Question 63 of 70 1.5 Points When 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1-butene is produced. When potassium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base
Answers
Answer: between 45% and 70%
Explanation:
taking this step by step, let us analyze this question carefully.
From the original image in the question, as can be seen in the second uploaded image, we observe that;
potassium hydroxide having being less bulky produces about 45% of 2-methyl-butane whereas potassium tert-butoxide having being bulky produces 70% of this product.
Given the potassium propoxide as the base and intermediate to that of potassium hydroxide and potassium tert-butoxide, it produces 2-methyl-1-butane in between 45% to 70%.
Cheers i hope this helped!!!!
Final answer:
Without additional information, it is not possible to determine the exact percentage of 2-methyl-1-butene produced using potassium propoxide as the base for the reaction of 2-bromo-2-methylbutane.
Explanation:
When 2-bromo-2-methylbutane is treated with a base to form a mixture of 2-methyl-2-butene and 2-methyl-1-butene, the specific base used influences the proportion of each product. The use of potassium hydroxide results in 45% yield of 2-methyl-1-butene, while potassium tert-butoxide increases this yield to 70%. If potassium propoxide were used as the base, without further experimental data or specific trends indicated in the question, it is not possible to accurately predict the exact percentage of 2-methyl-1-butene in the mixture. Generally, the bulkiness of the base can influence the product distribution in elimination reactions due to steric hindrance, but without additional information about the reaction conditions or the specific mechanism with potassium propoxide, we can only hypothesize about the potential outcome based on trends we see with other bases.
Related Questions
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be 0.839 V. S/he does not perform Part 2; instead, s/he uses the theoretical slope of the Nernst plot for an Ag/Ag+ concentration cell, as instructed by her/his TA; this value is −0.0591 V. (a) What is the concentration of Ag+(aq) in the saturated solution of AgI (i.e., [Ag+ ]dilute)? ([Ag+ ]conc = 1.0 ✕ 10−1 M.) M (b) Using [I− ] = 0.20 M, calculate the experimental Ksp. (c) Suppose s/he mistakenly uses 1.039 V as Ecell. How does this affect [Ag+ ]dilute? Will it be too high, too low, or unaffected?
Answers
Answer:
a) [Ag+]dilute = 6.363 × 10⁻¹⁶ M
b) 1.273 × 10⁻¹⁶
c) 2.629×10⁻¹⁹ M Thus; the value for [Ag+ ]dilute will be too low
Explanation:
In an Ag | Ag+ concentration cell ,
The anode reaction can be written as :
Ag ----> Ag+(dilute) + e- &:
The cathode reaction can be written as:
Ag+(concentrated) + e- ----> Ag
The Overall Reaction : is
Ag+(concentrated) -----> Ag+(dilute)
However, the Standard Reduction potential of cell = E°cell = 0
( since both cathode and anode have same Ag+║Ag )
Also , given that the theoretical slope is - 0.0591 V
Therefore; the reduction potential of cell ; i.e
Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )
0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )
log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963
[Ag+]dilute = [tex]\mathbf{10^{-14.1963} }[/tex] × 1.0 × 10⁻¹ M
[Ag+]dilute = 6.363 × 10⁻¹⁶ M
b)
AgI ----> Ag + (dilute) + I⁻
So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]
= 6.363 × 10⁻¹⁶ M × 0.20 M
= 1.273 × 10⁻¹⁶
c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :
Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )
1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )
log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804
[Ag+]dilute = [tex]\mathbf{10^{-17.5804} }[/tex] × 1.0 × 10⁻¹ M
[Ag+]dilute = 2.629×10⁻¹⁹ M
Thus, the value for [Ag+ ]dilute will be too low
what is number of unpaired
electron of Mg
Answers
Answer:
2 unpaired Electrons
Explanation:
Answer:
I was told 2 I'm not so sure, But if it's wrong I'm Sorry :( I'm having a sorta bad day...I'm trying to help everyone to make my troubles go away god bless you and have a wonderful day, ~~ Night.
Use molecular orbital theory to determine whether He2 2+ or He2 + is more stable. Use molecular orbital theory to determine whether He2 2+ or He2 + is more stable. The He2 + ion is more stable since it has a higher bond order (bond order = 1) than the He2 2+ ion (bond order = 1/2). The He2 + ion is more stable since it has a higher bond order (bond order = 2) than the He2 2+ ion (bond order = 1). The He2 2+ ion is more stable since it has a lower bond order (bond order = 1/2) than the He2 + ion (bond order = 1). The He2 2+ a lower bond order (bond order = 3/2) than the He2 + ion (bond order = 2). The He2 2+ ion is more stable since it has a higher bond order (bond order = 1) than the He2 + ion (bond order = 1/2).
Answers
Answer:
The He₂ 2+ ion is more stable since it has a higher bond order (bond order = 1) than the He₂ + ion (bond order = 1/2).
Explanation:
Molecular orbital of He₂⁺
[tex]1\sigma_{1s}^21\sigma(star)_{1s}^1[/tex]
There are two electrons in bonding and 1 electron in antibonding orbital
Bond order = [tex]\frac{(2-1)}{2}[/tex]
= [tex]\frac{1}{2}[/tex]
Molecular orbital of He₂⁺²
[tex]1\sigma_{1s}^21\sigma(star)_{1s}^0[/tex]
There are two electrons in bonding and 0 electron in antibonding orbital
Bond order = [tex]\frac{(2-0)}{2}[/tex]
= 1
So bond order of He₂⁺² is 1 which is more stable than He₂⁺ whose bond order is [tex]\frac{1}{2}[/tex] .
In comparing the He2 2+ and He2 + ions and viewing their stability through the lens of Molecular Orbital Theory, we see that the He2 + ion is more stable due to its higher bond order.
Explanation:Using Molecular Orbital Theory, He2 2+ and He2 + ions can be compared based on their bond order. Bond order is a measure of the stability of a bond. A higher bond order signifies greater stability. The He2 + ion has a bond order of 1, while the He2 2+ ion has a bond order of 0.5. Therefore, the major difference between He2 2+ and He2+ comes from their bond orders, and the He2+ ion is considered more stable because its bond order is higher.
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What causes the formation of ionic bonds?
Answers
When two or more atoms loss or gain electrons to produce an ion there is a formation of ionic bonds.
What do you mean by an ionic bond ?The bond is produced when an atom, specially a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to get the electron(s) to become a negative ion, or anion.
Ionic bonds affect a cation and an anion.Ionic bonds are produce only between metals and nonmetals.
The ionic bond is produce through the transportation of electrons from the metal atoms to the non-metal atoms. The metal atoms lose their valence electrons to attain a stable noble gas electron arrangement.
Thus, The strong electrostatic forces of attraction between the oppositely-charged ions is known as ionic bond.
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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l) In an industrial synthesis of urea, a chemist combines 142.1 kg of ammonia with 211.4 kg of carbon dioxide and obtains 171.4 kg of urea. Part A Determine the limiting reactant. Express your answer as a chemical formula.
Answers
Answer:
The limiting reactant is Carbon dioxide, [tex]CO_{2}[/tex]
Explanation:
The balanced reaction equation is:
[tex]2NH_{3} + CO_{2}[/tex] → [tex]CH_{4} N_{2} O + H_{2} O[/tex]
The mole ratio of ammonia to carbon dioxide is 2:1
142100/17g = 8358.8 mol of NH3
211400/44g = 4, 804.5 mole of CO2
Now:
4,804.5 mol of CO2 × [tex]\frac{2 mol NH_{3} }{1 molCO_{2} }[/tex] = 9,609 mol of NH3 present
8,358.8 mole of NH3 × [tex]\frac{1 moleCO_{2} }{2moles NH_{3} }[/tex] = 4,179.4 mol of CO2 present
NH3 needs 8, 358.8 moles but had 9, 609 moles⇒ excess reactant
CO2 needs 4, 804,5 mol but had 4, 179.4 moles⇒ limiting reactant (used up completely)
The limiting reactant is carbon dioxide, CO2.
Ammonia is the limiting reactant in the synthesis of urea from ammonia and carbon dioxide.
Explanation:The reaction between ammonia (NH3) and carbon dioxide (CO2) produces urea (CH4N2O) and water (H2O) according to the balanced equation: 2NH3(aq) + CO2(aq) → CH4N2O(aq) + H2O(l).
To determine the limiting reactant, we need to compare the amount of each reactant used to the amount of urea produced. From the given information, 142.1 kg of ammonia and 211.4 kg of carbon dioxide react to produce 171.4 kg of urea.
We can use the stoichiometry of the balanced equation to find the theoretical yield of urea from both reactants:
For ammonia: 1 mole of urea is produced from 2 moles of ammonia. The molar mass of ammonia is 17.03 g/mol, so the number of moles of ammonia in 142.1 kg is (142.1 kg) / (17.03 g/mol) = 8358.4 mol. Therefore, the theoretical yield of urea from ammonia is (8358.4 mol) / 2 = 4179.2 mol or 4179.2 mol × 60.06 g/mol = 250783.4 g.For carbon dioxide: 1 mole of urea is produced from 1 mole of carbon dioxide. The molar mass of carbon dioxide is 44.01 g/mol, so the number of moles of carbon dioxide in 211.4 kg is (211.4 kg) / (44.01 g/mol) = 4801.6 mol. Therefore, the theoretical yield of urea from carbon dioxide is 4801.6 mol × 60.06 g/mol = 288255.4 g.Since the actual yield of urea is 171.4 kg, which is less than both the theoretical yields from ammonia and carbon dioxide, the limiting reactant is ammonia (NH3).
Therefore, the limiting reactant is ammonia (NH3).
The starting materials for this reaction are ethylene glycol and terephthalic acid. The letters A and B represent organic groups that are unreactive. What reactive organic functional group does ethylene glycol (containing the A functional group) contain?
Answers
Answer:
Functional groups are group of atoms which are responsible for the reactivity of compound. In ethylene glycol, organic functional group is OH which is ALCOHOL.
Explanation:
Ethylene glycol, the starting material in the reaction, contains two reactive hydroxyl groups (-OH). These groups enable it to react with terephthalic acid, which has carboxylic acid functional groups, resulting in a polymerization reaction that forms a plastic named polyethylene terephthalate.
Explanation:Ethylene glycol, also known as 1,2-ethanediol, is an organic compound with two hydroxyl groups (-OH). These hydroxyl groups are the reactive functional groups in ethylene glycol, making it a type of alcohol. In the context of the reaction mentioned, ethylene glycol can react with terephthalic acid, another organic molecule with carboxylic acid (-COOH) as the functional group, which will yield a polymer called polyethylene terephthalate (PET).
PET is a type of plastic used commonly for making bottles, films, and fibers for clothing. The formation of plastics by reaction of alcohols (like ethylene glycol) and carboxylic acids (like terephthalic acid) is known as a polymerization reaction. In these reactions, many small monomer units (the alcohol and the acid, in this case) join together to form a large polymer molecule.
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why would defective collagen proteins affect an individuals joints?
Answers
Answer:
Defective collagen proteins affect an individuals joints, as collagen is an important component protein in body connective tissues
Explanation:
Collagen is a crucial protein in the extracellular matrix of various connective tissues in body (tendons, ligaments, muscles). It constitutes major protein component in mammals body.
Collagen Vascular disease is a group of diseases affecting 'collagen' connecting tissue. Apart from inheritance, they can con occur due to autoimmune disease. This implies that body's immune system fights against itself & mistakenly damages its own tissues. So, these diseases have a huge bearing on body joints components. Eg : Lupus , Arthritis are common collagen vascular diseases. Arthritis mostly affects adults above 30 years of age, lupus can be diagnosed in younger people above 15 years of age also.
Matt has found a book on various marine zones. A picture illustrates a zone that shows diffused penetration of sunlight in water, but no marine plants. There are a few marine fish though. Which zone is illustrated?
Answers
Answer:
the mesopelagic, dysphotic, or twilight zone
Explanation:
Marine zones are the divisions of the ocean. The ocean is divided into two basic parts; the pelagic or open ocean, and the benthic or sea floor.
The pelagic zone is further divided into five broad zones according to how far down sunlight penetrates and they are:
1) the epipelagic, euphotic, or sunlit zone: the top layer of the ocean where enough sunlight penetrates for plants to carry on photosynthesis.
2) the mesopelagic, dysphotic, or twilight zone: a dim zone where some light penetrates, but not enough for plants to grow.
3) the bathypelagic, aphotic, or midnight zone: the deep ocean layer where no light penetrates.
4) the abyssal zone: the pitch-black bottom layer of the ocean; the water here is almost freezing and its pressure is immense.
5) the hadal zone: the waters found in the ocean's deepest trenches.
(a) Your TA will give you a 1H NMR spectrum of the 3-nitroaniline product. Using your NMR knowledge and the special NMR section in the lab book (especially pages 48 and 49), assign the 4 peaks in the spectrum to each of the protons in the product - you will need to use coupling pattern and chemical shift to complete the assignment. (b) Explain your assignment, specifically how you differentiated between HA and Hc
Answers
Answer: provided in the answer segment
Explanation:
Below is a step by step process to analyzing this problem
Let us begin;
From 1H-NMR singlet at 5.80 ppm show N-H peak as shown in structure.
Here, H(A) hydrogen has no neighbor hydrogen so it appears integrated singlet at 7.38 ppm.
H(C) hydrogen has the next 2 neighbor hydrogen H(B) and H(D) so it appears as a triplet at region 7.23-7.29 ppm.
H(B) hydrogen has next to one neighbor hydrogen H(C) show doublet at 6.92-6.98 ppm.
H(D) hydrogen has next to one neighbor hydrogen H(C) show doublet at 7.28-7.32 ppm.
(b). From our basic chemistry knowledge, we know that benzene molecule is planer so H(A) is more deshielding because of two substituent groups than H(C), which makes the delta value of H(A) is greater than H(C).
cheers i hope this helped!!!!!
The 1H NMR spectrum of 3-nitroaniline compounds has four different protons which can be differentiated by their chemical shift and coupling pattern. HA is usually the most deshielded due to its closeness to the nitro group. The coupling pattern helps especially in differentiating protons like HA and HC.
Explanation:In 3-nitroaniline compound, the protons, designated as HA, HB, HC, and HD all show up distinctly in the 1H NMR spectrum. The key to differentiating them comes down to their chemical shift and their coupling pattern.
The most deshielded proton (highest chemical shift) is usually HA due to its proximity to the nitro group. HB and HC are the aromatic protons on the benzene ring, and their positions relative to the aniline (amino) group aids in their differentiation. Finally, HD is usually the most shielded proton (lowest chemical shift), as it is the amino proton.
Coupling helps to differentiate between otherwise closely overlapping protons. Due to the nature of the coupling constants involved, HA and HC will generally have distinct coupling patterns that allow them to be differentiated.
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(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g of C8H18 (a typical component of gasoline)? Page 253 (b) For part (a), the source of O2 is air, which is 78% N2, 21% O2, and 1.0% Ar by volume. Assuming all the O2 reacts, but no N2 or Ar does, what is the total volume (in L) of the engine’s gaseous exhaust?
Answers
Answer:
Part A
The volume of the gaseous product is [tex]V = 787L[/tex]
Part B
The volume of the the engine’s gaseous exhaust is [tex]V_e = 2178 \ L[/tex]
Explanation:
Part A
From the question we are told that
The temperature is [tex]T = 350^oC = 350 +273 =623K[/tex]
The pressure is [tex]P = 735 \ torr = \frac{735}{760} = 0.967\ atm[/tex]
The of [tex]C_8 H_{18} = 100.0g[/tex]
The chemical equation for this combustion is
[tex]2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}[/tex]
The number of moles of [tex]C_8 H_{18}[/tex] that reacted is mathematically represented as
[tex]n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }[/tex]
The molar mass of [tex]C_8 H_{18}[/tex] is constant value which is
[tex]M = 114.23 \ g/mole[/tex]
So [tex]n = \frac{100 }{114.23} }[/tex]
[tex]n = 0.8754 \ moles[/tex]
The gaseous product in the reaction is [tex]CO_2_{(g)}[/tex] and water vapour
Now from the reaction
2 moles of [tex]C_8 H_{18}[/tex] will react with 25 moles of [tex]O_2[/tex] to give (16 + 18) moles of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
So
1 mole of [tex]C_8 H_{18}[/tex] will react with 12.5 moles of [tex]O_2[/tex] to give 17 moles of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
This implies that
0.8754 moles of [tex]C_8 H_{18}[/tex] will react with (12.5 * 0.8754 ) moles of [tex]O_2[/tex] to give (17 * 0.8754) of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
So the no of moles of gaseous product is
[tex]N_g = 17 * 0.8754[/tex]
[tex]N_g = 14.88 \ moles[/tex]
From the ideal gas law
[tex]PV = N_gRT[/tex]
making V the subject
[tex]V = \frac{N_gRT}{P}[/tex]
Where R is the gas constant with a value [tex]R = 0.08206 \ L\cdot atm /K \cdot mole[/tex]
Substituting values
[tex]V = \frac{14.88* 0.08206 *623}{0.967}[/tex]
[tex]V = 787L[/tex]
Part B
From the reaction the number of moles of oxygen that reacted is
[tex]N_o = 0.8754 * 12.5[/tex]
[tex]N_o = 10.94 \ moles[/tex]
The volume is
[tex]V_o = \frac{10.94 * 0.08206 *623}{0.967}[/tex]
[tex]V_o = 579 \ L[/tex]
No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as
[tex]V_e = V_o * \frac{0.79}{0.21}[/tex]
Substituting values
[tex]V_e = 579 * \frac{0.79}{0.21}[/tex]
[tex]V_e = 2178 \ L[/tex]
The total volume of gaseous products produced from the combustion of 100g C8H18 is approximately 382 L. By considering the composition of air and that N2 and Ar do not react, you can calculate the total volume of the gaseous exhaust.
Upon seeing this question, the first step is to write the balanced chemical equation for the combustion of C8H18, which is ubiquitous in gasoline:
C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O
This looks complex, but the Ideal Gas Law simplifies everything. According to the stoichiometric ratio, 1 mole of C8H18 will produce 8 moles of CO2 and 9 moles of H2O when burnt, totalling 17 moles of gas. The molar mass of C8H18 is around 114 g/mol, so 100g is approximately 0.877 moles.
Therefore, upon combustion, 0.877 * 17 = 14.9 moles of gas are produced. Applying the Ideal Gas Law (PV=nRT), keeping in mind that 350°C is 623K and 1 atm is 760 torr, we have: (nR(623K)/735torr) = V. Plugging the numbers and making sure to keep the units consistent, we get V = 14.9 x 0.0821 x 623 / 0.967 = approximately 382 L.
Coming to part (b), considering the composition of air and assuming no N2 or Ar reacts, you find that for every L of O2 burned, 1L of air is consumed. Adding the total volume of gas from part (a) and the extra nitrogen and argon, you find the total volume of the engine's gaseous exhaust.
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Nonmetal oxides can react with water to form acids. For example, carbon dioxide reacts with water to form carbonic acid: CO2 H2O H2CO3. (1) Write an equation to show how diiodine pentaoxide reacts with water to form an acid. Do not include states. Balanced equation for reaction (smallest integer coefficients): (2) What is the name of the acid (include 'acid'):
Answers
Answer:
Explanation:
Diiodine pentoxide is I₂O₅
Reaction with water
I₂O₅ + H₂O = 2HIO₃
2 ) HIO₃ is called iodic acid.
Final answer:
Nonmetal oxides react with water to form acids. One example is the reaction of diiodine pentaoxide with water to produce iodic acid.
Explanation:
Nonmetal oxides react with water to form acids. One example is the reaction of diiodine pentaoxide with water to produce iodic acid.
(1) Equation:
Diiodine pentaoxide + Water → 2HI + HIO3
(2) Acid Name:
Iodic acid
Which statement is always true of the cathode in an electrochemical cell? Reduction occurs here. It is considered the "negative" electrode. Negative ions flow toward the cathode. Metal is plated out here. It is considered the "positive" electrode.
Answers
The cathode is where reduction occurs, it's regarded as the 'positive' electrode, and in electrochemical cells, it's where metal is plated out. Negative ions do not necessarily flow toward the cathode.
Explanation:In an electrochemical cell, some statements are always true for the cathode. The cathode is considered the "positive" electrode in the cell, it is where the reduction reaction (gain of electrons) occurs, and it is the location where metal is plated out (in electrolytic cells). Negative ions do not flow toward the cathode; ions flow based on their charge and the charge of the electrode. Positive ions (cations) move toward the cathode.
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Practice the Skill 21.15a When the following ketone is treated with aqueous sodium hydroxide, the aldol product is obtained in poor yields. In these cases, special distillation techniques are used to increase the yield of aldol product. Predict the aldol addition product that is obtained, and propose a mechanism for its formation. For the mechanism, draw the curved arrows as needed. Include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly in your products. Do not use abbreviations such as Me or Ph.
Answers
Answer:
OH OH ONa NaOH OH
Explanation:
See attached image
Match the terms in the first list with the characteristics in the second list.1. adsorption chromatography2. partition chromatography3. ion-exchange chromatography4. molecular exclusion chromatography5. affinity chromatography
A. Ions in mobile phase are attracted to counterions covalently attached to stationary phase
B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase
C. Solute equilibrates between mobile phase and surface of stationary phase
D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase
E. Different-sized solutes penetrate pores in stationary phase to different extents. Large solutes are eluted first.
Answers
Answer: Please See answers below
Explanation:
A. Ions in mobile phase are attracted to counterions covalently attached to stationary phase -----Ion Exchange Chromatography
B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase -----Affinity Chromatography
C. Solute equilibrates between mobile phase and surface of stationary phase -----Adsorption Chromatography
D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase --- Partition Chromatography
E. Different-sized solutes penetrate pores in stationary phase to different extents. Large solutes are eluted first. ---Molecular Exclusion Chromatography
A sample of a gas at room temperature occupies a volume of 40.0 L at a pressure of 262 torr.If the pressure changes to 1310 torr,with no change in the temperature or moles of gas,what is the new volume,V2
Answers
Answer:
a) p1v1= p2v2
852 x 40.0= 4260 x v2
v2= 8 L
b) 852 x 40 = p2 x 69.0
p2= 494 torr
While estimating the Caloric content of the chemical constituents present in the food item, the carbohydrate amount is discounted to some extent compared to the average value measured using bomb calorimetry. This is because the fiber content __________. Select the correct answer below: is not actually considered a carbohydrate is not digestible cannot be burned in the calorimeter is classified as a carbohydrate but stores energy like a fat.
Answers
Answer:
Fiber is "Not Digestible"
Explanation:
Carbohydrates that contain fiber cannot be completely digested. the indigestible components of fiber are measured in the calorimeter, but they are not accessible for energy in the human body.
While estimating the Caloric content of the chemical constituents present in the food item, the carbohydrate amount is discounted to some extent compared to the average value measured using bomb calorimetry. This is because the fiber content is not digestible.
How is calorie content measured?
At its most fundamental, a calorie is a measure of electricity. One Calorie (same as one kilocalorie, or 1,000 energy) is the quantity of power that is required to warm one kilogram of water 1 degree Celsius at sea stage. The strength content of meals become traditionally measured by the usage of bomb calorimetry.
As an alternative, the entire caloric price is calculated by using including the calories supplied by using the strength-containing nutrients: protein, carbohydrate, fats, and alcohol.
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Enter your answer in the provided box. Lead(II) nitrate is added slowly to a solution that is 0.0400 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol / L) required to initiate the precipitation of PbCl2.
(Ksp for PbCl2 is 2.40 × 10−4.)
Answers
Answer:
the concentration of Pb2+ ions is [tex][Pb^{2+}] } = 0.15 mol/L[/tex]
Explanation:
From the question we are told that
The concentration of [tex]Cl^{-}[/tex] is [tex]C_s = 0.0400 M[/tex]
The reaction is
[tex]Pb^{2+} + 2Cl^- ----> PbCl_2[/tex]
The solubility constant for this reaction is mathematically represented as
[tex]K_{sp} = [Pb^{2+}][Cl^{-}]^2[/tex]
Substituting values
[tex]2.4 0 * 10^{-4} = 0.0400^2 * [Pb^{2+}][/tex]
[tex][Pb^{2+}] } = \frac{2.4 0 * 10^{-4} }{0.0400^2 }[/tex]
[tex][Pb^{2+}] } = 0.15 mol/L[/tex]
One type of water softener is precipitation softening, also known as ion-exchange. Based on the reactants used (Na2CO3 and CaCl2), what ions would remain in the softened water that would be consumed by the homeowners? What could be some negative aspects of consuming these ions?
Answers
Answer:
Explanation:
In ion exchange softener , the minerals of hard water are in the form of ions of Mg⁺² and Ca⁺² . They are replaced by Na⁺ ion . The Na⁺ that resides on the bed of resin used in ion exchange softener make all heavy ions like Mg⁺² and Ca⁺² . In this way all the hard ions of hard water are removed and they are replaced by Na⁺ ion . hard ions are removed because they form insoluble compounds so they are precipitated out.
The ion that remains in soft water is Na⁺ ion.
The negative aspect of consuming this ion is that it is harmful for those suffering from heart ailment like heart pressure.
What happens to the temperature of the liquid in a cup of water as some of the water evaporates?
A the temperature increases because there are fewer molecules to share energy between
B the temperature decreases because new molecular bonds form
C the temperature decreases because the most energetic molecules escape
D the temperature increases because the vapor pressure increases
Answers
Answer: c: The temperature decreases because the most energetic molecules escape
Explanation:
Evaporation is surface phenomenon in which liquid molecules gain energy from surrounding molecules and thus these high energy molecules escape from the surface in the form of vapors thus leaving low energy molecules in the system. As the kinetic energy of the left molecules decreases, the temperature of the molecules decreases as kinetic energy is directly proportional to temperature. Thus low temperature results in cooling of the system.
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The carbon-oxygen bond in CO has a higher bond dissociation enthalpy than a carbon-oxygen bond in CO2. Which is the best explanation for this difference? Group of answer choices CO has a triple bond while each carbon-oxygen bond in CO2 is a double bond. CO has a lone pair on carbon while CO2 does not. CO is a polar molecule while CO2 is a nonpolar molecule. CO contains one carbon-oxygen bond while CO2 contains two carbon-oxygen bonds.
Answers
Answer:
CO HAS A TRIPLE BOND WHILE C-O BOND IN CO2 IS A DOUBLE BOND
CO HAS A LONE PAIR ON CARBON WHILE CO2 DOES NOT
Explanation:
Bond dissociation bond enthalpy or energy is the energy needed to break 1 mole of a divalent molecule into separate atoms mostly in the gaseous state.
The carbon and oxygen in carbon monoxide form a triple bond as carbon monoxide has 10 electrons in their outermost shell which results into six shared electrons in 3 bonding orbitals as against the double bond formed by other carbon compounds. Four electrons come from oxygen and the remaining two from carbon and due to this, two electrons from oxygen will occupy one orbital and this forms a dative bond. Also because of the triple bond, carbon monoxide is often regarded as a more stable compound than carbon dioxide with a double bond. This gives it its higher bond dissociation enthalpy value and more energy is needed to break it into its separate atoms. This is in conjunction with a larger bond length similar to the bong length in a triple bond. This makes it more stronger than the bond dissociation enthalpy of carbon dioxide having a double bond.
The cell potential for the cell Zn(s) + 2H+(? M) LaTeX: \longrightarrow⟶ Zn2+(1.3 M) + H2(g) (8 atm) is observed to be 0.68 V. What is the pH in the H+/H2 half-cell? Reduction potential for H2(g)/H+(aq) is 0.00 V, for Zn(s)/Zn2+(aq) is -0.76 V. Enter number to 2 decimal places.
Answers
Answer:
pH in the [tex]H_{2}/H^{+}[/tex] half cell is 0.84.
Explanation:
Oxidation: [tex]Zn(s)-2e^{-}\rightarrow Zn^{2+}(aq.)[/tex]
Reduction: [tex]2H^{+}(aq.)+2e^{-}\rightarrow H_{2}(g)[/tex]
-------------------------------------------------------
Overall: [tex]Zn(s)+2H^{+}(aq.)\rightarrow Zn^{2+}(aq.)+H_{2}(g)[/tex]
[tex]E_{cell}^{0}=E_{H^{+}\mid H_{2}}^{0}-E_{Zn^{2+}\mid Zn}^{0}[/tex] = (0.00 V) + (0.76 V) = 0.76 V
According to Nernst equation for this cell reaction at room temperature (298 K):
[tex]E_{cell}=E_{cell}^{0}-\frac{0.0592}{n}log\frac{[Zn^{2+}].P_{H_{2}}}{[H^{+}]^{2}}[/tex]
where, [tex]E_{cell}[/tex] is cell potential, n is number of electron exchanged, [tex]P_{H_{2}}[/tex] is pressure of [tex]H_{2}[/tex] in atm and species under third bracket represent molarity of the respective species.
So, [tex]0.68V=0.76V-\frac{0.0592}{2}log\frac{(1.3M)\times (8atm)}{[H^{+}]^{2}}V[/tex]
[tex]\Rightarrow[/tex] [tex][H^{+}]=0.1436M[/tex]
pH = [tex]-log[H^{+}][/tex] = -log(0.1436) = 0.84
Final answer:
The pH in the H+/H2 half-cell of the given galvanic cell is 8.5.
Explanation:
In the given galvanic cell, the reduction half-reaction is 2H+ (aq) + 2e → H₂(g), and the overall reaction is Zn(s) + 2H+ (aq) -> Zn²+ (aq) + H₂(g). The reduction potential for the H2(g)/H+(aq) half-reaction is 0.00 V, and for the Zn(s)/Zn²+(aq) half-reaction is -0.76 V.
To find the pH in the H+/H2 half-cell, we can use the Nernst equation:
Ecell = E°cell - (0.0592 V / n) * log[H+]
Using the given cell potential of 0.68 V and plugging in the values, we can calculate the pH in the H+/H2 half-cell to be 8.5.
What volume (mL) of a 15% (m/v) NaOH solution contains 120 g NaOH
Answers
Answer:
8.0*10^2
Explanation:
What volume (in milliliters) of a 15% (m/v) NaOH solution contains 120 g NaOH?
For 15% (m/v), the conversion factor can be 15%/100 mL or 100 mL/15%. In this case, 100 mL/15% is used because we need mL as our final expression. 100 mL/15% * 120 g NaOH= 800 mL NaOH. In scientific notation: 8.0*10² or 8.0e² mL NaOH.
To find the volume of a 15% (m/v) NaOH solution that contains 120 grams of NaOH, use the formula V = (mass of solute x 100) / %m/v, which results in 800 mL. Therefore, the correct answer is E) 800 mL.
To find the volume of a 15% (m/v) NaOH solution that contains 120 grams of NaOH, we will use the formula for mass/volume percentage concentration:
%m/v = (mass of solute/volume of solution) x 100%
Given:
Mass of NaOH (solute) = 120 g
Concentration of NaOH solution = 15% (m/v)
We need to find the volume of the solution (V).
Rearranging the formula to solve for V:
V = (mass of solute x 100%) / %m/v
Substituting the given values:
V = (120 g x 100%) / 15%
Converting the percentage:
V = (120 g x 100) / 15 = 12,000 / 15 = 800 mL
Therefore, 800 mL of a 15% NaOH solution contains 120 g NaOH. The correct answer is E) 800 mL.
What volume (mL) of a 15% (m/v) NaOH solution contains 120 g NaOH?
A) 18 mL
B) 0.13 mL
C) 13 mL
D) 120 mL
E) 800 mL
what is the mass of 1.25 moles of Na2O?
Answers
Answer:
Explanation:
I have the same question
IS PLUTO ROCKY OR GASY
Answers
Answer:
Rocky.
Explanation:
Jovian planets are giant gas balls not unlike the SUN although they have a small rocky central core. Pluto is a rock ice planet---more like Europa, a satellite of Jupiter. In fact, Pluto is probably the largest of the so-called KUIPER BELT objects
9. For each reaction listed, identify the proton donor (or acid) and the proton acceptor(or base).
Label each conjugate acid-base pair.
a. CH3COOH + H2O 2 H30° + CH3C00-
b. HCO3 + H2O = H2CO3 + OH
C. HNO3 + SO42- à HSO4 + NO3
Answers
Answer:
Explanation:
conjugate acid, based on Brønsted–Lowry acid–base theory, is a chemical compound that is formed by the reception of a proton by a base
a. CH₃COOH + H₂O ⇌ H₃0⁺ + CH₃C00-
Acid <> CH₃COOH
Base <> H₂O
Conjugate acid <> H₃0 +
Conjugate base <>CH₃C00-
b. HCO₃ + H₂O ⇌ H₂CO₃⁻ + OH⁻
Acid <> H₂O
Base <> HCO₃
Conjugate acid <> H₂CO₃⁻
Conjugate base <>OH⁻
C. HNO₃ + SO₄²⁻ ⇌ HSO₄⁻ + NO₃⁻
Acid <>HNO₃
Base <>SO₄²⁻
Conjugate acid <>HSO₄⁻
Conjugate base <>NO₃⁻
A Bronsted acid is reffered to as a proton donor while a Bronsted base is a proton acceptor
In the reaction CH3COOH + H2O -> H3O+ + CH3C00-, CH3COOH is the proton donor and H2O is the proton acceptor. In the reaction HCO3- + H2O -> H2CO3 + OH-, HCO3- is the proton donor and H2O is the proton acceptor. In the reaction HNO3 + SO42- -> HSO4- + NO3-, HNO3 is the proton donor and SO42- is the proton acceptor.
Explanation:a. In the reaction CH3COOH + H2O -> H30+ + CH3C00-, CH3COOH is the proton donor (acid) and H2O is the proton acceptor (base). The conjugate acid-base pairs are CH3COOH/CH3COO- and H3O+/H2O.
b. In the reaction HCO3- + H2O -> H2CO3 + OH-, HCO3- is the proton donor (acid) and H2O is the proton acceptor (base). The conjugate acid-base pairs are HCO3-/H2CO3 and H2O/OH-
c. In the reaction HNO3 + SO42- -> HSO4- + NO3-, HNO3 is the proton donor (acid) and SO42- is the proton acceptor (base). The conjugate acid-base pairs are HNO3/NO3- and HSO4-/SO42-
Compare the value you calculated for the number of sand grains on Earth’s beaches to the number of particles in a mole. What can you conclude about the size of a mole?
The number of particles in a mole is about ____(blank) the number of grains of sand on Earth’s beaches.
Answers
Final answer:
The number of particles in a mole is far greater than the number of grains of sand on Earth's beaches, showcasing the mole's significance in representing vast quantities of atoms or molecules in chemistry.
Explanation:
When comparing the number of grains of sand on Earth’s beaches to the number of particles in a mole, it becomes evident that a mole represents a vastly larger quantity. The number of particles in a mole is 6.022x10²³, which dwarfs the estimated 7.5x10¹¸ grains of sand on Earth's beaches. This expansive difference illustrates that the mole is an incredibly large unit, used to easily convey numbers of particles or atoms that would otherwise be incomprehensible in scale. A mole's enormity is further illustrated by examples such as a mole of paper sheets extending more than a million times the distance from the Earth to the Sun, or a mole of sand filling a cube about 32 kilometers on a side. Therefore, the number of particles in a mole is far greater than the number of grains of sand on Earth’s beaches, highlighting the mole's utility in chemistry for working with the huge numbers of atoms and molecules involved in even small chemical samples.
The number of particles in a mole is about [tex]\(7.5 \times 10^{18}\),[/tex] (blank) the number of grains of sand on Earth’s beaches.
The number of particles in a mole, also known as Avogadro's number, is approximately [tex]\(6.022 \times 10^{23}\).[/tex]
Comparing this to the estimated number of sand grains on Earth's beaches, which is on the order of [tex]\(7.5 \times 10^{18}\),[/tex] we can conclude that the number of particles in a mole is significantly larger than the number of sand grains on Earth's beaches.
In fact, the number of particles in a mole is approximately 80 million times greater than the number of sand grains on Earth's beaches.
This stark difference highlights the immense size of a mole, indicating that it represents an enormous quantity of particles.
Therefore, a mole encompasses an incredibly large number of particles, underscoring its importance in chemistry and the sciences.
What is the H* concentration in a solution with a
pH of 1.25?
x 10" Mn=
Answers
Answer:
[H+] = 0.0562 M
Explanation:
Message me for more help.
H⁺ concentration in a solution with a pH of 1.25 is 0.0562.
What is pH?pH may be defined as a measure of acidity of a substance.
More precisely, pH is defined as the negative logarithm of the hydrogen ion concentration.
pH = - log [H⁺]
The range of pH extends from zero to 14.
A pH value of 7 is neutral, because pure water has a pH value of exactly 7. Values lower than 7 are acidic; values greater than 7 are basic or alkaline.
pH measurements can be made using litmus paper or pH paper known to change colors around a certain pH value.
A universal indicator is a mixture of indicator solutions intended to provide a color change over a pH range of 2 to 10.
pH measurements can also be done with pH meters that give the exact value of pH.
Given,
pH = 1.25
We know that, pH = - log [H⁺]
[H⁺] = antilog [ -pH]
[H⁺] = antilog [ -1.25]
= 0.0562 M
Therefore, H⁺ concentration in a solution with a pH of 1.25 is 0.0562.
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What is the molarity (concentration) of 2.53 L of NaOH which exactly neutralizes 5.0L of 1.25 M HCl?
Answers
Answer:
The concentration of NaOH is 2.47 M
Explanation:
The equation of the reaction is
HCl (aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)
Given;
Concentration of acid CA= 1.25 M
Volume of acid VA= 5.0L
Concentration of base CB= ????
Volume of base VB= 2.53 L
Number of moles of acid NA= 1
Number of moles of base NB= 1
From;
CAVA/CBVB= NA/NB
CAVANB= CBVBNA
CB= CAVANB/VBNA
CB= 1.25 × 5.0 ×1 / 2.53×1
CB= 2.47 M
Therefore the concentration of NaOH is 2.47 M
Choose the selection which correctly characterizes all three of the following substances in terms of whether they are polar or nonpolar: SiH4 and BBr3 and SiF4 a) SiH4 is nonpolar and BBr3 is polar and SiF4 is nonpolar. b) SiH4 is nonpolar and BBr3 is polar and SiF4 is polar. c) SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is polar. d) SiH4 is polar and BBr3 is nonpolar and SiF4 is polar. e) SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is nonpolar.
Answers
Answer:
SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is nonpolar.
Explanation:
SiH4 is a non-polar compound. Though the Si–H bonds are polar, as a result of different electronegativities of Si and H. However, as there are 4 electron repulsions around the central Si atom, the polar bonds are arranged symmetrically around the central atom having a tetrahedral shape hence they cancel out making the compound nonpolar.
SiF4 is a nonpolar molecule because the fluorine atoms are arranged symetrically around the central silicon atom in a tetrahedral molecule with all of the regions of negative charge cancelling each other out just like in SiH4.
The 3 bromine atoms all lie in the same plane thus the geometry of the compound will be trigonal planar. The BBr3 will be non polar because the three B-Br bonds will cancel out each others' dipole moment given that they are in the same plane.
A non-polar chemical is SiH4. Despite the fact that the Si-H bonds are polar, this is because Si and H have differing electronegativities.
Thus, The polar bonds are symmetrically organized around the center Si atom, which has a tetrahedral form, and thus cancel out because there are four electron repulsions around it, rendering the combination nonpolar.
Because the fluorine atoms are symmetrically positioned around the center silicon atom in a tetrahedral molecule with all of the negative charge regions canceling each other out, SiF4 is a nonpolar molecule and electronegativities.
The geometry of the compound will be trigonal planar since all three bromine atoms are located in the same plane. The three B-Br bonds will cancel, making the BBr3 non-polar.
Thus, A non-polar chemical is SiH4. Despite the fact that the Si-H bonds are polar, this is because Si and H have differing electronegativities.
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2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would be produced?
how should i go about solving this equation
Answers
Answer:
1.822 g of magnesium hydroxide would be produced.
Explanation:
Balanced reaction: [tex]2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl[/tex]
Compound Molar mass (g/mol)
NaOH 39.997
[tex]MgCl_{2}[/tex] 95.211
[tex]Mg(OH)_{2}[/tex] 58.3197
So, 2.50 g of NaOH = [tex]\frac{2.50}{39.997}[/tex] mol of NaOH = 0.0625 mol of NaOH
4.30 g of [tex]MgCl_{2}[/tex] = [tex]\frac{4.30}{95.211}[/tex] mol of [tex]MgCl_{2}[/tex] = 0.0452 mol of [tex]MgCl_{2}[/tex]
According to balanced equation-
2 mol of NaOH produce 1 mol of [tex]Mg(OH)_{2}[/tex]
So, 0.0625 mol of NaOH produce [tex](\frac{0.0625}{2})[/tex] mol of NaOH or 0.03125 mol of NaOH
1 mol of [tex]MgCl_{2}[/tex] produces 1 mol of [tex]Mg(OH)_{2}[/tex]
So, 0.0452 mol of [tex]MgCl_{2}[/tex] produce 0.0452 mol of [tex]Mg(OH)_{2}[/tex]
As least number of moles of [tex]Mg(OH)_{2}[/tex] are produced from NaOH therefore NaOH is the limiting reagent.
So, amount of [tex]Mg(OH)_{2}[/tex] would be produced = 0.03125 mol
= [tex](0.03125\times 58.3197)[/tex] g
= 1.822 g
To find the grams of magnesium hydroxide produced, you need to balance the chemical equation and calculate stoichiometry.
To find the grams of magnesium hydroxide produced, we need to balance the chemical equation and calculate the stoichiometry.
The balanced chemical equation for the reaction is:
2NaOH + MgCl2 → Mg(OH)2 + 2NaCl
Using the molar masses of sodium hydroxide (NaOH), magnesium chloride (MgCl2), and magnesium hydroxide (Mg(OH)2), we can calculate the grams of magnesium hydroxide produced:
Calculate the number of moles of sodium hydroxide and magnesium chloride using their respective masses and molar masses.Use the stoichiometry of the balanced equation to determine the moles of magnesium hydroxide produced.Convert the moles of magnesium hydroxide to grams using its molar mass.After performing these calculations, we find that 4.59g of magnesium hydroxide would be produced.
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