The average bond length in C-C in benzene (C6H6) is between single and double bond lengths. This is due to which of the following effect?
1. Due to its refractive index2. Due to the possession of resonance3. Due to its H atoms4. Due to the H-bonds
Answer:
2. Due to the possession of resonance
Explanation:
In the benzene ring, the electrons that results in the bonds between the carbon atoms are delocalized. That is, they do not belong to a specific carbon atom. It is this unique feature that enables them to have a bond length between single and double bond lengths.
This feature is as a result of resonance.
The correct option is 2.
Carbon and oxygen combine to form the molecular compound CO2, while silicon and oxygen combine to form a covalent network solid with the formula unit SiO2. Explain the difference in bonding between the two group 4A elements and oxygen. g
Answer:
See explanation below.
Explanation:
Both carbon and silicon are members of group 4A(now group 14) i n the periodic table. Carbon is the first member of the group. CO2 is a gas while SiO2 is a solid. In SiO2, there are single bonds between silicon and oxygen and the geometry around the central atom is tetrahedral while in CO2, there are double carbon-oxygen bonds and the geometry around the central atom is linear. CO2 molecules are discrete and contain only weak vanderwaals forces.
Again, silicon bonds to oxygen via its 3p orbital while carbon bonds to oxygen via a 2p orbital. As a result of this, there will be less overlap between the pi orbitals of silicon and that of oxygen. This is why tetrahedral bonds are formed with oxygen leading to a covalent network solid rather than the formation of a silicon-oxygen pi bond. A covalent network solid is known to be made up of a network of atoms of the same or different elements connected to each other continuously throughout the structure by covalent bonds.
In SiO2, each silicon atom is surrounded by four oxygen atoms. Each corner is shared with another tetrahedron. SiO2 forms an infinite three dimensional structure and melts at a very high temperature.
Without doing any calculations (just compare the correct Ksp values) complete the following statements: 1. nickel(II) carbonate is MORE soluble than ___ . 2. nickel(II) carbonate is LESS soluble than ___ . A. CaSO3 B. BaSO4 C. Ni(OH)2 D. Ag2CrO4
Answer:
1. nickel(II) carbonate is MORE soluble than barium sulfate, nickel hydroxide and silver chromate because its Ksp is higher than those of such compounds, it means more ions will be dissolved.
2. nickel(II) carbonate is LESS soluble than calcium sulfite because its Ksp is lower than it of such compound, it means, less ions will be dissolved.
Explanation:
Hello,
In this case, for calcium sulfite, barium sulfate, nickel hydroxide and silver chromate Kps is 6.13x10-5, 1.53x10-9, 1.63x10-16 and 9.03x10-12 respectively. Now, since Ksp for nickel (II) carbonate is 1.43 x10-7, we can notice that:
1. nickel(II) carbonate is MORE soluble than barium sulfate, nickel hydroxide and silver chromate because its Ksp is higher than those of such compounds, it means more ions will be dissolved.
2. nickel(II) carbonate is LESS soluble than calcium sulfite because its Ksp is lower than it of such compound, it means, less ions will be dissolved.
Best regards.
What type of reaction is shown below? Check all
that apply.
AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)
synthesis
decomposition
combustion
single replacement
double replacement
Answer:
double replacement
Explanation:
Double replacement reactions are where both cations and anions of reactants switch.
For eg: AB + CD -----> AD + BC (Here we see both things are replaced)
Synthesis reactions is where two simple things make one complex thing.
So A + B --- > AB
Combustion is usually reaction where Oxygen is reactant but here we don't have that.
Single replacement only replaces either cation or anion.
So AB + C .------> AC + B
Answer:
Replacement
Explanation:
When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose from 25.00 °C to 31.75 °C. Determine the ∆Hrxn in kJ/mol for the reaction as written below. Assume that the density of the solutions is 1.000 g/mL, and that the specific heat capacity of the solutions is that of pure water. Must show your work (scratch paper) to receive credit.
Answer:
THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.
Explanation:
1. write out the variables given:
Mass of Calcium chloride = 13.6 g
Change in temperature = 31.75°C - 25.00°C = 6.75 °C
Density of the solution = 1.000 g/mL
Volume = 100.0 mL = 100.0 mL
Specific heat of water = 4.184 J/g °C
Mass of the water = unknown
2. calculate the mass of waterinvolved:
We must first calculate the mass of water in the bomb calorimeter
Mass = density * volume
Mass = 1.000 * 100
Mass = 0.01 g
3. calculate the quantity of heat evolved:
Next is to calculate the quantity of heat evolved from the reaction
Heat = mass * specific heat of water * change in temperature
Heat = mass of water * specific heat *change in temperature
Heat = 13.6 g * 4.184 * 6.75
Heat = 13.6 g * 4.184 J/g °C * 6.75 °C
Heat = 384.09 J
Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.
4. calculate the molar mass of CaCl2:
Next is to calculate the molar mas of CaCl2
Molar mass = ( 40 + 35.5 *2) = 111 g/mol
The number of moles of 13.6 g of CaCl2 is then:
Number of moles of CaCl2 = mass / molar mass
Number of moles = 13.6 g / 111 g/mol
Number of moles = 0.1225 mol
So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..
5. Calculate the enthalpy of solution in kJ/mol:
If 1 mole of CaCl2 is involved, the heat evolved is therefore:
Heat per mole = 384.09 J / 0.1225 mol
Heat = 3 135.43 J/mol
The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.
List the following compounds in order from strongest acid to weakest acid. Rank the acids from strongest to weakest.
CH2CHCH2COOH CH2CH2CH2COOH CH3CHCH2COOH CH3CH2CH2COOH
Strongest Weakest
Answer:
CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH
Explanation:
We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.
The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.
4
Select the correct answer.
Which of the following describes a compound? (Hint: Carbon and oxygen both appear on the periodic table.)
ОА.
a piece of pure carbon, containing only carbon atoms
OB. oxygen gas surrounding a solid piece of carbon
a substance made of two oxygen atoms for each carbon atom
OD
carbon and oxygen atoms mixed without being bonded together
(4Ga + 3S2 → 2Ga2S3)
1. How many moles of Sulfur are needed to react with 100.0 grams of Gallium?
Answer:
[tex]n_S=1.076molS[/tex]
Explanation:
Hello,
In this case, given the undergoing chemical reaction, we can see a 4:3 mole ratio between the consumed moles of gallium and sulfur respectively, therefore, the consumed moles of sulfur, from the 100.0 g of gallium (use its atomic mass) turn out:
[tex]n_{S}=100.0gGa*\frac{1molGa}{69.72gGa}*\frac{3molS}{4molS} \\\\n_S=1.076molS[/tex]
Best regards.
Describe the difference between ionic and molecular compounds. Give an example of each. Check all that apply. Check all that apply. A molecular compound is formed between a metal and a nonmetal (or polyatomic ions), and is held together through the attraction of opposite charges. An example is KCl. An ionic compound is usually formed between two or more nonmetals, and is held together through the sharing of electrons between atoms. An example is SO2. An ionic compound is formed between a metal and a nonmetal (or polyatomic ions), and is held together through the attraction of opposite charges. An example is KCl. A molecular compound is usually formed between two or more nonmetals, and is held together through the sharing of electrons between atoms. An example is SO2. An ionic compound is formed between two or more metals, and is held together through the attraction of opposite charges. An example is Na2Al.
Answer:
An ionic compound is formed between a metal and a nonmetal (or polyatomic ions), and is held together through the attraction of opposite charges. An example is KCl. A molecular compound is usually formed between two or more nonmetals, and is held together through the sharing of electrons between atoms. An example is SO2.
Explanation:
When we talk about ionic bonds, the first thing that must come to mind is the electrostatic attraction between oppositely charged ions. Hence, we know that metals form cations and nonmetals form anions, thus metals could transfer electrons to nonmetals to facilitate the formation of ionic bonds. Ionic bonds could also be formed by the combination of metals with polyatomic ions such as CaCO3. Always keep it in mind that ionic bonds are characterized by electrostatic attraction between any pair of oppositely charged ions.
Molecular compounds are formed by sharing of electrons between nonmetals. We find covalent or polar covalent bonds in molecular compounds such as SO2.
A buffer solution is 0.100 M in both HC7H5O2 and KC7H5O2 has a pH of 4.19. Which of the fo pH values would you expect from the addition of a small amount of a dilute solution of a strong base?
A. 5.79
b. 4.49
c. 3.69
d. 3.89
Argon gas has a boiling point of -197 °C. Which of the following diagrams best represents the
distribution of argon atoms in a steel sphere at -190 °C?
Answer:
(a)
Explanation:
Hello,
In this case, the temperature required to boil argon, it means, transform it from liquid to gas is -197 °C. In such a way, since the temperature inside the steel sphere is -190 °C, which is greater than the boiling point, we realize argon is gaseous, therefore, the molecules will be spread inside the sphere as they will be moving based on the kinetic theory of gases.
For that reason, answer is scheme (a).
Best regards.
If the charge on the chloride ion is -1, what is the charge on the magnesium ion in the compound MgCl2?
Answer:
+2
Explanation:
Because the charge of the chloride ion is negative, that means that the charge of the magnesium ion must be positive since cations and anions go together, not cation and cation nor anion and anion. Using the "reverse criss-cross method", since the subscript of Mg is 1, that means that this is the lowest whole number ratio so we don't need to worry about simplifying. Therefore, since the charge of Cl is 2, the answer is +2.
Consider an endothermic reaction C(s) CO2(g)2CO(g) If such a system at equilibrium is heated and compressed at the same time, what happens to the equilibrium position?
A. It will shift to the left
B. It will shift to the right
C. It will not change
D. There is not enough information to tell
Answer:
A. It will shift to the left
Explanation:
In the equilibrium:
C(s) ⇄ CO2(g)2CO(g)
The system will shift to the right if any change stimulate the production of gas -LeChatelier's principle-; in the same way, if a change doesn't favors the production of gas the system will shift to the left producing less gas.
The changes that increasing the pressure of the system, doesn't favors the gas production doing the system shift to the left.
A gas that is heated expands itsellf doing the pressure increases.
In the same way, if you compress the gas, the gas increases its pressure.
Thus, both changes increase pressure of the gas doing the system shift to the left.
A. It will shift to the leftWhat volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the surroundings
Answer:
[tex]V=27992L=28.00m^3[/tex]
Explanation:
Hello,
In this case, the combustion of methane is shown below:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:
[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]
Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:
[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]
Best regards.
The location of chlorine below fluorine on the periodic table implies which of the following?
a. Chlorine atoms have a larger radius than fluorine atoms.
b. Chlorine atoms have a lower mass than fluorine atoms.
c. Chlorine atones have fewer valence electrons than fluorine atoms.
d. Chlorine atoms are more reactive than fluorine atoms.
The element nitrogen forms a(n) _______ with the charge . The symbol for this ion is , and the name is . The number of electrons in this ion is .
Answer:
The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.
Explanation:
The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.
The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.
"Calculate the pH during the titration of 30.00 mL of 0.1000 M C3H7COOH(aq) with 0.1000 M LiOH(aq) after 29.9 mL of the base have been added. Ka of butanoic acid
Answer:
pH = 7.29
Explanation:
Ka of butanoic acid is 1.54x10⁻⁵
To obtain the pH of the solution you must use H-H equation for butanoic acid:
pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]
Where pKa is defined as -log Ka = 4.81
Now, you need to find [C₃H₇COO⁻] and [C₃H₇COOH] concentrations (Also, you can find moles of each substance and replace them in the equation.
Butanoic acid reacts with LiOH, producing C₃H₇COO⁻, thus:
C₃H₇COOH + LiOH → C₃H₇COO⁻ + H₂O + Li⁺
Moles of both reactants, C₃H₇COOH and LiOH are:
C₃H₇COOH = 0.0300L ₓ (0.1000mol / L) = 0.003000moles of C₃H₇COOH
LiOH = 0.0299L ₓ (0.1000mol / L) = 0.00299 moles of LiOH.
That means moles of C₃H₇COO⁻ produced are 0.00299 moles.
And moles of C₃H₇COOH that remains in solution are:
0.00300 - 0.00299 = 0.00001 moles of C₃H₇COOH
Replacing in H-H equation:
pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]
pH = 4.81 + log₁₀ [0.00299moles] / [0.00001moles]
pH = 7.29given the following quantities of reactants and the balance equations; assuming an 85.0% yield, determine how many grams of potassium aluminum sulphate dodecahydrate can be produced.
Answer:
[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=32.23gKAl(SO_4)_2\dot \ 12H_2O[/tex]
Explanation:
Hello,
In this case, we balance the given equations as shown below:
[tex]Al(s)+KOH(aq)+3H_2O(l)\rightarrow KAl(OH)_4(aq)+\frac{3}{2} H_2(g)\\\\KAl(OH)_4(aq)+2H_2SO_4(aq)\rightarrow KAl(SO_4)_2(aq)+4H_2O(l)\\\\KAl(SO_4)_2(aq)+12H_2O\rightarrow KAl(SO_4)_2\dot\ 12H_2O(aq)[/tex]
Now, with 3.00 grams of aluminium, 50.00 mL of water and 10.00 mL of 8.00M potassium hydroxide, the first step is to identify the limiting reactant by firstly computing the moles of all of them:
[tex]n_{Al}=3.00 gAl*\frac{1molAl}{27gAl}=0.111molAl\\ \\n_{KOH}=0.010L*8.00mol/L=0.08molKOH\\\\n_{H_2O}=50.00mL*\frac{1g}{1mL} *\frac{1mol}{18g}=2.78molH_2O[/tex]
Thus, we can notice that 0.111 mol of aluminium will consume 0.11. moles of potassium hydroxide and 2.78 moles of water will consume 0.927 moles of potassium hydroxide, for that reason, we can infer that since there are only 0.08 moles of potassium hydroxide, it is the limiting reactant, therefore, we compute the yielded moles of potassium aluminium hydroxide in the first reaction:
[tex]n_{KAl(OH)_4}=0.08molKOH*\frac{1molKAl(OH)_4}{1molKOH} =0.08molKAl(OH)_4[/tex]
Next, we compute the yielded moles of potassium aluminium sulfate in the second reaction assuming sulfuric acid is in excess:
[tex]n_{KAl(SO_4)_2}=0.08molKAl(OH)_4*\frac{1molKAl(SO_4)_2}{1molKAl(OH)_4}=0.08molKAl(SO_4)_2[/tex]
Finally, in the third reaction, we compute the yielded grams of potassium aluminum sulphate dodecahydrate by using its molar mass and its mole ratio with potassium aluminium sulfate:
[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}=0.08molKAl(SO_4)_2*\frac{1molKAl(SO_4)_2\dot \ 12H_2O}{1molKAl(SO_4)_2} *\frac{474.00gKAl(SO_4)_2\dot \ 12H_2O}{1molKAl(SO_4)_2\dot \ 12H_2O} \\\\m_{KAl(SO_4)_2\dot \ 12H_2O}=37.92gKAl(SO_4)_2\dot \ 12H_2O[/tex]
Which is the theoretical yield, thus, by using the percent yield the actual yielded mass turns out:
[tex]m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=0.85*37.92gKAl(SO_4)_2\dot \ 12H_2O\\\\m_{KAl(SO_4)_2\dot \ 12H_2O}^{actual}=32.23gKAl(SO_4)_2\dot \ 12H_2O[/tex]
Best regards.
Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.
H2(g) → 2H+(aq) + 2e- oxidation
Cu2+(aq) → + 2e-Cu(s) reduction
Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.
Answer:
H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)
Explanation:
In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.
In the reactions:
H₂(g) → 2H⁺(aq) + 2e⁻ oxidation
Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.
Cu²⁺(aq) + 2e⁻ → Cu(s) reduction
And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.
The sum of both reactions is:
H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)
Subtracting the electrons in both sides of the reaction:
H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)You use 10.0 mL of solution A, 10.0 mL of solution B, and 70.0 mL of water for your first mixture. What is the initial concentration of KIO3
Complete Question
The complete question is shown on the first uploaded image
Answer:
The initial concentration is [tex]C_f = 0.0022 \ M[/tex]
Explanation:
From the question we are told that
The volume of solution A is [tex]V_i = 10.0 mL[/tex]
The concentration of A is [tex]C_i = 0.0200 \ M[/tex]
The volume of solution B is [tex]V_B = 10.0mL[/tex]
The volume of water is [tex]V_{w } = 70.0 mL[/tex]
Generally the law of dilution is mathematically represented as
[tex]C_i * V_i = C_f * V_f[/tex]
Where [tex]C_f[/tex] is the concentration of the mixture
[tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as
[tex]V_f = 10 + 10 + 70[/tex]
[tex]V_f = 90mL[/tex]
So
[tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]
substituting values
[tex]C_f = \frac{0.0200 * 10 }{90}[/tex]
[tex]C_f = 0.0022 \ M[/tex]
Note the mixture obtained is [tex]KIO_3[/tex]
neeeeed helpppppppppp
Answer:
Option C. Will always.
Explanation:
A spontaneous reaction is a reaction that occurs without an external supply of heat.
This implies that spontaneous reaction will always occur as no external supply of heat is needed.
A 23.0% by mass LiCl solution is prepared in water. What is the mole fraction of the water? 0.113 0.887 0.873 0.127 0.911
Answer: The Answer should be B. 0.887 (:
Explanation:
The mole fraction of the solution is 0.887.
Mass percent of LiCl present = 23.0%
Mass of the LiCl can be obtained from;
23 = x/100 × 100/1
x = 23 g
Number of moles of LiCl = 23g/42 g/mol = 0.55 moles
Number of moles of water = (100 - 23) g/18 g/mol = 4.3 moles
Total number of moles = 0.55 + 4.3 = 4.85 moles
Mole fraction of water = 4.3 moles/4.85 moles = 0.887
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Suppose that the NaOH sitting on the shelf had absorbed 1 g of water for every 1 g of NaOH. How would this have affected your results
Answer:
The concentration of NaOH will be lower and the titration will be affected.
Explanation:
Hello,
In this case, sodium hydroxide is acknowledged as a highly hygroscopic substance, which means that is able to absorb water to its molecules. In such a way, in any measurement, if sodium hydroxide has absorbed water, the results will be wrong in terms of accuracy. More specifically, for concentration, if we have for example 30 grams of NaOH and we dissolve it a 100-mL solution, as it absorbed 30 grams of water, the total volume could be now approximated to 130 mL, thus, the concentration will change as follows:
[tex]M_1=\frac{30g/40g/mol}{0.1L}=7.5M\\ \\M_2=\frac{30g/40g/mol}{0.13L}=5.77M[/tex]
It causes the actual molarity to be decreased, it means that in a titration procedure, less acid would be used to neutralize it or more of it would be needed to neutralize a given acid.
Best regards.
We have that the NaOH sitting on the shelf had absorbed 1 g of water for every 1 g of NaOH will affect the Molarity of NaOH and its effectiveness
With the situation of NaOH sitting on the shelf having to absorbed 1 g of water for every 1 g of NaOH.
Means that for every g of NaOH collected 1/2g is water and 1/2g is actual NaOH.
Hence this will cause a change in the molarity of NaOH thereby causing Molarity to drop by half as well. Giveing the resultant Molarity to be somewhere around half the regular molarity.
This many cause a variation also in the results of titration
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Santalbic acid, a fatty acid isolated from the seeds of the sandalwood tree, is an unusual fatty acid that contains a carbon-carbon triple bond.What orbitals are used to form each of the three indicate single bongs in santalbic acid?
Answer and Explanation:
The explanation is described below:-
a. Bond that is a which is created by sp3 - sp2 orbitals
b. Bond that is b which is developed by sp2-sp orbitals
c. Bond that is c which is created by sp-sp3 orbitals
Sp2 is hybridised by a double bonded carbon, and a triple bonded carbon is hybridised. Both single bonded carbons are hybridised to sp3.
How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through a solution containing CrCl3
Answer:
483.27 minutes
Explanation:
using second faradays law of electrolysis
HELPPP.
Which of the following is a property of matter?
O A. It takes up space.
OB. It is everywhere.
O C. It is constantly changing.
O D. It cannot be divided.
Answer:
a
Explanation:
Matter can be volume or density. So, this concludes that it is when it takes up space.
Answer: A.
Explanation:
it takes up space
When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of Bacteria and Archaea
Answer:
the answer is monerans
Explanation:
When Carl Woese developed the modern system of classification, he broke the previous kingdom of Monera into the two kingdoms of Bacteria and Archaea.
What kingdom of Monera ?Some biologists believed it made sense to classify prokaryotes as belonging to their own kingdom, the Monera. That served as the foundation for Richard Whittaker and Lynn Margulis's five-kingdom proposal, which enhanced the Haeckel plan by include a kingdom of fungus.
Protists, protozoa, monera, fungi, and viruses have long been proposed as belonging to different kingdoms, but traditional evolutionists during the majority of the 20th century had given none of them any thought.
Later, the Monera kingdom was split into Eubacteria and Archaebacteria by Carl Woese . Moreover, he divided the five kingdoms into three domains: Eukaryotes, Archaea, and Bacteria.
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Your question is incomplete. But your complete question is as follows:
When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of _____ into Bacteria and Archaea.
Write the condensed electron configuration for the manganese atom. When writing your answers, do not add spaces and do not try to italicize the orbitals.
Answer:
[tex][Ar]4s^23d^5[/tex]
Explanation:
Hello,
In this case, we write the electron configuration of the manganese atom by noticing its atomic number is 25, so we fill the orbitals and levels up-to 25 electrons as shown below:
[tex]1s^22s^22p^63s^23p^64s^23d^5[/tex]
Moreover, for the condensed electron configuration, we consider the previous noble gas, that is argon, electron configuration which is:
[tex]1s^22s^22p^63s^23p^6[/tex]
By cause of its atomic number that is 18. In such a way, we combine argon's electron configuration with manganese's to obtain its condensed version:
[tex][Ar]4s^23d^5[/tex]
Regards.
What is the absolute magnitude of the rate of change for [NH3] if the
rate of change for [Hz] is 9.00 M/s in the reaction 2 NH3(g) → N2(g) +
3 H2(g)?
Answer:
[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:
[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]
In such a way, knowing the rate of formation hydrogen (H₂), we can know the rate of change of ammonia, that must be negative for consumption:
[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]
Nevertheless, the absolute magnitude will be positive:
[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]
Best regards.
Note the dynamic equilibrium in the opening photo which solution changes color when the pH of both solutions is increased explain?
Answer:
The colour of the orange solution becomes yellow.
Explanation:
1. Before adding NaOH
Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.
Both solutions are involved in the same equilibrium:
[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]
The first beaker contains mostly chromate ions with a few dichromate ions.
The position of equilibrium lies to the left and the solution is yellow.
The second beaker contains mostly dichromate ions with a few chromate ions.
The position of equilibrium lies to the right and the solution is orange.
2. After adding NaOH
According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
Beaker 1
If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).
The system responds by having the dichromate react with water to replace the H⁺.
At the same time, the system forms more of the yellow chromate ion.
The position of equilibrium shifts to the left.
However, the solution is already yellow, so you see no change in colour.
Beaker 2
The reaction is the same as in Beaker 1.
This time, however, as the dichromate ion disappears, do does its orange colour.
Also, the yellow chromate is being formed and its yellow colour appears .
The colour changes from orange to yellow.