Answers
Answer:
B. 1-amino-2-methylpentan-2-ol
Explanation:
In this case, the first step, we have the attack of the nucleophile cyanide ([tex] CN^-[/tex] produced by sodium cyanide to the carbon on the carbonyl group (C=O) producing a negative charge in the oxygen.
Then HCl protonates the molecule to produce a cyanohydrin. This cyanohydrin can be reduced by the action of hydrogen gas ([tex]H_2[/tex]) in the presence of a catalyst ([tex]Pd[/tex]), producing an amino group. With this in mind, the final molecule is: 1-amino-2-methylpentan-2-ol.
See figure 1 to further explanations
I hope it helps!
Related Questions
Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Answers
Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
In a reversible reaction, the endothermic reaction absorbs ____________ the exothermic reaction releases. A. less energy than B. None of these, endothermic reactions release energy C. the same amount of energy as D. more energy than
Answers
Answer: C. the same amount of energy as
Explanation:
A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back.
Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change.
[tex]A+B\rightleftharpoons C+D[/tex]
Thus if forward reaction is exothermic i.e. the heat is released , the backward reaction will be endothermic i.e. the heat is absorbed and in same amount.
The amount of energy released will be equal and opposite in sign to the energy absorbed in that reaction.
Answer:
C.) the same amount of energy as
Explanation:
I got it correct on founders edtell
Calculate the energy required to heat of 1.50 kg silver from -7.8 C to 15.0 C . Assume the specific heat capacity of silver under these conditions is .0235 J*g^-1*K^-1 . Be sure your answer has the correct number of significant digits.
Answers
Answer:
804 J
Explanation:
Step 1: Given data
Mass of silver (m): 1.50 kgInitial temperature: -7.8 °CFinal temperature: 15.0 °CSpecific heat capacity of silver (c): 0.0235J·g⁻¹K⁻¹Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.0235J·g⁻¹K⁻¹ × (1.50 × 10³g) × [15.0°C-(-7.8°C)]
Q = 804 J
g Use the References to access important values if needed for this question. A researcher took 2.592 g of a certain compound containing only carbon and hydrogen and burned it completely in pure oxygen. All the carbon was changed to 7.851 g of CO2, and all the hydrogen was changed to 4.018 g of H2O . What is the empirical formula of the original compound
Answers
Answer:
Empirical formula is: C₂H₅
Explanation:
The chemical equation of burning of a compound that conatins only Carbon and Hydrogen is:
CₓHₙ + O₂ → XCO₂ + n/2H₂O
That means the moles of CO₂ produced are the moles of Carbon in the compound and moles of hydrogen are twice moles of water. Empirical formula is the simplest ratio between moles of each element in the compound. Thus, finding molse of C and moles of H we can find empirical formula:
Moles C and H:
Moles C = Moles CO₂:
7.851g CO₂ ₓ (1mol / 44g) = 0.1784 moles CO₂ = Moles C
Moles H = 2 Moles H₂O
4.018g H₂O ₓ (1mol / 18.01g) = 0.2231 * 2 = 0.4417 moles H
Ratio C:H
The ratio between moles of hydrogen and moles of Carbon are:
0.4417 moles H / 0.1784 moles C = 2.5
That means there are 2.5 moles of H per mole of Carbon. As empirical formula must be given only in whole numbers,
Empirical formula is: C₂H₅For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.
Answers
Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
Two elements represents by the letter Q and R atomic number 9 and 12 respectively. Write the electronic configuration of R
Answers
Answer:
Atomic no = 12 = Mg
Explanation:
It is given that,
The atomic number of two elements that are represented by letter Q and R are 9 and 12.
We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.
For R, atomic number = 12
Its electronic configuration is : 2,8,2
It has two valance electrons in its outermost shell. The element is Magnesium (Mg).
Assuming 100% dissociation, which of the following compounds is listed incorrectly with its van't Hoff factor i? Al2(SO4)3, i = 4 NH4NO3, i = 2 Mg(NO3)2, i = 3 Na2SO4, i = 3 Sucrose, i = 1
Answers
Answer:
- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).
Explanation:
Hello,
In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that
- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).
- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).
- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).
- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).
Best regards.
Enter the balanced chemical equation for the reaction of each of the following carboxylic acids with KOH.Part Aacetic acidExpress your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part B2-methylbutanoic acid (CH3CH2CH(CH3)COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part C4-chlorobenzoic acid (ClC6H4COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).
Answers
Answer:
Explanation:
Answer in attached file .
Refer to the figure.
30. How many planes are shown in the figure?
31. How many planes contain points B, C, and E?
32. Name three collinear points.
3. Where could you add point G on plane N
so that A, B, and G would be collinear?
4. Name a point that is not coplanar with
A, B, and C.
5. Name four points that are coplanar.
BN
Answers
Answer:
30. 5 planes are shown
31. 1 plane
32. CEF
33. on line AB
34. E or F
35. ABCD or BCEF or CDEF or ACEF
Explanation:
30. Each of the surfaces of the rectangular pyramid is a plane. There are 5 planes.
__
31. 3 points define one plane only.
__
32. The only points shown on the same line segment are points E, F, and C.
__
33. If G is to be collinear with A and B, it must lie on line AB.
__
34. The only points shown that are not on plane N are points E and F. Either of those will do.
__
35. There are three planes that have 4 points shown on them. The four points that are on the same plane are any of ...
ABCDBCEFCDEFPlane ACEF is not shown on the diagram, but we know that those 4 points are also coplanar. (Any point not on line CE, together with the three points on that line, will define a plane with 4 coplanar points.)
How many atoms are in 65.0g of zinc?
Answers
from
1moles=iatom
Mole=mass÷avogardos
Where
Avogadro's= 6.02×10²³
So moles = 65.0÷6.02×10²³
Atoms of zinc = 391.6 ×10²³
The number of atoms present in the given mass of Zinc that is 65.0gm is [tex]5.99\times10^{ 23}[/tex].
Atoms are the basic building blocks of matter. They are the smallest units of an element that retain the chemical properties of that element.
Now, to determine the number of atoms in a given number of moles, we can use Avogadro's number, which is approximately [tex]6.022 \times10^{23}[/tex]atoms per mole.
First, we calculate the number of moles of zinc in 65.0g by dividing the given mass by the molar mass of zinc. The molar mass of zinc (Zn) is 65.38 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 65.0g / 65.38 g/mol ≈ 0.9942 mol
Next, multiply the number of moles by Avogadro's number to find the number of atoms.
Number of atoms =[tex]Number of moles \times Avogadro's number[/tex]
Number of atoms = [tex]0.9942[/tex]mol × [tex]6.022 \times10^{23}[/tex] atoms/mol
Therefore, approximately [tex]5.99\times10^{ 23}[/tex] atoms are present in 65.0g of zinc.
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The tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of the alkene that is produced.
Answers
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.
The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
I hope it helps!
An atom of 120In has a mass of 119.907890 amu. Calculate the mass defect (deficit) in amu/atom. Use the masses: mass of 1H atom
Answers
Answer:
a
Explanation:
answer is a on edg
A saturated sodium carbonate solution at 0°C contains 7.1 g of dissolved sodium carbonate per 100. mL of solution. The solubility product constant for sodium carbonate at this temperature is
Answers
Answer:
[tex]Ksp=1.2[/tex]
Explanation:
Hello,
In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):
[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]
Next, as its dissociation reaction is:
[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]
And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):
[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]
Best regards.
The surface temperature on Venus may approach 753 K. What is this temperature in degrees Celsius?
Answers
Answer:
461.85 degrees Celsius
In which of the following compounds does the carbonyl stretch in the IR spectrum occur at the lowest wavenumber?
a. Cyclohexanone
b. Ethyl Acetate
c. λ- butyrolactone
d. Pentanamide
e. Propanoyl Chloride
Answers
Answer:
a. Cyclohexanone
Explanation:
The principle of IR technique is based on the vibration of the bonds by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is a specific energy that generates a specific vibration. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.
Now, we must remember that the lower the wavenumber we will have less energy. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.
If we look at the structure of all the molecules we will find that in the last three we have heteroatoms (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of resonance structures which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.
The molecule that fulfills this condition is the cyclohexanone.
See figure 1
I hope it helps!
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
Answers
Answer:
[tex][H^+]=0.000123M[/tex]
[tex]pH=3.91[/tex]
Explanation:
Hello,
In this case, dissociation reaction for acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
For which the equilibrium expression is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Which in terms of the reaction extent [tex]x[/tex] could be written as:
[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]
Thus, solving by using a solver or quadratic equation we obtain:
[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]
And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:
[tex][H^+]=0.000123M[/tex]
Now, the pH is computed as follows:
[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]
Best regards.
What's the mass in grams of 0.442 moles of calcium bromide, CaBr2? The atomic
weight of Ca is 40.1 and the atomic weight of Br is 79.9.
A) 452.3 g
B) 53.04 g
C) 44.2 g
D) 88.4 g
Answers
Answer:
Below
Explanation:
Let n be the quantity of matter in the Calcium Bromide
● n = m/ M
M is the atomic weight and m is the mass
M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)
M = 40.1 + 2×79.9
● 0.422 = m/ (40.1+2×79.9)
●0.422 = m/ 199.9
● m = 0.422 × 199.9
● m = 84.35 g wich is 88.4 g approximatively
88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.
What do you mean by mass ?Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .
To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,
Let n be the quantity of matter in the Calcium Bromide
M is the atomic weight and m is the mass
n = m/ MM of CaBr2 is the sum of the atomic weight of its components
Mass of Ca = 40.1 , Mass of Br = 79.9
M = 40.1 + 2×79.9
0.422 = m/ (40.1+2×79.9)
0.422 = m/ 199.9
m = 0.422 × 199.9
m = 84.35 g which is 88.4 g approximatively .
Thus ,88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .
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In 1988, three gray whales were trapped in Arctic ice. Television crews captured the frantic
attempts of hundreds of people to save the whales. Eventually, a Soviet icebreaker and U.S.
National Guard helicopters arrived to help free the whales. The cost of the rescue mission
exceeded $5 million.
i. Write a scientific question related to the whale story. (1 point)
Answers
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many
Answers
The given question is incomplete.
The complete question is:
When methane is burned with oxygen, the products are carbon dioxide and water. If you produce 9 grams of water and 11 grams of carbon dioxide from 16 grams of oxygen, how many grams of methane were needed for the reaction?
Answer: 4 grams of methane were needed for the reaction
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
{tex]CH_4+2O_2\rightarrow CO_2+H_2O[/tex]
Given: mass of oxygen = 16 g
Mass of carbon dioxide = 11 g
Mass of water = 9 g
Mass of products = Mass of carbon dioxide + mass of water = 11 g +9 g = 20 g
Mass or reactant = mass of methane + mass of oxygen = mass of methane + 16 g
As mass of reactants = mass of products
mass of methane + 16 g= 20 g
mass of methane = 4 g
Thus 4 grams of methane were needed for the reaction
What is the final volume V2 in milliliters when 0.551 L of a 50.0 % (m/v) solution is diluted to 23.5 % (m/v)?
Answers
Answer:
[tex]V_2=1.17L[/tex]
Explanation:
Hello,
In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:
[tex]V_1C_1=V_2C_2[/tex]
Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:
[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]
Best regards.
What is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?
Answers
Answer:
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
Explanation:
When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.
what are the monomers of bakelite
Answers
Answer:
Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.
Answer: The monomers of bakelite are formaldehyde and phenol
Explanation:
place the following substances in Order of decreasing boiling point H20 N2 CO
Answers
Answer:
-195.8º < -191.5º < 100º
Explanation:
Water, or H20, starts boiling at 100ºC.
Nitrogen, or N2, starts boiling at -195.8ºC.
Carbon monoxide, or C0, starts boiling at -191.5ºC.
When we place these in order from decreasing boiling point:
-195.8º goes first, then -191.5º, and 100º goes last.
Answer:
therefore, N2, CO, H20
Decreasing boiling point
Explanation:
the bond existing in H2O is hydrogen bond
bond existing in N2 is covalent bond, force existing is dipole-dipole-interaction
bond existing in CO is covalent bond , force existing between is induced -dipole- induced dipole-interaction
hydrogen bond is the strongest , followed by dipole-dipole-interaction and induced -dipole- induced dipole-interaction
the stronger the bond , the higher the boiling point
therefore, N2, CO, H20
-------------------------------------->
Decreasing boiling point
The electrolysis of molten AlCl 3 for 2.50 hr with an electrical current of 15.0 A produces ________ g of aluminum metal.
Answers
A sample of ice absorbs 15.6kJ of heat as it undergoes a reversible phase transition to form liquid water at 0∘C. What is the entropy change for this process in units of JK? Report your answer to three significant figures. Use −273.15∘C for absolute zero.
Answers
Answer:
Entropy change of ice changing to water at 0°C is equal to 57.1 J/K
Explanation:
When a substance undergoes a phase change, it occurs at constant temperature.
The entropy change Δs, is given by the formula below;
Δs = q/T
where q is the quantity of heat absorbed or evolved in Joules and T is temperature in Kelvin at which the phase change occur
From the given data, T = 0°C = 273.15 K, q = 15.6 KJ = 15600 J
Δs = 15600 J / 273.15 K
Δs = 57.111 J/K
Therefore, entropy change of ice changing to water at 0°C is equal to 57.1 J/K
The entropy change of ice changing to water will be "57.1 J/K".
Entropy changeThe shift in what seems like a thermodynamic system's condition of confusion is caused by the transformation of heat as well as enthalpy towards activity. Entropy seems to be greater mostly in a network with a high quantity or measure of chaos.
According to the question,
Temperature, T = 0°C or,
= 273.15 K
Heat, q = 15.6 KJ or,
= 15600 J
We know the formula,
Entropy change, Δs = [tex]\frac{q}{T}[/tex]
By substituting the values, we get
= [tex]\frac{15600}{273.15}[/tex]
= 57.11 J/K
Thus the above answer is correct.
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What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL
Answers
Answer:
The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"
Explanation:
[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]
In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor and each other suspend
[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]
[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]
The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.
The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.
Cell potential changeThe cell potential has been changed when there has been a change in the potential of the half cells.
The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.
With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.
Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.
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2NO + 2H2 ⟶N2 + 2H2O What would the rate law be if the mechanism for this reaction were: 2NO + H2 ⟶N2 + H2O2 (slow) H2O2 + H2 ⟶2H2O (fast)
Answers
Answer:
rate = [NO]²[H₂]
Explanation:
2NO + H2 ⟶N2 + H2O2 (slow)
H2O2 + H2 ⟶2H2O (fast)
From the question, we are given two equations.
In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.
This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.
This means our rate law is;
rate = [NO]²[H₂]
Explain why only the lone pairs on the central atom are taken into consideration when predicting molecular shape
Answers
Answer:
Lone pairs cause more repulsion than bond pairs
Explanation:
A lone pair takes up more space around the central atom than bond pairs of electrons. This is because, a lone pair is attracted to only one nucleus while bond pairs are attracted to two nuclei.
Hence the repulsion between lone pairs is far greater than the repulsion between bond pairs or repulsion between a lone pair and a bond pair. The presence of a lone pair therefore distorts a molecule away from the ideal shape predicted on the basis of the valence shell electron pair repulsion theory.
Lone pairs are found to decrease the observed bond angles in a molecule.
Come up with a definition for density
Answers
Density measures how tightly packed particles are.
If particles are tightly packed together, they will be more dense.
If they are loosely together, they will be less dense.
However, a common mistake is thinking that if something
is more dense it means that it's heavier.
However, that's not the case.
It has to do with how particles are packed in an object.
A piece of plastic sinks in oil but floats in water. Place these three substances in order from lowest density to greatest density.
Answers
Answer:
[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]
Explanation:
Hello,
In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:
[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]
Best regards.
