Answer:
B. they are unsaturated
Explanation:
Alkanes are long chain hydrocarbons with only single bonds, alkenes are hydrocarbons with at least one double bond and alkynes are hydrocarbons with at least one triple bond.
Alkanes, alkenes and alkynes are all hydrocarbons. Therefore statement D. is incorrect
Hydrocarbons are compounds containing carbon and hydrogen only
Neither of the three are aromatic compounds therefore statement A is incorrect
Saturated hydrocarbons are where all the bonds between atoms are single bonds. Unsaturated hydrocarbonds are when at least there's one double or triple bond.
Alkanes are saturated and Alkynes and alkenes are unsaturated.
therefore statement B is correct where alkenes and alkynes are unsaturated but alkanes are not
Final answer:
Alkenes and alkynes are unsaturated hydrocarbons due to the presence of carbon-carbon double and triple bonds, respectively, differentiating them from saturated alkanes.
Explanation:
The correct answer to the question is B: Alkenes and alkynes are unsaturated hydrocarbons. This is because they contain double or triple carbon-carbon bonds respectively, which means they have fewer hydrogen atoms attached to the carbon backbone compared to alkanes, which contain only single bonds and are therefore saturated. The term saturated indicates that a molecule contains the maximum possible number of hydrogen atoms whereas unsaturated indicates the presence of double or triple bonds, which replace some hydrogen atoms.
Aromatic compounds, like benzene, are indeed hydrocarbons but they are classified by their distinct ring structure with delocalized electrons, which is not a characteristic of alkenes or alkynes. Alkenes and alkynes are not aromatic simply because they have unsaturated bonds. Alkanes are also hydrocarbons; however, they are saturated, which is a term not applicable to alkenes or alkynes.
We could avoid a large increase in temperature if greenhouse emissions peaked by the year of ?
To curb the increase in Earth's temperature due to greenhouse gases, we need to reach 'net zero' carbon emissions by 2050 or sooner. The problem arises largely from burning fossil fuels and deforestation, leading to increased CO2 in the atmosphere. Continued trajectory could double Earth's CO2 levels from pre-industrial times by the end of the century.
Explanation:To avoid a large increase in temperature due to greenhouse gas emissions, it is crucial that these emissions peak as soon as possible. According to the Union of Concerned Scientists, we should aim for 'net zero' carbon emissions by 2050 or even sooner. This would mean that no more carbon enters the atmosphere than is removed.
Greenhouse gases, primarily carbon dioxide, are largely a result of burning fossil fuels, such as coal and oil. Alongside this, the destruction of tropical forests which absorb CO2 from the atmosphere exacerbates the problem.
If this trajectory continues, Earth's CO2 levels are predicted to double from pre-industrial levels by the end of this century, leading to potentially catastrophic climactic changes. Hence, mitigation strategies to stabilize carbon emissions constitute a major part of our global strategy towards climate change.
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Fernanda looked at a sample of paint under a microscope. At right is the sketch she made of what she saw. Label A points to the solid particles of pigment that give the paint its color. The solid particles are the . Label B represents the liquid surrounding the solid particles. This liquid is called the for this mixture. This paint does not settle over time, so it is an example of a .
Answer:dispersed state
dispersed medium
colloid
Answer:
Label A points to the solid particles of pigment that give the paint its color. The solid particles are the (dispersed state.)
Label B represents the liquid surrounding the solid particles. This liquid is called the (Dispersed Medium)
This paint does not settle over time, so it is an example of a (Colloid)
Explanation:
Fernanda looked at a sample of paint under a microscope. At right is the sketch she made of what she saw.
Identify whether each process is either oxidation or reduction:
Al → Al3+ + 3 e-
S2- → S + 2 e-
Cu → Cu2+ + 2 e-
Show ALL work.
Answer:
Al → Al³⁺ + 3 e⁻, is an oxidation process.
S²⁻ → S + 2 e⁻, is an oxidation process.
Cu → Cu²⁺ + 2 e⁻, is an oxidation process.
Explanation:
The oxidation-reduction reaction contains a reductant and an oxidant (oxidizing agent).The oxidation process is the process in which electrons are lost and produce positively charged ions.The reduction process is the process in which electrons is gained and negatively charge ions are produced.So,
Al → Al³⁺ + 3 e⁻, is an oxidation process.Al loses 3 electrons and produce Al³⁺c
S²⁻ → S + 2 e⁻, is an oxidation process.S²⁻ loses 2 electrons and produce S.
Cu → Cu²⁺ + 2 e⁻, is an oxidation process.Cu loses 2 electrons and produce Cu²⁺.
All the given processes (Al becoming Al₃+, S₂- becoming S, and Cu becoming Cu₂+) are examples of oxidation, as each involves the loss of electrons.
Explanation:To determine whether each process is oxidation or reduction, we will look at the change in electron configuration for each reaction.
Al → Al₃+ + 3 e-: Here, Aluminum (Al) is losing three electrons to form Al₃+. This is an oxidation process because oxidation is characterized by the loss of electrons.S₂- → S + 2 e-: In this process, the sulfide ion (S₂-) is losing two electrons to become elemental sulfur (S). This is also an oxidation process, as it involves the loss of electrons.Cu → Cu₂+ + 2 e-: Copper (Cu) loses two electrons to form Cu₂+. Similar to the previous examples, this is an oxidation process because it involves the electron loss.Thus, all the given processes are examples of oxidation, where each element or ion loses electrons.
Explain what is meant by sp3 hybridization
Final answer:
The process of sp³ hybridization involves the mixing of an s orbital with three p orbitals to form four sp³ hybrid orbitals. This type of hybridization occurs in molecules like methane. On the other hand, sp² hybridization involves the mixing of an s orbital with two p orbitals to form three sp² hybrid orbitals, which occurs in molecules like boron trifluoride and ethene. Lastly, sp hybridization is the mixing of an s orbital with a single p orbital to form two sp hybrid orbitals, which occurs in molecules like beryllium hydride.
Explanation:
sp³ hybridization is a process where an s orbital and three p orbitals mix to form a set of four sp³ hybrid orbitals. These hybrid orbitals are arranged in a tetrahedral geometry, with each lobe of the hybrid orbitals pointing towards one of the corners of the tetrahedron. This type of hybridization occurs in molecules such as methane (CH₄).
sp² hybridization is the mixing of an s orbital and two p orbitals to form a set of three sp² hybrid orbitals. These hybrid orbitals are arranged in a trigonal planar geometry, with each lobe of the hybrid orbital pointing towards one corner of the triangle. This type of hybridization occurs in molecules such as boron trifluoride (BF₃) and ethene (C₂H₄).
sp hybridization is the mixing of an s orbital with a single p orbital to form a set of two sp hybrid orbitals. These hybrid orbitals are oriented linearly, with each lobe of the hybrid orbital pointing in opposite directions. This type of hybridization occurs in molecules such as beryllium hydride (BeH₂).
In the case of ethene (C₂H₄), each carbon atom undergoes sp² hybridization to form three sp² hybrid orbitals. One of these hybrid orbitals forms a bond with the identical hybrid orbital on the other carbon atom, resulting in the formation of a double bond. The remaining two hybrid orbitals form bonds with the 1s orbitals of two hydrogen atoms. The unhybridized 2pz orbitals on each carbon atom form another bond by overlapping sideways with each other.
The concept of orbital hybridization allows atomic orbitals to combine and form hybrid orbitals that have different energy and orientation compared to the constituent orbitals. In the case of carbon, different combinations of s and p orbitals can produce four sp³, three sp², or two sp hybrid orbitals.
An sp³ hybrid orbital can also hold a lone pair of electrons. For example, in ammonia (NH₃), the nitrogen atom is surrounded by three bonding pairs and a lone pair of electrons. The nitrogen atom undergoes sp³ hybridization, with one hybrid orbital occupied by the lone pair.
Calculate ℰ° values for the galvanic cells described below. (a) cr3+(aq) + cl2(g) equilibrium reaction arrow cr2o72-(aq) + cl -(aq) v (b) io3-(aq) + fe2+(aq) equilibrium reaction arrow fe3+(aq) + i2(aq)
Answer:
[tex]\boxed{\text{(a) 0.00 V; (b) 0.424 V}}[/tex]
Explanation:
We must look up the standard reduction potentials for the half-reactions.
ℰ°
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771
(a) Cr³⁺/Cl₂
We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.
ℰ°/V
2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27
2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00
(b) Fe²⁺/IO₃⁻
ℰ°/V
Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424
The ℰ° values for the cells are [tex]\boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}[/tex]
Answer:
Answer:
\boxed{\text{(a) 0.00 V; (b) 0.424 V}}
Explanation:
We must look up the standard reduction potentials for the half-reactions.
ℰ°
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O 1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.35827
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
Fe³⁺ + e⁻ ⇌ Fe²⁺ 0.771
(a) Cr³⁺/Cl₂
We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.
ℰ°/V
2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻ -1.36
Cl₂ + 2e⁻ ⇌ 2Cl⁻ 1.358 27
2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻ + 6Cl⁻ + 14H⁺ 0.00
(b) Fe²⁺/IO₃⁻
ℰ°/V
Fe²⁺ ⇌ Fe³⁺ + e⁻ -0.771
2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O 1.195
10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O 0.424
The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}
Explanation:
its right trust
In the reaction of sodium with bromine, explain which atom is reduced.
I need you to show ALL the steps because I don't understand.
Answer:
The reduced atom is Br.
Explanation:
The oxidation-reduction reaction contains a reductant and an oxidant (oxidizing agent). The oxidation process is the process in which electrons are lost and produce positively charged ions. The reduction process is the process in which electrons is gained and negatively charge ions are produced. In the reaction of chlorine with calcium:2Na + Br₂ → 2NaBr,
Na loses 1 electrons and is oxidized to Na⁺. (Na → Na⁺ + e).
Br₂ gains 2 electrons and is reduced to 2Br⁻. (Br₂ + 2e → 2Br⁻).
So, the reduced atom is Br.
4.32×10^26 atoms of francium would make how many moles of francium sulfide
Answer:
4.32e+26
Explanation:
Answer:
4.32e+26
Explanation:
Which of these is the percent of error in evaluating the molecular mass of a compound if the experimental value was 105.2 amu and the known value was 107.5 amu? f 2.1% g 4.2% h 3.3% j 1.0%
Answer:
The percent error, % error, is 2.1% (option f)Explanation:
1) Data:
a) Experimental value, m₁ = 105.2 amu
b) Known value, m = 107.5 amu
b) % error = ?
2) Formulae:
a) absolute error = | experimental value - known value|
b) % error = [ absolute value / known value ] × 100
3) Solution:
a) absolute error = | m₁ - m | = | 105.2 amu - 107.5 amu | = 2.3 amu
b) % error = [ 2.3 amu / 107.5 amu ] × 100 = 2.1% ← answer
The two priorities for the guest quarters are privacy and cleanliness.
True
False
trueee hope it helpsss
The given statement is true. Privacy and cleanliness are considered as two priorities for the guest quarters.
Explanation:Quarters are place where people stay as guest for some duration of time. People consider many priorities while booking hotels or quarters for their stay because as they are going to live in it for some span of time.
The first two priorities include privacy and cleanliness. They look out for both the external as well as internal cleanliness. External cleanliness includes surroundings, laws, etc., whereas the inner cleanliness includes right from the articles present in the room till the toilet of the room. As they are living in another individual living area they obviously demand for some sort of privacy.
when 2.85 moles of chlorine reacts with excess tin, how many moles of tin choloride are formed?
chlorine by twenty øne piløts
Write a nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90.
Answer:
235/92U+10n→144/54Xe+90/38Sr+2/10n
Explanation:
The nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90 is represented by;235/92U+10n→144/54Xe+90/38Sr+2/10n
In nuclear fission reactions a heavy nuclide is split into two light nuclides and is coupled by the release of energy.Distinguish between a solution in general and an aqueous solution
Answer:
What distinguish a solution in general from an aqueous solution is the solvent. A solution in general may contain any solvent, which may be solid, liquid or gas, while an aqueous solution is formed with water as solvent.Explanation:
A solution in general is a homogeneous mixture in which a substance, named solute, is dissolved, in other substance, name solvent.
Solutions may be in solid, liquid or gas state. There are many kind of solvents. Usually, in a lab you work with liquid solutions. Some liquid solvents are: ethanol, glycerin, hexane, benzene, and water, among many others.
Aqueous solution is a solution where the solvent is water. Of course, the solute may be any one: NaCl, sugar, ethanol, an acid, a base, a salt.
What distinguish a solution in general and an aqueous solution is the solvent.
Which of the following statements regarding glucose is FALSE?a) Glucose is the main component of starch and glycogen.b) Glucose is also called blood sugar.c) Glucose is the most common disaccharide in our diet.d) Glucose is the most important carbohydrate fuel for the human body.
Glucose is the most common disaccharide in our diet is the false statement regarding glucose.
What is the chemical formula of glucose?The chemical formula of glucose is written as C₆H₁₂O₆.
Starch and glycogen is the highly branched compound or the polymers, which can be formed from the fundamental monomer unit of glucose. Glucose is also present in the blood of as a blood sugars and used as a source of energy or we can say that fuel for the human body. Glucose is a monosaccharide, it means it will not able to break down in following sub units.
Hence, option (c) is false because glucose is a monosaccharide.
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The false statement about glucose is (C), that is Glucose is the most common disaccharide in our diet. Glucose is actually a monosaccharide. Common disaccharides include sucrose, lactose, and maltose.
To determine which statement regarding glucose is FALSE, let's analyze the given options:
a) Glucose is the main component of starch and glycogen. This is true, as both starch and glycogen are polysaccharides composed of glucose units.b) Glucose is also called blood sugar. This is true, because glucose circulates in the blood as a primary energy source.c) Glucose is the most common disaccharide in our diet. This statement is false. Glucose is a monosaccharide, not a disaccharide. Common disaccharides include sucrose, lactose, and maltose.d) Glucose is the most important carbohydrate fuel for the human body. This is true, as it is the primary energy source for cellular functions.After an atomic bomb has been dropped, the damage continues to get worse.
Why is this?
A. Because a fission reaction has no products
B. Because people get sick from exposure to high levels of radiation
C. Because atomic bombs continue going off
D. Because fusion reactions occur afterward
Answer:
C. Because people get sick from exposure to high levels of radiation
Explanation:
plants take in carbon dioxide from the air through holes in the
a. xylem
b. spongy layer
c. palisade layer
d. vascular bundle
Answer:
c. palisade layer
Which sediment type requires the least amount of energy to be eroded
Answer:
SEDIMENTARY ROCK
Explanation:
Iodine, I2, has many uses, including the production of dyes, antiseptics, photographic film, pharmaceuticals, and medicinal soaps. It forms when chlorine, Cl2, reacts with iodide ions in a sodium iodide solution. Which of the following half-reactions for this oxidation‑reduction reaction describes the oxidation, and which one describes the reduction? Cl2 + 2e‒ → 2Cl‒ 2I‒ → I2 + 2e‒
Answer:
2I ⁻ → I₂ + 2e⁻ describes the oxidation.Cl₂ + 2e⁻ → 2Cl ⁻ describes the reduction.Explanation:
Oxidation-reduction reaction is the simulaneous oxidation and reduction of the substances and is represented by two half-reactions.
The oxidation half-reaction is the loss of electrons, with the consequent increase in the oxidation state by the oxidized substance.
In this case, the process that shows the loss of electrons is:
2I⁻ → I₂ + 2e⁻That reaction shows:
Two I⁻ ions lose two electrons (one each) to be oxidized to I₂.The change in the oxidation number is from -1 to 0.Hence this half-reaction is the oxidation reaction.On the other hand, the reduction half-reaction is the gain of electrons, with the consequent reduction of the oxidation state by the reduced substance.
In this case, the process that shows the gain of electrons is:
Cl₂ + 2e⁻ → 2Cl⁻That reaction shows:
Two Cl atoms gain two electrons (one each) to be reduced to Cl⁻.The change in the oxidation number is from 0 to - 1.Hence, this half-reaction is the reduction reaction.Summary:
2I ⁻ → I₂ + 2e⁻ describes the oxidation.Cl₂ + 2e⁻ → 2Cl ⁻ describes the reduction.
To balance a chemical equation it may be necessary to adjust the
Answer:
To balance a chemical equation it may be necessary to adjust the coefficients.Explanation:
The coefficients of a chemical equation are the numbers that you put in front of each reactant and product. They are used to balance the equation and comply with the law of mass conservation.
By adjusting the coefficients you obtain the relative amounts (moles) of each product and reactant, i.e. the mole ratios.
Here an example.
The first information is what is called a word equation. E.g. nitrogen and hydrogen react to form ammonia:
Word equation: hydrogen + nitrogen → ammoniaSkeleton equation: H₂ + N₂ → NH₃This equation shows the chemical formulae but it is not balanced. The law of mass conservation is not observed.
So, in order to comply with the law of mass conservation you adjust the coefficients as follow.
Balanced chemical equation: 3H₂ + N₂ → 2NH₃As you see, it was necessary to modify the coefficients. Now the law of conservation of mass is observed and you get the mole ratios:
3 mol H₂ : 1 mol N₂ : 2 mol NH₃
The substance reduced in a hydrogen-oxygen fuel cell is _____. oxygen hydrogen water hydrogen peroxide\
Answer: Oxygen
Explanation:
Answer: The substance which gets reduced in the hydrogen-oxygen fuel cell is oxygen.
Explanation:
A reduced substance undergoes reduction reaction. This is defined as the reaction in which an atom gains electrons.
A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.
The reactions which occur in hydrogen-oxygen fuel cell are:
At cathode: [tex]H_2+2OH^-\rightarrow 2H_2O+2e^-[/tex]
At anode: [tex]\frac{1}{2}O_2+H_2O+2e^-\rightarrow 2OH^-[/tex]
As, the oxygen is gaining electrons from the above reaction. Thus, it is undergoing reduction reaction and is getting reduced.
Hence, the substance which gets reduced in the hydrogen-oxygen fuel cell is oxygen.
Rutherfordium-257 was synthesized by bombarding Cf-249 with C-12. Write a nuclear equation for this reaction.
Final answer:
The nuclear equation for the synthesis of rutherfordium-257 by bombarding californium-249 with carbon-12 is 24998Cf + 126C → 257104Rf + 410n. It shows the reactants, products, and the conservation of mass and atomic numbers in the reaction.
Explanation:
The student is asking for a nuclear equation for the synthesis of rutherfordium-257, which is created by bombarding californium-249 (Cf-249) with carbon-12 (C-12). The balanced nuclear equation takes into account the conservation of mass and atomic numbers. The californium-249, which has an atomic number of 98 and a mass number of 249, reacts with carbon-12, which has an atomic number of 6 and a mass number of 12.
The equation is:
24998Cf + 126C → 257104Rf + 410n
In this reaction, rutherfordium-257 has an atomic number of 104 and a mass number of 257. To conserve the mass number and atomic numbers on both sides of the equation, we need to add four free neutrons (n) with a mass number of 1 and an atomic number of 0 to the products side. This completes the balanced nuclear equation for the formation of rutherfordium-257.
When light shines on a sample, each element emits specific wavelengths producing a unique fingerprint called its ______ spectra. A) infrared B) line C) raman D) ultraviolet
Answer:
B.
Explanation:
Answer:
B) line is the correct answer.
Explanation:
When light shines on a sample, each element emits specific wavelengths producing a unique fingerprint called its line spectra.
The line spectrum is electromagnetic spectra consist of discrete spectra lines.
When the atoms are excited they emit ray of specific wavelengths that correspond to various colors. The emitted ray can be seen as a range of colored line, this range of colored lines is termed as line spectra.
lines spectra are usually utilized to recognize atoms and molecules. Each element has individual line emission spectra.
Calculate the hydrogen-ion concentration [H+] for the aqueous solution in which [OH-] is 1 x 10-2 mol/L. Is this solution acididc, basic, or neutral? Show your work.
Answer:
[H⁺] = 1.0 x 10⁻¹² M.
Explanation:
∵ [H⁺][OH⁻] = 10⁻¹⁴.
[OH⁻] = 1 x 10⁻² mol/L.
∴ [H⁺] = 10⁻¹⁴/[OH⁻] = (10⁻¹⁴)/(1 x 10⁻² mol/L) = 1.0 x 10⁻¹² M.
∵ pH = - log[H⁺] = - log(1.0 x 10⁻¹² M) = 12.0.
∴ The solution is basic, since pH id higher than 7 and also the [OH⁻] > [H⁺].
What type of a reaction occurs when potassium metal reacts with fluorine gas?
Exothermic Synthesis
Answer:
the answer is Exothermic Synthesis
You are presented with a mixture of iron beads and iron filings. How can you separate them?
Use a magnet
Dissolve the fillings in water
Use a screen
Distillation
NEED HELP ASAP
Answer:
By using a screen
Answer:
Use a screen
Explanation:
A student wrote a chemical equation as shown.
2H2S + 3O2 → H2O + SO2
What should the student do to balance the equation?
Add a coefficient 4 before H2O and a 2 before SO2.
Add a coefficient 2 before H2O and a 4 before SO2.
Add a coefficient 5 before H2O and a 4 before SO2.
Add a coefficient 2 before H2O and a 2 before SO2.
Answer:
A. Add a coefficient 4 before h2o and add a 2 before so2
Explanation:
Answer : The correct option is, Add a coefficient 2 before [tex]H_2O[/tex] and a 2 before [tex]SO_2[/tex].
Explanation :
Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
If the amount of atoms of each type on the left and right sides of a reaction differs then to balance the equation by adding coefficient in the front of the elements or molecule or compound in the chemical equation.
The coefficient tell us about that how many molecules or atoms present in the chemical equation.
The given chemical reaction is,
[tex]2H_2S+3O_2\rightarrow H_2O+SO_2[/tex]
This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen, sulfur and oxygen atoms are not balanced.
In order to balance the chemical equation, the coefficient '2' is put before the [tex]H_2O[/tex] and [tex]SO_2[/tex] and we get the balanced chemical equation.
The balanced chemical reaction will be,
[tex]2H_2S+3O_2\rightarrow 2H_2O+2SO_2[/tex]
Bohr's atomic model differed from Rutherford's because it explained that electrons exist in specified energy levels surrounding the nucleus. electrons are embedded within the rest of the atom like plum pudding. an atom is an indivisible sphere. electrons circle the nucleus.
Electrons exist in specified energy levels surrounding the nucleus.
Answer:
electrons exist in specified energy levels surrounding the nucleus.
how to identify oxidation half-reaction and reduction half-reaction in a redox equation?
explain
Answer:
Explanation:
A redox reaction is a reaction in which there is an exchange of electrons between different species. As one specie gains electrons the other loses it and this furnishes the reaction.
Redox reactions are made up of half-reactions which shows how a specie gains electrons and what happens to it. It also shows what happens to a specie that loses electrons.
To understand and be able to identify what an oxidation and reduction half reaction is, we need to know what oxidation and reduction entails.
Oxidation is the loss of electron by an atom. It also entails addition of oxygen to a specie, removal of hydrogen from a specie. This leads to increase in oxidation number of the specie.
Reduction is just the opposite of oxidation and electrons are gained by an atom here.
To identify an oxidation or reduction half in a redox reaction, we look at the species closely and we check for the following:
1. Changes in oxidation number
2. Wether a species loses or gains electrons.
3. Addition /removal of oxygen and hydrogen.
Let's take for example:
Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺
By inspection, we look at the species involved:
Fe: Fe²⁺ → Fe³⁺
The oxidation number increases from 2+ to 3+
There is a loss of an electron for this to occur
We see that this is an oxidation half
Mn: MnO₄⁻ → Mn²⁺
The oxidation number of Mn changes from +7 to +2:
X+ (-2x4) = - 1
X - 8 = - 1
X = 7
This implies that 5 electrons were gained.
There is also a loss of oxygen.
This makes the reaction a reduction half.
This simple way is used to identify reduction and oxidation halves.
To identify oxidation and reduction half-reactions in a redox equation, remember that oxidation involves the loss of electrons and reduction involves the gain of electrons. Balance the reaction by ensuring electrons lost equal electrons gained. Iron and hydrogen serve as examples of a reducing agent and an oxidizing agent, respectively.
To identify oxidation and reduction half-reactions in a redox equation, it's essential to understand the basic principle that oxidation involves the loss of electrons, while reduction involves the gain of electrons.
An oxidation half-reaction will show electrons as products (on the right side), indicating that the species is losing electrons. Conversely, a reduction half-reaction will have electrons as reactants (on the left side), signifying the gain of electrons by the species.
Example of Identifying Half-Reactions
Consider a redox process where Iron (Fe) is oxidized, and Hydrogen (H₂) is reduced. The oxidation half-reaction can be represented as Fe -> Fe²⁺ + 2e⁻, showing iron losing two electrons. For the reduction process, H⁺ + e⁻ -> H₂ illustrates hydrogen gaining electrons.
To ensure the reaction is balanced, the number of electrons lost in the oxidation process must be equal to those gained in the reduction process. This might require multiplying one or both half-reactions by a certain number to balance the electrons.
In the overall reaction, the species that gets oxidized (loses electrons) acts as the reducing agent, and the species that gets reduced (gains electrons) acts as the oxidizing agent. Therefore, in our example, iron acts as the reducing agent, and hydrogen acts as the oxidizing agent.
Joe wants to prepare a nitrogen containing compound. Which compound can he prepare?
A. Methane
B. Water
C. Ammonia
D. Hydrochloric acid
Answer- The correct answer is ammonia.
Explanation- All other compounds that are formed do not have any essence of Nitrogen in it. Whereas, Methane is a compound with carbon and hydrogen as elements while water has hydrogen and oxygen as its elements.
Lastly Hydrogen and chlorine combined together to form hydrochloric acid. Hence the only compound left with nitrogen is ammonia and has nitrogen and hydrogen as its basic elements.
Answer:
The answer is C. Ammonia
Explanation: I got it right on the test.
Have a good day :)
Which metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe?
manganese The metal that is used as a sacrificial electrode to prevent the rusting of iron is manganese.
Which is the strongest oxidizing agent? lithium; li+ + e– → li e0 = –3.05 v hydrogen; 2h+ + 2e– → h2e0 = 0.00 v fluorine; f2 + 2e– → 2f–e0 = +2.87 v?
Answer:
Fluorine; f2 + 2e– → 2f–e0 = +2.87 v
Explanation:
Oxidizing agents remove electrons from other atoms to complete a stable outer octet. In this case, Flourine is an oxidizing agent.Hydrogen and lithium are reducing agents because they can lose electrons more easily than accept them. We can also use the electrode potential or e.m.f to determine whether an element is a strong oxidizing agent or not. An element with the largest positive e.m.f is the strongest oxidizing agent as it indicates that its more electronegative.