs become irritated and constricted in people with asthma. This can make it difficult for air to enter and exit the lungs, which can cause symptoms including coughing, wheezing, and breathing problems.
Being less able to enter the lungs and reach the body's cells is one-way oxygen molecules can interact differently with the systems and cells in people with asthma. It is more difficult for oxygen to enter the lungs when the airways are constricted, which can result in a reduction in the amount of oxygen reaching the body's cells. This may cause symptoms including trouble breathing and difficulty engaging in physical activity.
In people with asthma, oxygen molecules can also affect the cells and systems in different ways by inducing an immunological reaction. When allergens or pollutants are breathed by asthmatic patients, their immune systems may be overly sensitive to them and may respond inappropriately. This may result in the release of inflammatory substances, which may further constrict the airways and hinder the absorption of oxygen by the lungs.
Inflammation and immune system overactivity can generally disrupt how oxygen molecules interact with the body's processes and cells in people with asthma, which can cause breathing problems and other symptoms.
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In a gas chromatogram, the retention time of compounds A and B are 1.25 min and 1.45 min, respectively. If both of these compounds have similar structures, which of the compounds has a higher boiling point?
In a gas chromatogram, the retention time of compounds A and B are 1.25 min and 1.45 min, respectively. If both of these compounds have similar structures, Compound A has a higher boiling point.
Gas chromatogram is a widely used technique in analytical chemistry for the separation and investigation of compounds that may evaporate without disintegrating (GC). The purity of a substance can be assessed by GC, as can the components of a mixture. Preparative chromatography can make use of GC to separate pure compounds from a mixture. Other names for gas chromatography include vapor-phase chromatography (VPC) and gas-liquid partition chromatography (GLPC). These alternative names and the accompanying acronyms are frequently used in scientific literature. Gas chromatography is a technique for separating compounds in mixtures by injecting a gaseous or liquid sample into a mobile phase, which is usually referred to as the carrier gas, and passing the gas through a stationary phase. The mobile phase is often composed of an inert gas or an unreactive gas, such as helium, argon, nitrogen, or hydrogen. The stationary phase is a minuscule layer of viscous liquid on a surface of solid particles on an inert solid support in a piece of glass or metal tubing known as a column.
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The Solubility Of Fe (OH)2 In Water At 25 °C Is Measured To Be 5.2x 10-4 Fe(OH)2 Use This Information To Calculate K Round Your Answer To 2 Significant Digits.
The Solubility Of Fe(OH)₂ In Water At 25 °C Is Measured To Be 5.2x 10⁻⁴ Fe(OH)₂ Ksp value is 5.6 × 10⁻¹⁰.
The chemical reaction is given as follows :
Fe(OH)₂ ⇄ Fe²⁺ + 2OH⁻
Ksp expression for the given reaction is given as follows :
Ksp = [Fe²⁺] [OH⁻]²
the solubilty S = 5.2x 10⁻⁴
so , the Ksp = S × (2S)²
= 4S³
Ksp = 4 ×S³
= 4 × ( 5.2x 10⁻⁴ )³
= 4 × 140.6 × 10⁻¹²
= 5.6 × 10⁻¹⁰
The solubility product value for Fe(OH)₂ that is Ksp value is 5.6 × 10⁻¹⁰.
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Correct the volume of 2.90 L of a gas at –12 °C to the volume occupied at 25 °C. Remember to convert between °C and K.
The volume occupied at 25 °C. Remember to convert between °C and K. V2 = 3.311 L is the correct answer.
What is volume?
Volume is the measure of the amount of space occupied by an object or a substance. It is measured in three dimensions: length, width, and height. In the International System of Units (SI), it is measured in cubic meters (m3). Volume is also commonly measured in liters (L), gallons (gal), or cubic feet (ft3). Volume can also refer to the amount of a substance or object, such as the volume of a gas or the volume of a liquid.
Temperature and Volume equation
Given,
The volume at -12°C = 2.90L
Temperature, T1 = -12°C
= (-12°C + 273)
= 261K
And, Volume at 25°C = ?
(25°C + 273)K
= 298K
Now, by using this equation:-
V1/T1 = V2/T2
V2 = V1 x T2/T1
V2 = 2.90L x 298K/261K
V2 = 3.311L
Hence, The volume at 25°C will be 3.311L
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An organism with the genotype of AaXx can produce gametes containing _________ if the two genes are unlinked.
a. either Aa or Xx
b. either AX, Ax, aX, ax
c. AaXx
d. AX or ax
e. None of the above.
An organism with the genotype of AaXx can produce gametes containing either AX, Ax, aX, ax if the two genes are unlinked .
An organism's genotype is the complete set of its genetic material. [1] Genotype can also be used to refer to the alleles or variants that an individual has at a particular gene or locus. The number of alleles an individual can have for a particular gene depends on the number of copies of each chromosome (also called polyploidy) found in that species. Diploid species like humans have two complete sets of chromosomes. This means that each individual has her two alleles for a particular gene. If both alleles are the same, the genotype is said to be homozygous. If the alleles are different, the genotype is called heterozygous.
Genotypes contribute to the phenotype, observable traits and characteristics of an individual or organism. The extent to which genotype influences phenotype depends on the trait.
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i can not figure out this question to save my life
Specific heat capacity of a substance is the amount of heat required to raise the temperature by one degree Celsius of one gram of a substance. Therefore, the total enthalpy of the given reaction is -781.44KJ.
What is Enthalpy?Enthalpy term is basically used in thermodynamics to show the overall energy that a matter have. Mathematically, Enthalpy is directly proportional to specific heat capacity of a substances.
Mathematically,
q = n ×ΔH
where
q = amount of heat
n = no of moles
ΔH = enthalpy
n = w / M.M
w = given mass
M.M = molar mass
4CH[tex]_3[/tex]NH[tex]_2[/tex]+2H[tex]_2[/tex]O[tex]\rightarrow[/tex]3CH[tex]_4[/tex]+CO[tex]_2[/tex]+4NH[tex]_3[/tex] ΔH=-138.8KJ
number of moles of methyl amine=175/31.06
=5.63moles
enthalpy for reaction of one mole of methyl amine=-138.8KJ
enthalpy for reaction of 5.63 mole of methyl amine=-138.8KJ×5.63
=-781.44KJ
Therefore, the total enthalpy of the given reaction is -781.44KJ.
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Determine the structure of C6H10, which upon oxidation gives acetic and isobutyric acid
Answer:CH
3
CH
2
CH
2
C≡C−CH
3
Explanation:
using curved arrow notation, write lewis acid/base equations for each of the following. remember to place formal charge on the appropriate atoms. a. aici3- b. phyp: bf3 bh3
AlCl 3 is Lewis base,PH3 is Lewis base,BH3 is Lewis acid ,BF3 is Lewis acid.
The crucial step is AlCl 3 accepting a chloride ion lone-pair to generate AlCl 4 and the highly acidic, or electrophilic, carbonium ion. Lewis bases: RCl +AlCl 3 → R + + AlCl 4−, etc. [ edit] The highest occupied molecular orbital (HOMO) of an atomic or molecular species that is strongly localised is known as a Lewis base.Lewis acid borane (BH3) has three hydrogen atoms and one boron atom in its molecule. The Lewis structure of borane features a single connection connecting each hydrogen atom to boron (BH3). Only three bonds surround the boron atom, and there are no lone pairs on the atom itself. We will learn how to draw the BH3 Lewis structure in this lesson.Only six electrons orbit the boron atom, according the Lewis structure of BF3. As a result, the boron atom's octal is incomplete. Borane BF3 is therefore regarded as a Lewis acid.A highly poisonous gaseous chemical is phosphorus. Three sigma bonds and one lone pair are present around the phosphorus atom in the Lewis structure of phosphine (PH3). Hydrogen and phosphorous atoms have no charges. The trigonal pyramidal shape of PH3 is a fundamental shape.
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zones in lakes and oceans are delineated by depth, distance from shore, or light penetration. different zones have different physical and chemical characteristics, such as temperature and salinity. label the diagrams of zonation in lakes and oceans. drag the labels to their correct targets on the diagrams below.
The diagrams of zonation in lakes and oceans with correct targets are :
a) fresh plant may be rooted here
b) littoral zone
c) limnetic zone
d) alternately dry and submerged
e) extends to edge of the continental shelf
f) generally low nutrient levels
g) light penetrates here
h) no photosynthesis occurs here
i) benthic zone
There are zones which divided the lakes. they are separated by the water from the top to bottom and the side to the side. that is named as : the littoral zone , the limnetic zone, the profundal zone , the benthic zone and the euphotic zone . the waters temperature will be affected the density. there are several zones in the ponds and the lakes. these are dived by the water column.
The limnology divides the lakes in the three types of zones ., that is the limnetic zone, the littoral zone and the benthic zone.
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