Answers
Answer:
Most stable: 2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane : Least stable
Explanation:
Treatment of NaOH with (R)-3-bromo-2,3-dimethylpentane results in the elimination of HBr. Each H atoms present on each [tex]\beta[/tex]-carbon atoms can be eliminated result in the formation of four possible products: (1) 2,3-dimethylpent-2-ene, (2) (E)-3,4-dimethylpent-2-ene, (3) (Z)-3,4-dimethylpent-2-ene and (4) 2-methyl-3-methylenepentane.
The stability of these alkenes depends on the number of hyperconjugative H atoms present with respect to the double bond. In accordance with this, 2,3-dimethylpent-2-ene is the most stable alkene (11-hyperconjugative H atoms). Then, 3,4-dimethylpent-2-ene is the second most stable alkene (7-hyperconjugative H atoms). Among (E)-3,4-dimethylpent-2-ene and (Z)-3,4-dimethylpent-2-ene, (E)-3,4-dimethylpent-2-ene is more stable due to it's less sterically hindered structure. 2-methyl-3-methylenepentane is the least stable alkene (3-hyperconjugative H atoms).
So, decreasing order of stability of alkenes from most stable to least stable:
2,3-dimethylpent-2-ene > (E)-3,4-dimethylpent-2-ene > (Z)-3,4-dimethylpent-2-ene > 2-methyl-3-methylenepentane
Three bromo As just a result of the nucleophilic substitution mechanism, 2,4 dimethylpentane generates a racemic mix containing both the r and s forms of the molecule.
In the first, sluggish phase, a two-degree cation with a positive charge is produced at carbon 3, and then it undergoes rearrangement to a 3-degree carbocation at carbon 2 either from the side. So because carbocation only has 6 electrons in its outermost shell, it is sp2 hybridized and thus planar in structure. In step two then occurs, as well as the attack of the -OH group, as the -OH (hydroxyl) group can strike from either side (top or bottom), leading to the formation of a racemic mixture of 2,4 dimethyl pentane-2-ol.This quantity of alpha-hydrogens in an alkyl group can be used to determine its stability. The greater the number of alpha-hydrogens inside an alkene, the overall larger the number of hyperconjugated structures, and thus the greater the stability. Due to the obvious symmetrical structure of a trans-isomer, whenever the amount of alpha-hydrogens is the same, the trans-alkene isomer is much more stable than that of the cis-alkene isomer.Please find the attached file.
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Related Questions
29. Which alcohol combines with carboxylic acid to produce the ester called ethyl butanoate?
A) butan-2-ol
B) propan-1-ol
C) butan-1-ol
D) ethanol
Answers
Answer:
The answer is option D.
ethanol
Hope this helps you
230g sample of a compound contains 136.6g carbon, 26.4g hydrogen, and 31.8g nitrogen. What is masspercentif oxygen
Answers
Answer:
15.3 %
Explanation:
Step 1: Given data
Mass of the sample (ms): 230 gMass of carbon (mC); 136.6 gMass of hydrogen (mH): 26.4 gMass of nitrogen (mN): 31.8 gStep 2: Calculate the mass of oxygen (mO)
The mass of the sample is equal to the sum of the masses of all the elements.
ms = mC + mH + mN + mO
mO = ms - mC - mH - mN
mO = 230 g - 136.6 g - 26.4 g - 31.8 g
mO = 35.2 g
Step 3: Calculate the mass percent of oxygen
%O = (mO / ms) × 100% = (35.2 g / 230 g) × 100% = 15.3 %
2) 2.5 mol of an ideal gas at 20 oC under 20 atm pressure, was expanded up to 5 atm pressure via; (a) adiabatic reversible and (b) adiabatic irreversible process. Calculate the values of w, q, ΔU, ΔH for each process. (Cv = 5 cal / mol.K ≈ 5/2 R; R ≈ 2 cal / mol.K) (Please find the desired values by making the corresponding derivations
Answers
Answer:
a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)
workdone (w) = -8442.6 J ≈ -8.443 KJ
heat transferred (q) of the ideal gas = - w
q = 8.443 KJ
b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0
the workdone(w) in the ideal gas= - 4567.5 J ≈ - 4.57 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Explanation:
given
mole of an ideal gas(n) = 2.5 mol
Temperature (T) = 20°C
= (20°C + 273) K = 293 K
Initial pressure of the ideal gas(P₁) = 20 atm
Final pressure of the ideal gas(P₂) = 5 atm.
2) (a)for adiabatic reversible process,
note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.
Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]
= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]
= 6090.01 J × [-1.3863]
= -8442.6 J ≈ -8.443 KJ
So, the work done (w) of ideal gas = -8.443 KJ
For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-8.443 KJ)
q = 8.443 KJ
heat transfer (q) of the ideal gas = 8.443 KJ
(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.
Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )
= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]
= - 6090.01 J × 0.75
= - 4567.5 J ≈ - 4.57 KJ
∴work done(w) of an ideal gas = - 4.57 KJ
For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-4.5675 KJ)
q = 4.5675 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
How many grams are in 5.87 x 10^21 molecules of sulfur?
Answers
Answer:
0.312g
Explanation:
From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 molecules. This means that 1mole of sulphur also contains 6.02x10^23 molecules
1mole of sulphur = 32g
If 1 mole(i.e 32g) of sulphur contains 6.02x10^23 molecules
Then, Xg of sulphur will contain 5.87x10^21 molecules i.e
Xg of sulphur = (32x5.87x10^21)/6.02x10^23 = 0.312g
just saying this is not my work, thank Eduard22sly
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What element is primarily used in appliances to make electronic chips
A. Silicon (Si)
B. Nickel (Ni)
C. Copper (Cu)
D. Selenium (Se)
Answers
Answer:
Option A
Explanation:
Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.
Answer:
A. Silicon (Si)
Explanation:
Silicon (Si) is primarily used as a semiconductor material to make electronic chips.
What does the period number tell about the energy levels occupied by
electrons in an atom?
A. The period number tells how many electrons are in the highest
energy level of the atom.
B. The period number tells which is the highest energy level occupied
by the electrons.
C. The period number tells how many electrons are in each sublevel
of the atom.
D. The period number tells how many energy sublevels are occupied
in the atom.
Answers
Answer: B. The period number tells which is the highest energy level occupied by the electrons
Explanation:
The period number ( denoted by 'n' ) is the outer energy level that is occupied by electrons in an atom. The period number that an element is in, is the number of energy levels that the element has.When we move across a period from left to right in a periodic table the number of electrons in atoms increases within the same orbit.Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.
hence, the correct option is B. The period number tells which is the highest energy level occupied by the electrons.
The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.
What are energy levels ?The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.
When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.
Therefore, option B , second option is correct.
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what are the similarities between amorphous solid and crystalline solid
Answers
Answer:
solid dont know
Explanation:
so sorry ask another
14. Based on your previous observations, predict the impact of changing the number of moles of a gas sample on the volume of the gas sample (if pressure and temperature are held constant). What effect would changing the number of moles of a gas sample have on the temperature of a gas sample (if pressure and volume are held constant)? Explain
Answers
Answer:
Number of moles of gas is directly proportional to the volume of the gas
Number of moles of the gas is directly proportional to the temperature of the gas
Explanation:
According to Avogadro's law, changing the number of moles of a gas changing the volume of the gas also since the volume of a gas is directly proportional to the number of moles of the gas.
Hence from Avogadro's law; V= kn where k is a proportionality constant, V is the volume of the gas and n is the number of moles of the gas.
Changing the number of moles will also lead to a change in the temperature of the gas, since volume is directly proportional to the number of moles of the gas and volume is also directly proportional to temperature (Charles law), it the follows that number of moles of the gas is directly proportional to its temperature.
Choose the most appropriate indicator for the titration of a weak acid with NaOH, where the expected equivalence point of the titration is at pH 8.8.
a. methyl orange, pH range 3.2-4.4
b. methyl red, pH range 4.8 6.0
c. bromothymol blue, pH range 6.0-7.6
d. phenolphthalein, pH range 8.2-10.0
e. alizarin yellow R. pH range 10.1-12.0
Answers
Answer:
D phenolphthalein,pH range 8.2-10.0
How many atoms of hydrogens are found in 3.21 mol of
C3H8?
Answers
Answer:
1.55 × 10²⁵ atoms of H
Explanation:
3.21mol C₃H₈ × 8mol H × (6.022×10²³)
4.2 mol of oxygen and 4.0 mol of NO are introduced to an evacuated 0.50 L reaction vessel. At a specific temperature, the equilibrium 2NO(g) + O 2(g) 2NO 2(g) is reached when [NO] = 1.6 M. Calculate K c for the reaction at this temperature.
Answers
Explanation:
At 593K a particular decomposition’s rate constant had a value of 5.21×10−4 and at 673K the same reaction’s rate constant was 7.42×10−3. It was noticed that when the reactant’s initial concentration was 0.2264 M (with a 593K reaction temperature), the initial reaction rate was identical to the initial rate when the decomposition was run at 673K with an initial reactant concentration of 0.05999 M. Recall that rate laws have the form rate = k [A]x and, showing work, determine the order of the decomposition reaction.
2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = –118 kJ Calculate the heat when 250.0 mL of 0.500 M HCl is mixed 500.0 mL of 0.500 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0 oC and that the final mixture has mass of 750.0 g and a specific heat capacity of 4.18 J oC–1g–1, calculate the final temperature (in oC) of the mixture.
Answers
Answer:
Heat = 7375J
Final temperature of the mixture = 27.35°C
Explanation:
In the reaction:
2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2H₂O(l) ΔH = –118 kJ
When 2 moles of HCl reacts with excess of Ba(OH)₂ there are released 118kJ.
In the reaction, moles of HCl and Ba(OH)₂ that reacts are:
Moles HCl = 0.250L ₓ (0.500 moles / L) = 0.125 moles HCl
Moles Ba(OH)₂ = 0.500L ₓ (0.500 moles / L) = 0.250 moles Ba(OH)₂
For a complete reaction of 0.125 moles of HCl you need:
0.125 mol HCl ₓ (1 mole Ba(OH)₂ / 2 moles HCl) = 0.0625 moles Ba(OH)₂
As you have 0.250 moles of Ba(OH)₂, this reactant is in excess
2 moles of HCl that react release 118kJ, 0.125 moles of HCl release:
0.125 moles HCl ₓ (118kJ / 2 moles) = 7.375kJ =
7375JThe heat released can be obtained with the formula:
Q = C×m×ΔT
Where Q is heat, C specific heat of the solution, m its mass and ΔT change in temperature.
Replacing:
Q = C×m×ΔT
7375J = 4.18J/g°C×750.0g×ΔT
2.35°C = ΔT
As ΔT = Final T - Initial T:
2.35°C = Final T - 25.0°C
27.35°C = Final temperature of the mixture
Green plants use light from the Sun to drive photosynthesis. Photosynthesis is a chemical reaction in which water and carbon dioxide chemically react to form the simple sugar glucose and oxygen gas . What mass of simple sugar glucose is produced by the reaction of 4.9 of carbon dioxide?
Answers
Answer:
3.3 g of glucose, C6H12O6.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
6CO2 + 6H2O —> C6H12O6 + 6O2
Next, we shall determine the mass of CO2 that reacted and the mass of C6H12O6 produced from the balanced equation.
This is illustrated below:
Molar mass of CO2 = 12 + (2x16) = 44 g/mol
Mass of CO2 from the balanced equation = 6 x 44 = 264 g
Molar mass of C6H12O6 = (12x6) + (12x1) + (16x6) = 180 g/mol
Mass of C6H12O6 from the balanced equation = 1 x 180 = 180 g
From the balanced equation above,
264 g of CO2 reacted to produce 180 g of C6H12O6.
Finally, we shall determine the mass of C6H12O6 produced by reacting 4.9 g of CO2 as follow:
From the balanced equation above,
264 g of CO2 reacted to produce 180 g of C6H12O6.
Therefore, 4.9 g of CO2 will react to produce = (4.9 x 180)/264 = 3.3 g of C6H12O6.
Therefore, 3.3 g of glucose, C6H12O6 were obtained from the reaction.
1. Define the Law of Conservation of Mass (via text). Now that you’ve defined this law, explain what it means in your own words using an example.
Answers
Explanation:
The law of conservation of mass states that mass can neither be created nor be destroyed.
Explanation in own words = this means that in this universe no one can create or destroy mass.
No physical or chemical force.
Which of these substances has the highest pOH? 0.10 M HCl, pH = 1 0.001 M HNO3, pH = 3 0.01 M NaOH, pH = 12 The answer is 0.10 M HCI, pH=1
Answers
Answer:On these combined scales of pH and pH it can be shown that because for water when pH = pH = 7 that pH + pH = 14. This relationship is useful in the inter conversion of values. For example, the pH at a 0.01 M solution of sodium hydroxide is 2, the pH of the same solution must be 14-2 = 12.
Explanation:
The 0.10M HCI, pH = 1 solution has the highest pOH. Therefore, option (1) is correct.
What is the pOH?pOH of a solution can be determined from the negative logarithm of the hydroxide ions concentration in the solution.
The mathematically pOH of the solution can be expressed as:
pOH = -log [OH⁻] ..............(1)
Where [OH⁻] represents the concentration of hydroxide ions in an aqueous solution.
Given, the pH = 1 of HCl
pH + pOH = 14
1 + pOH = 14
pOH = 14 - 1
pOH = 13
Given, the pH = 3 of HNO₃
pH + pOH = 14
3 + pOH = 14
pOH = 14 - 3
pOH = 11
Given, the pH = 12 of NaOH = 0.01 M
pH + pOH = 14
12 + pOH = 14
pOH = 14 - 12
pOH = 2
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Write the electron configuration when Sulfur gains two electrons
Answers
Answer:
Explanation:
If sulfur gains 2 electrons then two electrons should be added to it electronic configuration.
Molarity of NaOH: From the following data calculate molarity of NaOH. Molar mass of KHP is 204.23 g/mol. Show calculation. Mass of Erlenmeyer flask + KHP 84.847 g Mass of Erlenmeyer flask 84.347 g Mass of KHP ??? Final buret reading 12.25 mL Initial buret reading 0.50 mL Volume of NaOH added ???
Answers
Answer:
Explanation:
Mass of Erlenmeyer flask + KHP = 84.847 g
Mass of Erlenmeyer flask = 84.347 g
Mass of KHP = .5 g
moles of KHP = .5 / 204.23
= 2.448 x 10⁻³ moles
moles of NaOH reacted = 2.448 x 10⁻³
Final buret reading = 12.25 mL
Initial buret reading = 0.50 mL
Volume of NaOH added=
How many Liters of 0.968M solution can be made if 0.581 moles of solute are added? Group of answer choices 0.600 L 60 mL 0.562 L 1.00 L
Answers
Answer:
0.6L
Explanation:
The formula of molarity is molSolute/litreSolution
[tex]0.968M=\frac{0.581}{LitreSolution} \\\\LitreSolution=\frac{0.581}{0.968} \\LitreSolution=0.6L[/tex].
Suppose there is 1.00 L of an aqueous buffer containing 60.0 mmol of formic acid (pKa=3.74) and 40.0 mmol of formate. Calculate the pH of this buffer.
Answers
Answer:
pH = 3.56
Explanation:
The pH of the buffer producing from the mixture of formic acid and formate ion can be found using H-H equation:
pH = pka + log [A⁻] / [HA]
pH = 3.74 + log [Formate] / [formic acid]
Where [] represents molar concentrations -or moles- of formate and formic acid in the solution.
Replacing knowing moles of formic acid = 0.0600 and moles formate = 0.0400:
pH = 3.74 + log [Formate] / [formic acid]
pH = 3.74 + log [0.0400] / [0.0600]
pH = 3.56
Sulfuric acid is commonly used as an electrolyte in car batteries. Suppose you spill some on your garage floor. Before cleaning it up, you wisely decide to neutralize it with sodium bicarbonate (baking soda) from your kitchen. The reaction of sodium bicarbonate and sulfuric acid is
Answers
Answer:
The mass of NaHCO3 required is 235.22 g
Explanation:
*******
Continuation of Question:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?
********\
The question requires us to calculate the mass of NaHCO3 to neutralize the acid.
From the balanced chemical equation;
1 mol of H2SO4 requires 2 mol of NaHCO3
1.4 would require x?
Upon solving for x we have;
x = 1.4 * 2 = 2.8 mol of NaHCO3
The relationship between mass and number of moles is given as;
Mass = Number of moles * Molar mass
Mass = 2.8 mol * 84.007 g/mol
Mass = 235.22 g
The blending of one s atomic orbital and three p atomic orbitals produces ________.
A three sp3
B four sp3
C three sp
D four sp2
E four sp
Answers
Answer:
B. four sp3
Hope that helps.
We have that for the Question "The blending of one s atomic orbital and three p atomic orbitals produces?"
Answer:
Option B = four [tex]sp^3[/tex]
Explanation:
When 1 s orbital blends with 3 p orbitals, they form a tetrahedrical shaped figure with each being a [tex]SP^3[/tex] orbital.. A total of 4 orbitals
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Why is it important that the primary standard chemical be non-hygroscopic and pure? Why is it important to dry the primary standard to a constant weight?
Answers
Answer:
It is extremely important for the primary standard chemical to be non – hygroscopic and pure and to also have a constant weight because you don't want any moisture or any impurities to alter the stoichiometric point in the reaction
It is important that the primary standard chemical be non-hygroscopic and pure to calculate the exact calculation of the reaction.
What is non hygroscopic chemicals?Non hygroscopic chemicals are those compounds which will not absorb water or mositure from the outside.
If we take any substance which are hygroscopic in nature and during the chemical reaction if they absorb water content or moisture then the mass of that substance will alter and changes all the calculation of the reaction.
So, to maintain the stability of calculation we use non hygroscopic materials.
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If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Answers
Answer:
1.3 mL
Explanation:
First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:
1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.
It should have 2 significant figures, because 1.2kg has 2 and we are dividing.
The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.
What is volume?Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.
Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.
This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.
Here, it is given that mass of olive oil is 1.2kg.
We know that,
Density of olive oil = 0.917kg/l.
Volume = mass/density
Volume = 1.2/0.917.
Volume = 1.30 lit.
Volume = 1300mL.
Thus, the volume of olive oil will be 1300 mL.
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Use 1-Butanol as the only organic compound, design a method to synthesize 5-Nonanone. You may use any other inorganic reagents. Any organic reagents have to be made from 1-butanol.
Answers
Answer:
See attached picture.
Explanation:
Hello,
In this case, starting by 1-butanol, we can make it react with hydrogen bromide (1st step) in order to yield 1-bromobutane. Next, the formed alkyl halide is treated magnesium in the presence of an ether in order to yield butyl magnesium bromide which is a Grignard reagent (2nd step). Finally, by adding carbon dioxide, water and extra hydrogen chloride, a carbonyl group can be formed between two butyl radicals in order to form the 5-nonanone (3rd step) as shown on the attached picture.
Best regards.
A base solution contains 0.400 mol of OH–. The base solution is neutralized by 43.4 mL of sulfuric acid. What is the molarity of the sulfuric acid solution?
Answers
Answer:
Molarity of the sulfuric acid solution is 4.61M
Explanation:
The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:
2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻
That means, 2 moles of base react with 1 mole of sulfuric acid.
If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:
0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄
As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:
0.200 moles H₂SO₄ / 0.0434L = 4.61M
Molarity of the sulfuric acid solution is 4.61MCopper was one of the earliest metals used by humans, because it can be prepared from a wide variety of copper minerals, such as cuprite (Cu2O), chalcocite (Cu2S), and malachite [Cu2CO3(OH)2]. Balance the following reactions for converting these minerals into copper metal. Place a coefficient in each gray box.
(a) Cu2O(s) + C(s) rightarrow Cu(s) + CO2(g)
(b) Cu2O(s) + Cu2 S(s) rightarrow Cu(s) + SO2(g)
(c) Cu2 CO3 (OH)2(s) rightarrow CuO(s) + CO2(g) + H2O(g)
Use the left and rightarrow keys to move the cursor out of a superscript or subscript in the module.
Answers
Answer:
a. 2 Cu₂O(s) + C(s) → 4Cu(s) + CO₂(g)
b. 2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)
c. Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)
Explanation:
A reaction is balanced when you have the same amount of atoms in reactants and products.
In the reactions:
(a) Cu₂O(s) + C(s) → Cu(s) + CO₂(g)
As a general rule, you first balance oxygen and hydrogen. In products you have 2 oxygens, then:
2 Cu₂O(s) + C(s) → Cu(s) + CO₂(g)
Carbon is balanced yet. Thus, you need just to balance Cu:
2 Cu₂O(s) + C(s) → 4Cu(s) + CO₂(g)
(b) Cu₂O(s) + Cu₂S(s) → Cu(s) + SO₂(g)
Balancing oxygen:
2Cu₂O(s) + Cu₂S(s) → Cu(s) + SO₂(g)
Sulfur is balanced yet. Now you just need to balance Cu:
2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g)
(c) Cu₂CO₃(OH)₂(s) → CuO(s) + CO₂(g) + H₂O(g)
This reaction is different because the reactant is a chemical with a lot of atoms. we will first balance Cu:
Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)
Balancing copper, oxygen, hydrogen and carbon are balanced:
Cu₂CO₃(OH)₂(s) → 2 CuO(s) + CO₂(g) + H₂O(g)
When two molecules of methanol (CH3OH) react with oxygen, they combine with three O2 molecules to form two CO2 molecules and four H2O molecules. How many H2O molecules are formed when 94 methanol molecules react
Answers
Answer:
188
Explanation:
For every 2 molecules of methanol reacted, 4 molecules of water are formed. Use this relationship to solve.
2/4 = 94/x
2x = 376
x = 188
188 molecules of water will be formed.
Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. Use the theoretical yield of liquid iron and the mass or iron ingots to calculate the percent yield of the reaction.
Answers
Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles[/tex]
[tex]\text{Moles of} CO=\frac{260}{28}=9.3moles[/tex]
[tex]Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)[/tex]
According to stoichiometry :
1 mole of [tex]Fe_2O_3[/tex] require 3 moles of [tex]CO[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] will require=[tex]\frac{3}{1}\times 2.8=8.4moles[/tex] of [tex]CO[/tex]
Thus [tex]Fe_2O_3[/tex] is the limiting reagent as it limits the formation of product and [tex]CO[/tex] is the excess reagent.
As 1 mole of [tex]Fe_2O_3[/tex] give = 2 moles of [tex]Fe[/tex]
Thus 2.8 moles of [tex]Fe_2O_3[/tex] give =[tex]\frac{2}{1}\times 2.8=5.6moles[/tex] of [tex]Fe[/tex]
Mass of [tex]Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g[/tex]
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield
[tex]\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%[/tex]
Therefore, the percent yield is, 91.8%
A cell was prepared by dipping a Cu wire and a saturated calomel electrode into 0.10 M CuSO4 solution. The Cu wire was attached to the positive terminal of a potentiometer and the calomel electrode was attached to the negative terminal.(a) Write a half-reaction for the Cu electrode. (Use the lowest possible coefficients. Omit states-of-matter.)
(c) Calculate the cell voltage.
Answers
Answer:
(a) Cu²⁺ +2e⁻ ⇌ Cu
(c) 0.07 V
Explanation:
(a) Cu half-reaction
Cu²⁺ + 2e⁻ ⇌ Cu
(c) Cell voltage
The standard reduction potentials for the half-reactions are+
E°/V
Cu²⁺ + 2e⁻ ⇌ Cu; 0.34
Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241
The equation for the cell reaction is
E°/V
Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu; 0.34
2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻; -0.241
Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂; 0.10
The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation
(ii) Calculations:
T = 25 + 273.15 = 298.15 K
[tex]Q = \dfrac{\text{[Cl}^{-}]^{2}}{ \text{[Cu}^{2+}]} = \dfrac{1}{0.1} = 10\\\\E = 0.10 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(10)\\\\=0.010 -0.01285 \times 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}[/tex]
Arrange the following elements in order of decreasing first ionization energy: S, Ca, F, Rb, and Si.
Rank from largest to smallest. To rank items as equivalent, overlap them.
Answers
Answer:
The concentration of energy needed to withdraw an electron from an atom’s mole in the gas phase is known as the ionization energy of an atom. It is more accurately termed as the first ionization energy. The ionization energy upsurges from left to right through a period and from top to bottom in the groups.
Of the given elements S, Ca, F, Rb, and Si, the S, and Si belong to the third period, and the atomic radius of S is less in comparison to Si, F belongs to the second period, Rb belongs to the fifth period, and Ca belongs to the fourth period. Thus, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.
Considering the definition of ionization energy,
Ionization energy, also called ionization potential, is the necessary energy that must be supplied to a neutral, gaseous, ground-state atom to remove an electron from an atom. When an electron is removed from a neutral atom, a cation with a charge equal to +1 is formed.
You should keep in mind that the electrons of the last layer are always lost, because they are the weakest attracted to the nucleus.
In a group, the ionization energy increases upwards because when passing from one element to the bottom, it contains one more layer of electrons. Therefore, the valence layer electrons, being further away from the nucleus, will be less attracted to it and it will cost less energy to pluck them.
In the same period, in general, it increases as you shift to the right. This is because the elements in this way have a tendency to gain electrons and therefore it will cost much more to tear them off than those on the left which, having few electrons in the last layer will cost them much less to lose them.
Taking into account the above, the decreasing order of first ionization energy, that is, from largest to smallest is F > S > S > Ca > Rb.
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