The energy change (ΔE) for the decomposition of hydrogen peroxide is -628 kJ/mol, indicating an exothermic reaction.
To calculate the energy change (ΔE) for the decomposition of hydrogen peroxide, we need to determine the total bond energies of the reactants and products and then calculate the difference between the two.
Reactants:
2 [tex]H_{2} O_{2}[/tex](g)
Products:
2 [tex]H_{2} O[/tex](g)
[tex]O_{2}[/tex](g)
First, let's calculate the bond energy for the reactants:
H-H bonds:
2 H-H bonds in[tex]H_{2} O_{2}[/tex] = 2 * 436 kJ/mol = 872 kJ/mol
O-H bonds:
4 O-H bonds in[tex]H_{2} O_{2}[/tex] = 4 * 463 kJ/mol = 1852 kJ/mol
O-O bonds:
1 O-O bond in [tex]H_{2} O_{2}[/tex] = 1 * 146 kJ/mol = 146 kJ/mol
Total bond energy of the reactants:
872 kJ/mol (H-H) + 1852 kJ/mol (O-H) + 146 kJ/mol (O-O) = 2870 kJ/mol
Next, let's calculate the bond energy for the products:
H-H bonds:
4 H-H bonds in [tex]H_{2} O[/tex]= 4 * 436 kJ/mol = 1744 kJ/mol
O=O bonds:
1 O=O bond in [tex]O_{2}[/tex]= 1 * 498 kJ/mol = 498 kJ/mol
Total bond energy of the products:
1744 kJ/mol (H-H) + 498 kJ/mol (O=O) = 2242 kJ/mol
Finally, calculate the energy change (ΔE) by taking the difference between the total bond energy of the products and the total bond energy of the reactants:
ΔE = Total bond energy of products - Total bond energy of reactants
= 2242 kJ/mol - 2870 kJ/mol
= -628 kJ/mol
Therefore, the energy change (ΔE) for the decomposition of hydrogen peroxide is -628 kJ/mol. The negative sign indicates that the reaction is exothermic, meaning it releases energy.
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The force of attraction that holds two atoms together within a molecule is
A
thermonuclear bonding
B
chemical bonding
crystal bonding
D
an atomic number
How are the environment of a swamp in a rain forest similar
50 POINTS PLEASE ANSWER CORRECTLY
Two balls collide on a pool table. Before the collision, ball 1 is traveling with a speed of 4 m/s, and ball 2 is at rest. After the collision both balls are in motion.
What has happened in this collision?
A. There was no change in ball 2's velocity, therefore momentum was not conserved.
B. Ball 2's velocity decreased, and it gained some of ball 1's momentum.
C. Ball 1's velocity decreased, and it gained momentum.
D. Ball 2's velocity increased, and it gained some of ball 1's momentum.
Answer:
D
Explanation:
With the collision, obviously Ball 2's velocity increased, while Ball 1 slowed down a little bit due to the impact. This decrease in velocity caused a decrease in Ball 1's Momentum. Satisfying both conditions, option D is right.
Why is sublimation an effective technique for the isolation of pure caffeine?
Select one or more:
O The impurities of caffeine synthesis are expected to decompose at high temperatures.
O The purified caffeine will vaporize and can be collected as a solid, leaving impurities behind.
O When heated, caffeine will sublime at a lower temperature than it melts.
O The synthesis of caffeine tends not to produce impurities, so a more thorough purification technique is not needed.
Sublimation is an effective technique for the isolation of pure caffeine because: The purified caffeine will vaporize and can be collected as a solid, leaving impurities behind.
So, the answer is B.
Sublimation is a chemical technique that is used to isolate the pure form of a substance from impure or mixed form. It is the phase transition of a solid directly to a gas without passing through a liquid phase.
It is an effective technique for the isolation of pure caffeine due to the following reasons:During sublimation, the purified caffeine will vaporize and can be collected as a solid, leaving impurities behind.
The caffeine is heated, and the heat causes the solid to vaporize directly from the solid phase to the gas phase, skipping the liquid phase. Thus, it separates the caffeine from impurities and provides pure caffeine. Therefore, option B is correct.
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Balance the following redox equations by the half-reaction method: a. Mn^2+ + H_2O_2 rightarrow MnO_2 + H_2O (in basic solution) b. Bi(OH)_3 + SnO_2^2- rightarrow SnO_3^2- + Bi (in basic solution) c. Cr_2O_7^2- + C_2O_4^2- rightarrow Cr^3+ + CO_2 (in acidic solution) d. ClO_3^-+ Cl^- rightarrow Cl_2 + Cl_O_2 (in acidic solution) e. Mn^2+ + BiO_3^- rightarrow Bi^3+ + MnO_4^- (in acidic solution)
The balanced equation is:
(a) 2Mn^2+ + 2H2O2 + 4OH^- → 2MnO_2 + 4H2O.
(b)3SnO2^2- + 6OH^- + 2Bi(OH)3 → 3SnO3^2- + 2Bi + 9H2O. (c)14Cr2O7^2- + 7C2O4^2- + 22H2O → 4Cr^3+ + 14CO2 + 28H+ + 28e^-. (d)2ClO3^- + 16H^+ + 3Cl^- → 3Cl2 + 8H2O
(e)10BiO3^- + 60H^+ + 12Mn^2+ → 10Bi^3+ + 30H2O + 12MnO4^-
a. In the balanced redox equation Mn^2+ + H_2O_2 → MnO_2 + H_2O (in basic solution), the half-reactions are:
Reduction: Mn^2+ → MnO_2
Oxidation: H_2O_2 → H_2O
To balance the reduction half-reaction, we need to add four OH^- ions to the left side: Mn^2+ + 4OH^- → MnO_2 + 2H2O + 2e^-
To balance the oxidation half-reaction, we add four OH^- ions to the right side and water molecules to balance the oxygen atoms: H2O2 + 4OH^- → 2H2O + 2e^-
Now, multiply the reduction half-reaction by two and the oxidation half-reaction by one to equalize the electrons:
2(Mn^2+ + 4OH^- → MnO_2 + 2H2O + 2e^-)
H2O2 + 4OH^- → 2H2O + 2e^-
Finally, add the two half-reactions together and cancel out the common species: 2Mn^2+ + 2H2O2 + 4OH^- → 2MnO_2 + 4H2O
b. The balanced redox equation Bi(OH)3 + SnO2^2- → SnO3^2- + Bi (in basic solution) can be balanced by following these steps:
Reduction: SnO2^2- → SnO3^2-
Oxidation: Bi(OH)3 → Bi
To balance the reduction half-reaction, add two OH^- ions to the left side: SnO2^2- + 2OH^- → SnO3^2- + H2O + 2e^-
To balance the oxidation half-reaction, add six OH^- ions to the right side: Bi(OH)3 → Bi + 3H2O + 3e^-
Multiply the reduction half-reaction by three and the oxidation half-reaction by two to equalize the electrons:
3(SnO2^2- + 2OH^- → SnO3^2- + H2O + 2e^-)
2(Bi(OH)3 → Bi + 3H2O + 3e^-)
Combine the two half-reactions and cancel out the common species:
3SnO2^2- + 6OH^- + 2Bi(OH)3 → 3SnO3^2- + 2Bi + 9H2O
c. In the acidic solution, the balanced redox equation Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 can be balanced as follows:
Reduction: Cr2O7^2- → Cr^3+
Oxidation: C2O4^2- → CO2
To balance the reduction half-reaction, add seven H2O molecules to the right side: Cr2O7^2- → 2Cr^3+ + 7H2O + 14e^-
To balance the oxidation half-reaction, add eight H^+ ions to the left side:
C2O4^2- + 2H2O → 2CO2 + 4H+ + 4e^-
Multiply the reduction half-reaction by two and the oxidation half-reaction by seven to equalize the electrons:
2(Cr2O7^2- → 2Cr^3+ + 7H2O + 14e^-)
7(C2O4^2- + 2H2O → 2CO2 + 4H+ + 4e^-)
Combine the two half-reactions and cancel out the common species:
14Cr2O7^2- + 7C2O4^2- + 22H2O → 4Cr^3+ + 14CO2 + 28H+ + 28e^-
d. In the acidic solution, the balanced redox equation ClO3^- + Cl^- → Cl2 + ClO2 can be balanced as follows:
Reduction: ClO3^- → ClO2
Oxidation: Cl^- → Cl2
To balance the reduction half-reaction, add two H^+ ions to the right side:
ClO3^- + 2H^+ → ClO2 + H2O + 2e^-
To balance the oxidation half-reaction, add two H2O molecules to the left side:
2Cl^- → Cl2 + 2e^-
Multiply the reduction half-reaction by two and the oxidation half-reaction by one to equalize the electrons:
2(ClO3^- + 2H^+ → ClO2 + H2O + 2e^-)
Cl^- → Cl2 + 2e^-
Combine the two half-reactions and cancel out the common species:
2ClO3^- + 16H^+ + 3Cl^- → 3Cl2 + 8H2O
e. In the acidic solution, the balanced redox equation Mn^2+ + BiO3^- → Bi^3+ + MnO4^- can be balanced as follows:
Reduction: BiO3^- → Bi^3+
Oxidation: Mn^2+ → MnO4^-
To balance the reduction half-reaction, add six H^+ ions to the left side:
BiO3^- + 6H^+ → Bi^3+ + 3H2O + 6e^-
To balance the oxidation half-reaction, add eight H^+ ions to the right side:
Mn^2+ → MnO4^- + 8H^+ + 5e^-
Multiply the reduction half-reaction by five and the oxidation half-reaction by two to equalize the electrons:
5(BiO3^- + 6H^+ → Bi^3+ + 3H2O + 6e^-)
2(Mn^2+ → MnO4^- + 8H^+ + 5e^-)
Combine the two half-reactions and cancel out the common species:
10BiO3^- + 60H^+ + 12Mn^2+ → 10Bi^3+ + 30H2O + 12MnO4^-
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(06.05 LC)
Why is it important to maintain records of the procedures of an experiment?
points)
Multiple Choice
(b) identify both of the brønsted-lowry conjugate acid-base pairs in the neutralization reaction above. for each pair, label the acid and the base.
In the neutralization reaction, the Brønsted-Lowry conjugate acid-base pairs are the acid HCl and the base OH- as well as the acid H₂O and the base Cl-.
Which species act as acid and base?The neutralization reaction involves the formation of two Brønsted-Lowry conjugate acid-base pairs. The acid-base pairs are HCl (acid) and OH⁻ (base), as well as H₂O (acid) and Cl⁻ (base). In this reaction, the acid donates a proton (H+) to the base, resulting in the formation of water and a salt. HCl donates a proton to OH- to form water, while H₂O donates a proton to Cl⁻ to form hydrochloric acid (HCl). The transfer of protons between the acid and base creates the conjugate acid-base pairs.
Brønsted-Lowry acid-base theory describes the transfer of protons (H+) between acids and bases. Acids are proton donors, while bases are proton acceptors. In a neutralization reaction, an acid reacts with a base to form a salt and water.
The acid donates a proton to the base, forming a conjugate base, while the base accepts the proton, forming a conjugate acid. In the given neutralization reaction, HCl is the acid and OH- is the base, forming water. Simultaneously, H₂O acts as an acid, donating a proton to Cl⁻, which acts as a base, forming HCl. Understanding conjugate acid-base pairs is essential in comprehending acid-base reactions and their equilibrium.
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Which type of fuel do you predict might have the
greatest impact on carbon dioxi de levels in the
atmosphere?
Please answer this. If your just gonna give me a fake answer please just don’t
Answer:
If the choices were coal, petroleum, and natural gas, I would say that petroleum has the greatest impact on the CO₂ levels in the atmosphere. I chose petroleum because of transportation, such as cars, trucks, planes, and more.
can we observe a lunar eclipse during a new moon phase? explain your answer.
please help me thx
subject is science
Answer: There is no eclipse. However, two or four times a year, the Moon travels through some portion of the Earth's penumbral or umbral shadows, resulting in one of the three types of eclipses mentioned above. When the Moon crosses between the Earth and the Sun, this occurs. This is only possible when the Moon is in its New Moon phase.
Explanation:
HOPE THIS HELPS!!! : )
what is the energy that can transform from one thing to another
Answer:
Energía geotérmica (calor → energía eléctrica) Motores térmico, como el motor de combustión interna utilizado en automóviles o el motor de vapor (calor → energía mecánica) Energía térmica oceánica (calor → energía eléctrica) Represas hidroeléctricas (energía potencial gravitacional → energía eléctrica)
Explanation:
Answer:
One type of energy can change into another type of energy. Energy transformation means the changing of energy from one type to another, e.g. from kinetic energy to electrical energy, or from potential energy to kinetic energy.
Explanation:
1. How many moles are in 24.6 grams of NaCl?
If someone could give an explanation along with an answer that would be appreciated :)
how many electrons are in the cyclobutadiene molecule shown below?
There are 20 electrons in the cyclobutadiene molecule shown in the image.
Cyclobutadiene is a molecule that contains 4 carbon atoms and 4 hydrogen atoms. The chemical formula for cyclobutadiene is C4H4. It is a highly reactive molecule and is unstable in its pure form. It has a square planar shape, and each carbon atom is bonded to two neighboring carbon atoms and one hydrogen atom.
Each carbon atom in cyclobutadiene has four valence electrons, and each hydrogen atom has one valence electron. Valence electrons are the outermost electrons in an atom and are involved in chemical bonding. The number of valence electrons in an atom is determined by the group number in the periodic table.
Carbon is in group 14, so it has 4 valence electrons. Hydrogen is in group 1, so it has 1 valence electron. The total number of valence electrons in cyclobutadiene is calculated by adding the valence electrons of all the atoms.
4 carbon atoms x 4 valence electrons per carbon atom = 16 valence electrons4 hydrogen atoms x 1 valence electron per hydrogen atom = 4 valence electrons Total number of valence electrons = 16 + 4 = 20Therefore, there are 20 electrons in the cyclobutadiene molecule shown in the image.
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Name the compound and determine the number of moles present. SHOW YOUR WORK.
60.0g Na2SO4
Answer:
60.0g Na2SO4
Na2SO4 molar mass 142.04g
divide 60.0 by 142.02
And you get the answer .422moles Na2SO4
let me know if you need more explanation
1+1
[tex]1 + 1[/tex]
11
1. Acetylsalicylic acid is also known as aspirin- C9H8O4 – see figure. a. Is acetylsalicylic acid a STRONG or WEAK acid? Circle the acidic proton. b. What is the pH of a 0.20 M solution of acetylsalicylic acid, given that the K, at 25 °C is 3.0 x 10-4? Justify any assumptions you make. c. % ionization is defined as the ratio of the ionized acid concentration to the initial acid concentration x 100%. Determine the % ionization for the solution of acetylsalicylic acid in b. above
a. Acetylsalicylic acid is a weak acid because it dissociates only partially in water.
b. The expression for Ka for acetylsalicylic acid is
Ka = [H+][C9H7O4O-]/[C9H8O4].
The ionization of acetylsalicylic acid is,
C9H8O4 + H2O ⇌ H3O+ + C9H7O4O-
Initial 0.20 M 0 0
Change - x + x + x
Equilibrium 0.20 - x x x
The value of Ka for acetylsalicylic acid is 3.0 × 10-4.
Ka = [H+][C9H7O4O-]/[C9H8O4]= 3.0 × 10-4
[H+]^2/[C9H8O4] = 3.0 × 10^-4 × [0.20 - x]
The ionization constant expression is solved for the value of x using the quadratic equation,
0.60 x 10-4 = x^2
x = 0.0077 M
[H+] = 7.8 × 10-4
The percentage of ionization is,
% ionization = (x/[C9H8O4]) × 100
= (0.0077/[0.20]) × 100
= 3.85%
Therefore, the % ionization for the solution of acetylsalicylic acid is 3.85%.
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What is the molarity of a solution that contains 0.202 mol KCl in 7.98 L solution?
Explanation:
molarity = no. of moles of solute/solution in litres
molarity =0.202/7.98
=0.025 M
These diagrams show two atoms of fluorine and an atom of magnesium.
On the left, a purple circle labeled F is shown twice, both times surrounded by 2 concentric circles. The inner circle has 2 small green spheres. The outer circle has 7 small green spheres. On the right, a purple circle labeled M g surrounded by 3 concentric circles. The inner circle has 2 small green spheres. The middle circle has 8 small green spheres. The outer circle has 1 small green sphere.
Which shows the correct steps in the formation of an ionic bond between these atoms?
A magnesium atom accepts six electrons from the fluorine atoms → Each fluorine atom donates three of the electrons → The magnesium atom becomes a -2 ion → Each fluorine atom becomes a +1 ion
A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a -2 ion → Each fluorine atom becomes a +1 ion
A magnesium atom accepts two electrons from the fluorine atoms → Each fluorine atom donates one of the electrons → The magnesium atom becomes a -2 ion → Each fluorine atom becomes a +1 ion
A magnesium atom donates two electrons to the fluorine atoms → Each fluorine atom accepts one of the electrons → The magnesium atom becomes a +2 ion → Each fluorine atom becomes a -1 ion
Answer: D
Explanation:
took test
Answer:
D
Explanation:
got it right
Computer says I put 2 things wrong. Where I made a mistake?
Using the Lewis concept of acids and bases, identify the Lewis acid and base in each of the following reactions:
Fe(NO3)3(s)+6H2O(l)?Fe(H2O)63+(aq)+3NO3?(aq)
NH3(g)+HCl(g)?NH4Cl(s)
I put them like that:
Lewis acid: Fe(NO3)3
Lewis base: H2O
Neither: HCl, NH3
Lewis acid: Fe(NO3)3Lewis base: H2OThe Lewis acid is H+ and the Lewis base is NH3.
The Lewis concept of acids and bases, Lewis acids are those that have an incomplete valence shell and accept electrons from a Lewis base. A Lewis base has at least one electron pair available to form a covalent bond with a Lewis acid, filling its valence shell. The Lewis acid and base in each of the following reactions are given below:Fe(NO3)3(s) + 6H2O(l) ⟶ Fe(H2O)63+(aq) + 3NO3?(aq)The Lewis acid is Fe3+ and the Lewis base is H2O.NH3(g) + HCl(g) ⟶ NH4Cl(s)The Lewis acid is H+ and the Lewis base is NH3.Neither HCl nor NH3 is a Lewis acid or base in this particular reaction. Therefore, you made a mistake in your answer. The correct answers are as follows:Lewis acid: Fe(NO3)3Lewis base: H2OThe Lewis acid is H+ and the Lewis base is NH3.
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Which of the following compounds will be more soluble in pentane (C5H12)? Why?
A) pentanol (CH3CH2CH2CH2CH2OH)
B) benzene (C6H6)
C) acetic acid (CH3CO2H)
D) ethyl methyl ketone (CH3CH2COCH3)
Option B is correct. benzene (C6H6). Pentane is an aliphatic hydrocarbon with only nonpolar C-C and C-H bonds, making it a nonpolar solvent. Its chemical formula is C₅H₁₂.
What is BenzeneBenzene (C₆H₆) is also a nonpolar compound, consisting of a hexagonal ring of carbon atoms with hydrogen atoms attached. The carbon-carbon (C-C) bonds and carbon-hydrogen (C-H) bonds in benzene are nonpolar, resulting in a nonpolar molecule.
On the other hand, pentanol (C₅H₁₁OH), acetic acid (C₂H₄O₂), and ethyl methyl ketone (C₄H₈O) contain polar functional groups such as hydroxyl (-OH), carboxyl (-COOH), and carbonyl (-CO-) groups, respectively. These polar functional groups create polarity within the molecules.
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With a(n) _____, the results are analyzed as if you had separate experiments at each level of the other independent variable
With a simple main effect, the results are analyzed as if you had separate experiments at each level of the other independent variable.
What is simple main effect?In the realm of factorial ANOVA, an enlightening endeavor known as a simple main effect analysis arises. Within this statistical examination, the intricate interplay of two or more independent variables upon a dependent variable is meticulously unraveled.
When a notable interplay between these independent variables materializes, the pursuit of comprehension beckons the astute pursuit of simple main effects analyses, illuminating the essence of the interaction.
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You are given two flasks of equal volume. One contains H2 at 0°C and 1 atm while the other contains CO2 at and 2 atm. Which of the following quantities will be the same for both flasks?
a. average molecular speed
b. density
c. average molecular kinetic energy
d. number of molecules present
The number of molecules present is the only quantity that will be the same for both flasks. The quantity that will be the same for both flasks is:
d. number of molecules present
The number of molecules present in each flask is determined by the ideal gas equation, which states that the number of molecules (n) is proportional to the product of the pressure (P), volume (V), and the inverse of the temperature (T):
PV = nRT
Since the two flasks have the same volume, the number of molecules present will be the same if the temperature and pressure are also the same.
However, for the other quantities listed:
a. average molecular speed: The average molecular speed of a gas is directly proportional to the square root of the temperature. Since the temperatures in the two flasks are different (0°C and the other temperature is not specified), the average molecular speeds will be different.
b. density: The density of a gas is determined by its molar mass and pressure. Since [tex]H_2[/tex] and [tex]CO_2[/tex] have different molar masses, their densities will be different even if the pressures are the same.
c. average molecular kinetic energy: The average molecular kinetic energy of a gas is directly proportional to the temperature. Since the temperatures in the two flasks are different, the average molecular kinetic energies will be different.
Therefore, the number of molecules present is the only quantity that will be the same for both flasks.
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You have a hydrate of compound X that is 60.26g. You heat the hydrate, allow it to cool, and then remass it. The new mass is 23.0g. The molar mass of compound X as an anhydrate is 100 g/mol.
Answer:
100gmol
Explanation:
think back to our hypotheses of chapter 12 regarding melting points (that some force kept the material together). given those hypotheses, what would you predict about the forces that hold atoms of cesium together in the solid metal compared to the forces that hold atoms of lithium together? explain why.
Based on the hypotheses discussed in Chapter 12 regarding melting points and the forces that hold materials together, we can make a prediction about the forces holding atoms of cesium and lithium together in their solid metal forms.
One hypothesis suggests that stronger forces between atoms result in higher melting points. Another hypothesis proposes that metals are held together by metallic bonds, where positively charged metal ions are surrounded by a sea of delocalized electrons.
Considering these hypotheses, we can infer that cesium atoms would be held together by stronger forces compared to lithium atoms in their solid metal forms. This is because cesium is located further down the periodic table, belonging to Group 1 (alkali metals), whereas lithium is in Group 2 (alkaline earth metals). As we move down a group in the periodic table, the atomic radius generally increases, leading to weaker forces of attraction between atoms.
Therefore, the larger atomic size of cesium compared to lithium would result in weaker interatomic forces, making cesium's solid metal form have a lower melting point compared to lithium.
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Fossil Fuels are an ________ resource that people are hoping to replace with ______ resources.
Answer:
fossil fuels are an nonrenewable resource that people are hoping to replace with renewable resources
Explanation:
How can wind produce erosion?
Wind can cause plants to split a rock apart.
Wind can pick up rocks and then drop it into smaller pieces.
Wind blowing sand against a rock can reduce the rock's size.
Wind can roll rock near moving water.
Answer:
It would be C. Wind blowing against a rock can reduce the rock´s size.
Explanation:
A 148.20 sample of an unknown substance was heated from 25°C to 40°C. In the process, the substance absorbed 5683 J of energy. What is the
specific heat of the substance.
Answer: 2556.455 J/(kg.°C)
Explanation:
Im assuming the 148.20 is in grams
The equation is Q=mcΔT
Q= Heat (J)
m= Mass
c= SHC (specific heat capacity)
ΔT= Temp change
If there is a band in the W2 lane of the Western result, what could you conclude about the physical protein structure of rGFP present in this band? If so what would the MW be?
The MW of rGFP in the band can be determined by comparing the mobility of the band with that of protein standards run on the same gel.
The Western blotting technique is utilized to identify and detect specific proteins in a sample of tissue or extract. If there is a band in the W2 lane of the Western blot, one can conclude that rGFP is present in that band. The physical protein structure of rGFP could not be inferred from the presence of a band in the W2 lane of the Western blot. This requires additional analysis such as X-ray crystallography, nuclear magnetic resonance, or cryo-electron microscopy to analyze protein structure. MW is the molecular weight which can be determined using a molecular weight marker that runs in a parallel lane to the protein extract on the gel. In the Western blotting method, SDS-PAGE is typically used to separate proteins based on their size. The SDS-PAGE gel is calibrated with protein markers that have a known molecular weight. Hence, the MW of rGFP in the band can be determined by comparing the mobility of the band with that of protein standards run on the same gel.
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(GIVING BRAINLIEST) Which is NOT an example of a good source of complex carbohydrates?
peas
beans
whole-wheat bread
white bread
Answer: White bread
Explanation:
Write a short procedure (include type of glassware, calculations, etc.) for making 25.00 mL of a 0.0250 M NaF solution from 5. a. Solid NaF b. 0.100 M NaF solution
To prepare a 25.00 mL solution of 0.0250 M NaF, you can follow either of the following procedures:
Short procedure (include type of glassware, calculations, etc.) for making 25.00 mL of a 0.0250 M NaF solutiona. Using Solid NaF:
Weigh the appropriate amount of solid NaF (calculated based on desired concentration) using an analytical balance.Transfer the weighed NaF into a clean and dry 25.00 mL volumetric flask.Add distilled water to the flask up to the mark on the neck and swirl gently to dissolve the solid completely.Mix the solution thoroughly by inverting the flask several times.Adjust the solution level to the mark using a dropper or pipette if needed.Cap the flask and label it with the concentration and date.b. Using 0.100 M NaF Solution:
Calculate the volume of 0.100 M NaF solution needed to prepare 25.00 mL of a 0.0250 M solution using the dilution formula.Measure the calculated volume (6.25 mL) of the 0.100 M NaF solution using a graduated cylinder, pipette, or burette.Transfer the measured solution into a clean and dry 25.00 mL volumetric flask.Add distilled water to the flask up to the mark on the neck and swirl gently to ensure proper mixing.Follow steps 5-9 from the procedure using solid NaF to complete the preparation.Both procedures involve careful measurement and mixing of the NaF compound with distilled water in a volumetric flask. The resulting solutions will have a concentration of 0.0250 M NaF.
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Please help with ALL!
Answer:
A. Particle emitted = alpha particle; Protons lost = 2; Change in mass = -4
B. Particle emitted = alpha particle; Protons lost = 2; Change in mass = -4
C. Particle emitted = beta particles; protons gained = 1; Change in mass = 0
D. Particle emitted = alpha particle; Protons lost = 2; Change in mass = -4
E. Particle emitted = gamma ray; protons gained = 0; Change in mass = 0
F. Particle emitted = alpha particle; Protons lost = 2; Change in mass = -4
Explanation:
Radioactivity is the spontaneous disintegration of nucleus of an atom resulting in emission of particles and radiation unyil a stable atomic nuclei is obtained. Substances that are exhibit radioactivity are known as radioactive substances and are usually isotopes of stable atoms.
The types of radiation emitted by a radioactive substances include alpha rays, beta rays, and gamma rays.
Alpha rays or particles are fast moving positively charged particles with a mass number of four and atomic number of two. Thus, each alpha particle is a helium nucleus.
Beta rays or particles are vary fast moving stream of electrons having a mass number of zero and a charge of -1.
Gamma rays are not particles but are electromagnetic waves similar to visible light and X-rays but with shorter wavelengths. They have no mass or charge.
An atom emitting alpha particles has its mass reducing by 4 units and its atomic number decreasing by 2.
An atom emitting beta particles as it decays has no change in mass but its atomic number increases by 1.
From the table:
A. Particle emitted = alpha particle; Protons lost = 2; Change in mass = -4
B. Particle emitted = alpha particle; Protons lost = 2; Change in mass = -4
C. Particle emitted = beta particles; protons gained = 1; Change in mass = 0
D. Particle emitted = alpha particle; Protons lost = 2; Change in mass = -4
E. Particle emitted = gamma ray; protons gained = 0; Change in mass = 0
F. Particle emitted = alpha particle; Protons lost = 2; Change in mass = -4