Answer:
step 1 is trimolecular while step 2 is bimolecular.
Explanation:
Molecularity of an elementary reaction is defined the number of molecules that come together to react in a given elementary (single-step) reaction. It is equal to the sum of stoichiometric coefficients of reactants in that elementary reaction. Depending on the number of molecules that come together in an elementary step, a reaction can be designated as; unimolecular, bimolecular or trimolecular.
The kinetic order of any elementary reaction or reaction step is equal to its molecularity, and the rate equation of an elementary reaction can easily be determined by inspection, from the molecularity.
For a complex (multistep) reaction, the kinetic order of reaction is not determined from the molecularity since molecularity only describes elementary reactions or steps.
From our discussion above we can see that, step 1 is trimolecular while step 2 is bimolecular.
When methanol, CH3OH,CH3OH, is burned in the presence of oxygen gas, O2,O2, a large amount of heat energy is released. For this reason, it is often used as a fuel in high performance racing cars. The combustion of methanol has the balanced, thermochemical equation CH3OH(g)+32O2(g)⟶CO2(g)+2H2O(l)ΔH=−764 kJ CH3OH(g)+32O2(g)⟶CO2(g)+2H2O(l)ΔH=−764 kJ How much methanol, in grams, must be burned to produce 665 kJ665 kJ of heat?
To find the amount of methanol required to generate 665 kJ of heat, calculate the heat produced per gram of methanol burned. The balanced equation helps determine the amount needed. In this case, burning 27.9 grams of methanol will produce 665 kJ of heat.
To calculate the amount of methanol needed to produce 665 kJ of heat, we first need to determine the amount of heat produced per gram of methanol burned. From the balanced equation CH₃OH(g) + 3/2O₂(g) ⟶ CO₂(g) + 2HO(l), we see that 764 kJ of heat is released per 1 mol of methanol burned. This corresponds to 32 g of methanol. Therefore, to produce 665 kJ of heat, we would need to burn:
(665 kJ) * (32 g / 764 kJ) = 27.9 g of methanol.
So, 27.9 grams of methanol must be burned to produce 665 kJ of heat.
3. A volume of 90 mL of 0.2 M HBr neutralizes
a 60 mL sample of NaOH solution. What is
the concentration of the NaOH solution?
Answer:
0.3M
Explanation:
Step 1:
Data obtained from the question. This include the followingb:
Volume of acid (Va) = 90mL
Concentration of acid (Ca) = 0.2M
Volume of base (Vb) = 60mL
Concentration of base (Cb) =....?
Step 2:
The balanced equation for the reaction. This is given below:
HBr + NaOH —> NaBr + H2O
From the balanced equation above,
The mole ratio of the acid (nA) = 1
The mole ratio of the base (nB) = 1
Step 3:
Determination of the concentration of the base, NaOH.
The concentration of the base can be obtained as follow:
CaVa /CbVb = nA/nB
0.2 x 90 / Cb x 60 = 1
Cross multiply
Cb x 60 = 0.2 x 90
Divide both side by 60
Cb = 0.2 x 90 /60
Cb = 0.3M
Therefore, the concentration of the base, NaOH is 0.3M
Answer:
[tex]M_{NaOH}=0.3M[/tex]
Explanation:
Hello,
In this case, since we are talking about a neutralization reaction, the moles of acid must equal the moles of base as shown below:
[tex]n_{HBr}=n_{NaOH}[/tex]
Thus, in terms of molarities we've got:
[tex]M_{HBr}V_{HBr}=M_{NaOH}V_{NaOH}[/tex]
This is possible since HBr reacts with NaOH in a 1:1 molar ratio:
[tex]HBr+NaOH\rightarrow NaBr+H_2O[/tex]
Hence, for the given concentration and volume of hydrobromic acid and the volume of sodium hydroxide, we compute its concentration as shown below:
[tex]M_{NaOH}=\frac{M_{HBr}V_{HBr}}{V_{NaOH}} =\frac{90mL*0.2M}{60mL} \\\\M_{NaOH}=0.3M[/tex]
Best regards.
Periodic table
Problem
Given the following reaction:
\qquad \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}CH
4
+2O
2
→CO
2
+2H
2
Ostart text, C, H, end text, start subscript, 4, end subscript, plus, 2, start text, O, end text, start subscript, 2, end subscript, right arrow, start text, C, O, end text, start subscript, 2, end subscript, plus, 2, start text, H, end text, start subscript, 2, end subscript, start text, O, end text
How many grams of \text{CO}_2CO
2
start text, C, O, end text, start subscript, 2, end subscript will be produced from 12.0 \text{ g}12.0 g12, point, 0, start text, space, g, end text of \text{CH}_4CH
4
start text, C, H, end text, start subscript, 4, end subscript and 133 \text{ g}133 g133, start text, space, g, end text of \text{O}_2O
2
start text, O, end text, start subscript, 2, end subscript?
grams (round to three significant figures)
Answer:
33g of CO2
Explanation:
Step 1:
The balanced equation for the reaction.
CH4 + 2O2 —> CO2 + 2H2O
Step 2:
Determination of the masses of CH4 and O2 that reacted and the mass of CO2 produced from the balanced equation.
This is illustrated below:
Molar Mass of CH4 = 12 + (4x1) = 16g/mol
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 2 x 32 = 64g
Molar Mass of CO2 = 12 + (2x16) = 44g/mol.
Summary:
From the balanced equation above,
16g of CH4 reacted with 64g of O2 to produce 44g of CO2.
Step 3:
Determination of limiting reactant. This is illustrated below:
From the balanced equation above,
16g of CH4 reacted with 64g of O2.
Therefore, 12g of CH4 will react with = (12 x 64)/16 = 48g of O2.
From the above illustration, a lesser mass of O2 i.e 48g than what was given i.e 133g is required to react completely with 12g of CH4.
Therefore, CH4 is the limiting reactant and O2 is the excess reactant.
Step 4:
Determination of the mass of CO2 produced from the reaction.
Here, we shall use the limiting reactant, because it will give the maximum yield of CO2 as all of it is used up in reaction.
This is illustrated as follow:
From the balanced equation above,
16g of CH4 reacted to produce 44g of CO2.
Therefore, 12g of CH4 will react to produce = (12 x 44)/16 = 33g of CO2.
From the calculations made above, 33g of CO2 will be produced from the reaction of 12g of CH4 and 133g of O2.
How many grams of HF are needed to react with 3.0 moles of Sn? *
Answer:
120g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
Sn + 2HF —> SnF2 + H2
Next, we shall determine the number of mole of HF needed to react with 3 moles of Sn.
From the balanced equation above, 1 mole of Sn reacted with 2 moles of HF.
Therefore, 3 moles of Sn will react with = 3 x 2 = 6 moles of HF.
Finally, we shall convert 6moles of HF to grams
This is illustrated below:
Number of mole of HF = 6moles
Molar Mass of HF = 1 + 19 = 20g/mol
Mass of HF =..?
Mass = number of mole x molar Mass
Mass of HF = 6 x 20
Mass of HF = 120g
Therefore, 120g of HF is needed to react with 3 moles of Sn
does a mirror breaks up light into all the colors of the rainbow
Mirrors reflect light without separating it into colors, unlike prisms or water droplets which cause dispersion, leading to the formation of rainbows. Dispersion occurs when light passes through a medium that slows down colors differentially. Historical scientists like al-Farisi and Theodoric of Freiberg demonstrated this with water droplets.
The answer is no; a typical mirror reflects light without separating it into its constituent colors. This property allows a mirror to create an image with the same color composition as the original scene.
Now, the separation of white light into its constituent colors—an effect known as dispersion—occurs when light passes through a medium that causes different colors to travel at different speeds. This is how prisms or water droplets in the atmosphere create beautiful rainbows. It relies on the refraction of light, which bends the light as it moves from one medium to another, like air into water, and does so at varying degrees depending on the wavelength (color) of the light. However, while mirrors can exhibit total reflection due to refraction, they do not inherently disperse white light.
Historically, figures like Kamal al-Din al-Farisi and Theodoric of Freiberg conducted experiments which confirmed that water droplets are responsible for decomposing white light into the colors of the rainbow. Their work built on foundational optics principles from Ibn al-Haytham.
Pure R,R-hydrobenzoin (side product of meso-hydrobenzoin) has a melting point of 148°C. a) What do you expect the melting point of pure S,S-hydrobenzoin (another side product of meso-hydrobenzoin) to be? b) Do you expect the racemic mixtures of these enantiomers to have the same or different melting points as the pure enantiomers?
(a) The melting point of S-hydroxybenzoic should be 140°C.
(b) The Racemic mixture must have a diverse melting point in comparison to the pure enantiomers.
What is Meso-hydroxybenzoic?
(a) When The melting point of S, S-hydroxybenzoic should be 140°C. Since S, S-hydrozoan, and also R, R-hydroxybenzoic is enantiomeric pair their melting point and then boiling point should be the same.
(b) The different melting points,
When The racemic mixture must have a different melting point in comparison to the pure enantiomers.
considering a racemic mixture, a particular enantiomer possesses a greater affinity for its variety of molecules than for those of the different enantiomer
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Final answer:
The pure S,S-hydrobenzoin is expected to have a melting point similar to the R,R-hydrobenzoin at 148°C. Racemic mixtures containing both enantiomers typically have different and lower melting points compared to the pure enantiomers because they disrupt the crystal lattice.
Explanation:
Enantiomers and Melting Points
The melting point of the pure R,R-hydrobenzoin is 148°C. For the pure S,S-hydrobenzoin, which is the enantiomer of R,R-hydrobenzoin, we would expect the melting point to be very similar, if not identical, because enantiomers typically have the same physical properties including melting points.
Racemic Mixtures and Melting Points
When it comes to racemic mixtures, which contain equal amounts of both enantiomers, they often have melting points that are different, and usually lower, than those of the pure enantiomers. This is because the presence of both enantiomers in a racemic mixture can disrupt the crystal lattice, leading to a less orderly structure that requires less energy (heat) to disrupt.
Which organ(s) would be responsible for this trait? Select two options.
Answer:
The ovaries and the testes
Explanation:
The organs responsible for the development of secondary sex characteristics would be the ovaries and the testes.
Both ovary and testes are associated with the sex organs in the body and are responsible for the synthesis of hormones such as androgen and estrogen. While the latter (estrogen) is produced by the ovary, the former (androgen) is produced by the testes.
More specifically, both androgen and estrogen are hormones responsible for the growth of hairs in areas such as the armpit and the pubic regions of the body.
The correct options are the ovaries and the testes.
Which statement describes both homogeneous mixtures and heterogeneous mixtures?
A.
Their components can be distinguished visually.
B.
Their components are combined in a fixed ratio.
C.
Their components can be separated by physical processes.
D.
They always contain a liquid component.
E.
They are considered to be pure substances.
Final answer:
Homogeneous mixtures and heterogeneous mixtures can be distinguished visually and their components can be separated by physical processes.
Explanation:
A. Their components can be distinguished visually.
C. Their components can be separated by physical processes.
Both homogeneous mixtures and heterogeneous mixtures can be distinguished visually and their components can be separated by physical processes. In a homogeneous mixture, the components are uniformly distributed and cannot be visually distinguished. Examples of homogeneous mixtures include salt dissolved in water and air. In contrast, a heterogeneous mixture has components that are not uniform and can be visually distinguished. Examples of heterogeneous mixtures include sand and water and mixed nuts. Both types of mixtures can be separated into their individual components through physical processes such as filtration or evaporation.
What type of reaction is this?
AB - A+B
Single Replacement
Double Replacement
Decomposition
Combustion
Synthesis
A box of jello has a mass of 250 g. How many boxes must be bought to have 1 Kg of jello? *
2
3
4
6
Answer:
Number of boxes = 4
Explanation:
Given:
Mass of one box of jello = 250 grams
Total quantity want to purchase = 1 kg = 1 × 1,000 gram = 1,000 grams
Find:
Number of boxes in 1,000 grams = ?
Computation:
Number of boxes = Total quantity want to purchase / Mass of one box of jello
Number of boxes = 1,000 / 250
Number of boxes = 4
Therefore, 4 boxes of jello must be purchase to get 1 kg of Jello.
Calculate the concentration of IO−3 in a 1.65 mM Pb(NO3)2 solution saturated with Pb(IO3)2 . The Ksp of Pb(IO3)2 is 2.5×10−13 . Assume that Pb(IO3)2 is a negligible source of Pb2+ compared to Pb(NO3)2 . [IO−3]= M A different solution contains dissolved NaIO3 . What is the concentration of NaIO3 if adding excess Pb(IO3)2(s) produces a Pb2+ concentration of 5.60×10−6 M ?
Answer:
See explaination
Explanation:
1) Pb(NO3)2 => Pb2+ + 2 NO3-
[Pb2+] = [Pb(NO3)2] = 7.56 mM = 7.56 x 10-3 M
Pb(IO3)2 <=> Pb2+ + 2 IO3-
Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13
7.56 x 10-3 x [IO3-]2 = 2.5 x 10-13
[IO3-] = 5.75 x 10-6 M ≈ 5.8 x 10-6 M
(2) [Pb2+] = 1.7 x 10-6 M
Ksp = [Pb2+][IO3-]2 = 2.5 x 10-13
1.7 x 10-6 x [IO3-]2 = 2.5 x 10-13
[IO3-] = 3.83 x 10-4 M
[IO3-]from Pb(IO3)2 = 2 x [Pb2+]
= 2 x 1.7 x 10-6 = 3.4 x 10-6 M
[IO3-]from NaIO3 = [IO3-] - [IO3-]from Pb(IO3)2
= 3.83 x 10-4 - 3.4 x 10-6
= 3.80 x 10-4 M
NaIO3 => Na+ + IO3-
[NaIO3] = [IO3-]from NaIO3
= 3.80 x 10-4 M ≈ 3.8 x 10-4 M
Answer:
1) Concentration of IO−3 in a 1.65 mM Pb(NO3)2 solution:
[tex].[IO^{-3}]=1.228*10^{-5} M[/tex]
2) Concentration of NaIO3:
[tex].[NAIO3][/tex]=[tex]2*10^{-4} M[/tex]
Explanation:
1) Concentration of IO−3 in a 1.65 mM Pb(NO3)2 solution:
The reaction will be:
[tex]Pb(NO3)2[/tex] ⇒ [tex]Pb^{+2} +2NO^{-3}[/tex]
[tex]Concentration\ of\ Pb^{+2}=Pb(NO3)2=1.65*10^{-3}\ M[/tex]
Now,
[tex]Pb(IO3)2[/tex] ⇄ [tex]Pb^{+2}+2IO^{-3}[/tex]
Ksp=Concentration of [tex]Pb^{+2}[/tex] * (Concentration of [tex]2IO^{-3}[/tex])^2
[tex]Ksp=[Pb^{+2}][/tex]*[tex][2IO^{-3}]^2[/tex]
[tex]2.5*10^{-13}=1.65*10^{-3}*[IO^{-3}]^2\\.[IO^{-3}]^2=\frac{2.5*10^{-13}}{1.65*10^{-3}} \\.[IO^{-3}]^2=1.51*10^{-10} M\\.[IO^{-3}]=1.228*10^{-5} M[/tex]
2) Concentration of NaIO3:
Now,
[[tex]Pb^{+2}[/tex]]=[tex]5.60*10^{-6} M[/tex]
[tex]Ksp=[Pb^{+2}]*[2IO^{-3}]^2[/tex]
[tex]2.5*10^{-13}=5.60*10^{-6} *[IO^{-3}]^2\\.[IO^{-3}]^2=\frac{2.5*10^{-13}}{5.60*10^{-6}} \\.[IO^{-3}]^2=4.46*10^{-8} M\\.[IO^{-3}]=2.112*10^{-4} M[/tex]
Again:
[tex]Concentration\ of\ IO^{-3} from\ Pb(IO3)2 = 2* Concentration\ of\ Pb^{+2}\\.[IO^{-3}]_{From\ Pb(IO3)2}=2*5.60*10^{-6}\\.[IO^{-3}]_{From\ Pb(IO3)2}=1.12*10^{-5} M[/tex]
Calculating the concentration of [tex]IO^{-3}[/tex] from NaIO3:
[tex]Concentration\ of\ IO^{-3}_{[From\ NaIO3}]=[IO^{-3}]-Concentration\ of\ IO^{-3}_{[From\ Pb(IO3)2}]\\Concentration\ of\ IO^{-3}_{[From\ NaIO3}]=2.112*10^{-4}-1.12*10^{-5}\\Concentration\ of\ IO^{-3}_{[From\ NaIO3}]=2*10^{-4} M[/tex]
Reaction:
[tex]NaIO3=>Na^{+}+IO^{-3}[/tex]
Concentation of NaIO3= Concentration of [tex]IO^{-3}[/tex]
[tex].[NAIO3][/tex]=[tex]2*10^{-4} M[/tex]
Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H2O).
Upper C upper O (g) plus 3 upper H subscript 2 (g) double-headed arrow upper C upper H 4 (g) plus upper H subscript 2 upper O (g).
The reaction is at equilibrium at 1,000 K. The equilibrium constant of the reaction is 3.90. At equilibrium, the concentrations are as follows.
[CO] = 0.30 M
[H2] = 0.10 M
[H2O] = 0.020 M
What is the equilibrium concentration of CH4 expressed in scientific notation?
.0059
5.9 x 10-2
0.059
5.9 x 102
Answer:
5.9x10^-2 M
Explanation:
Step 1:
Data obtained from the question. This includes the following:
Concentration of CO, [CO] = 0.30 M
Concentration of H2, [H2] = 0.10 M
Concentration of H2O, [H2O] = 0.020 M
Equilibrium constant, K = 3.90
Concentration of CH4, [CH4] =..?
Step 2:
The balanced equation for the reaction. This is given below:
CO(g) + 3H2(g) <=> CH4(g) + H2O(g)
Step 3:
Determination of the concentration of CH4.
The expression for equilibrium constant of the above equation is given below:
K = [CH4] [H2O] / [CO] [H2]^3
3.9 = [CH4] x 0.02/ 0.3 x (0.1)^3
Cross multiply to express in linear form
[CH4] x 0.02= 3.9 x 0.3 x (0.1)^3
Divide both side by 0.02
[CH4] = 3.9 x 0.3 x (0.1)^3 /0.02
[CH4] = 5.9x10^-2 M
Therefore, the equilibrium concentration of CH4 is 5.9x10^-2 M
Answer:
B
Explanation:
Why is the endocrine system important?
Select the appropriate statements and arrange them in a logical order to explain the bonding in BeF2. i. Each of the singly occupied hybrid orbitals on Be overlaps with a singly occupied 2p orbital on an F atom. ii. The unhybridized 2p orbitals on Be interact with the 2p orbitals on the F atoms to form pi bonds. iii. The 2s and one of the 2p orbitals on Be combine to form two sp hybrid orbitals. iv. A 2s electron is promoted to a 2p orbital on Be. v. The 2s electrons in Be are promoted to 2p orbitals. vi. Each of the singly occupied 2p orbitals on Be overlaps with a singly occupied 2s orbital on a F atom.a.v, vi, ii b.v,vi c.iii, i, ii d.iv, iii, i e.iv, iii, i, ii
The bonding in BeF2 involves the promotion of a 2s electron to a 2p orbital on Be, followed by sp hybridization of the Be orbitals, and overlap with the F atom 2p orbitals to form two sigma bonds. So the correct statements are iv, iii, and i.
Explanation:To explain the bonding in BeF2, we must select the appropriate statements and arrange them in a logical order. The correct sequence of statements is:
iv. A 2s electron is promoted to a 2p orbital on Be.iii. The 2s and one of the 2p orbitals on Be combine to form two sp hybrid orbitals.i. Each of the singly occupied hybrid orbitals on Be overlaps with a singly occupied 2p orbital on an F atom.Firstly, a 2s electron in Be is promoted to a 2p orbital (Statement iv), creating two singly occupied orbitals. These orbitals then undergo hybridization to form two equivalent sp hybrid orbitals (Statement iii), which are oriented at a 180° angle to each other. Each sp hybrid orbital overlaps with a singly occupied 2p orbital on a fluorine atom to form a sigma bond (Statement i), involving the pairing up of the Be valence electrons with the unpaired electron on each F atom. This results in the formation of two identical Be-F sigma bonds, giving BeF2 a linear molecular geometry.
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when aqueous solutions of sodium hydroxide and iron (II) chloride react, which substance will precipitate? 2NaOH + FeCl2 = 2NaCl + Fe(OH)2
Answer:
Fe(OH)2
Explanation:
Iron hydroxides are chemical compounds that appear as precipitates after alkalizing solutions containing iron salts, both in their valence or degree of oxidation (III) and (II).
Hydroxide, in both cases, is a gelatinous colloid of difficult filtration that can be considered a hydrated oxide, iron (II) hydroxide is whitish in color and requires more alkalinization, above 7.
Show the curved-arrow mechanism for the Claisen condensation of ethyl ethanoate treated with ethoxide ion. Include all formal charges and nonbonding electrons. In each step, draw only the species that react in that step. If an enolate resonance form is possible, draw only the carbanionic form. If a CHx group is being deprotonated, draw all H\'s on the reacting site. Tip: always omit ethanol byproducts.
Answer:
See explaination
Explanation:
Please kindly check out the attached files for the curved-arrow mechanism for the Claisen condensation of ethyl ethanoate treated with ethoxide ion.
The Claisen condensation of ethyl ethanoate with ethoxide ion involves the formation of a β-keto ester. The reaction proceeds through a step-by-step mechanism, where the ethoxide ion acts as a strong base and abstracts a proton from the α-carbon of ethyl ethanoate, leading to the formation of an enolate ion. The enolate ion then attacks the carbonyl carbon of another molecule of ethyl ethanoate, resulting in the formation of a β-keto ester.
Explanation:The Claisen condensation is a reaction that involves the combination of two ester molecules in the presence of a strong base, resulting in the formation of a β-keto ester. In this specific case, the Claisen condensation involves the reaction of ethyl ethanoate (ester) with ethoxide ion (strong base). Let's go through the step-by-step mechanism of the reaction:
First, the ethoxide ion (CH3CH2O−) acts as a strong base, abstracting a proton from the α-carbon of the ethyl ethanoate. This leads to the formation of an enolate ion (negative charge on the α-carbon).In the second step, the enolate ion attacks the carbonyl carbon of another molecule of ethyl ethanoate. This results in the formation of a new carbon-carbon bond and the formation of a β-keto ester.It is important to note that in both steps, the species involved in the reaction are the ethoxide ion and the ethyl ethanoate molecules. The formal charges and nonbonding electrons can be indicated on the atoms involved. Also, it is mentioned to omit ethanol byproducts from the drawings.
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1. Copper (__) is an element on the periodic table.
CU (both letters capitalized)
Cu (first letter capitalized)
cU (second letter capitalized)
cu (both letters lowercase)
Answer: Cu
Explanation: It is Cu because the origin of the word Copper comes from the latin word "Cuprum".
Which represents an empirical formula?
A. C2H4
B. B2H6
C. Al2O3
D. C6H6
Answer:
C. Al2O3
Explanation:
The empirical formula is an expression that represents the simplest proportion in which the atoms that form a chemical compound are present. It is therefore the simplest representation of a compound.
For options A, B and D the empirical formulas are as follows:
A. CH2
B. BH3
D. CH
Disaccharide A is a reducing sugar that forms one equivalent of D-galactose and one equivalent of D-glucose on hydrolysis with aqueous acid. Reaction of A with iodomethane and Ag2O yields an octamethyl derivative, which can be hydrolyzed with aqueous acid to give one equivalent of 2,3,4,6-tetra-O-methyl-D-glucopyranose and one equivalent of 2,3,6-tri-O-methyl-D-galactopyranose. If A contains an beta-glycoside link, what is its structure
Answer: see the graphic
The gas phase decomposition of dimethyl ether at 500 °C CH3OCH3(g) → CH4 (g) + H2 (g) + CO (g) is first order in CH3OCH3 with a rate constant of 4.00×10-4 s-1 How much time in seconds is required for 85.2% of the CH3OCH3 initially present in a reaction flask to be converted to product at this temperature? (enter a numerical value, don't worry about the units, put exponents as e#. For example, 4.00×10-4 would be 4.00e-4)
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
To find the time it would take for 85.2% of the dimethyl ether to decompose in a first order reaction with a rate constant of 4.00e-4 s-1, we use the first order reaction equation. Substituting the known values, we solve for the time to be approximately 4142 seconds.
Explanation:The question asks for the time required to decompose 85.2% of an initial amount of dimethyl ether. This is a first order reaction with the rate constant provided as 4.00e-4 s-1.
In first order reactions, we can use the following equation to determine the time:
t = -1/k × ln([A]t/[A]0)
where:
t is the time (which we are solving for)k is the rate constant[A]0 is the initial concentration[A]t is the concentration at time tSince we’re looking for the time it takes for 85.2% of the substance to decompose, [A]t is 100% - 85.2% = 14.8% = 0.148 [A]0. Substituting the known values, we get:
t = -1/(4.00e-4 s-1) × ln(0.148) ≈ 4142 seconds
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Place the following in order of increasing magnitude of lattice energy
CaO, MgO, SrS
Answer:
SrS<CaO<MgO
Explanation:
Lattice energy is defined as a measure of the energy contained in the crystal lattice of a compound. It can be seen as the energy that would be released if the component ions were brought together from infinity.
Lattice energy depends on size of the ions. The smaller the ions, the greater the lattice energy. When we look at all the actions in the substances listed, we will notice that the order of decrease in cation size is Sr^2+>Ca^2+>Mg^2+. The lattice energy increases in the this order.
Hence MgO will have the greatest lattice energy of all the species listed.
The order of increasing magnitude of lattice energy is; SrS < CaO < MgO
Definition:
Lattice energy is simply defined as the energy required to convert one mole of an ionic solid into gaseous ionic constituents.
Lattice energy is dependent on the size of the constituent atoms.
The greater the distance between the nucleus of the constituent atoms, the lesser is the lattice energy.
Therefore, the order of increasing magnitude of lattice energy is;
SrS < CaO < MgORead more;
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True statements:1) deltaH for endothermic reaction is positive3) when the energy is transferred as heat from system to surroundings, deltaH is negative.6) A combustion reaction is exothermic .False statements:2) deltaH for exothermic reaction is positive. actually it is negative4) when the energy is transferred as heat from surroundings to system, deltaH is negative. actually it is positive5) the evaporation of water is an exothermic process. actually it is an endothermic process.
Answer:
Delta H for endothermic reaction is positive-True. This is because an endothermic reaction absorbs heat energy, therefore more energy is retained inside the product of the reaction system than the reactants, as the value of deltaH is greater than Zero.
Delta H for an exothermic reaction is positive. This is false. Because in an exothermic reaction heat is liberated to the surrounding environment. therefore the value of thus the outer environment contains more energy than the internal environments, thus the enthalpy of the reactants is greater than that of the products.
when the energy is transferred as heat from system to surroundings, deltaH is negative. True . This is true because the surrounding environment gain heat energy, (positive)while the system loses it,(negative) therefore delta H is negative.
when the energy is transferred as heat from surroundings to system, deltaH is negative. False. This is positive, because now the environments loses heat, (negative) while the systems gains heat,( positive) therefore delta H of the system is positive. endothermic
the evaporation of water is an exothermic process-False, This is an endothermic reaction in which water molecules need to gain heat energy from the surrounding environments to increase the average kinetic energy of collusion to escape the intermolecular forces to escape as steam.
Combustion reaction is exothermic. True., because heat energy is transferred to the surrounding from the internal system. The energy needed for the formation of new bonds in the products is higher than the energy for breaking of original bonds in the reactants. Thus more heat is liberated.
Explanation:
What is the pH of solution if the concentration of [H3O+] = 0.000558?
3.25
7.45
5.58
1.00
Answer:
3.25
Explanation:
pH = -log[H3O+]
-log(0.000558)=3.25
The energy content of food is typically determined using a bomb calorimeter. Consider the combustion of a 0.22-g sample of butter in a bomb calorimeter having a heat capacity of 2.67 kJ/°C. If the temperature of the calorimeter increases from 23.5°C to 26.1°C, calculate the energy of combustion per gram of butter. Energy of combustion = kJ/g
Answer: The energy of combustion of butter is 31.5 kJ/g
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=C\times \Delta T[/tex]
Q = Heat absorbed by calorimeter =?
C = heat capacity of calorimeter = [tex]2.67kJ/^0C[/tex]
Initial temperature of the calorimeter = [tex]T_i[/tex] = [tex]23.5^0C[/tex]
Final temperature of the calorimeter = [tex]T_f[/tex] = [tex]26.1^0C[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=(26.1-23.5)^0C=2.6^0C[/tex]
Putting in the values, we get:
[tex]Q=2.67kJ/^0C\times 2.6^0C=6.94kJ[/tex]
As heat absorbed by calorimeter is equal to heat released by combustion of butter
[tex]Q=q[/tex]
Heat released by 0.22 g of butter = 6.94 kJ
Heat released by 1g of butter = [tex]\frac{6.94}{0.22}\times 1=31.5kJ[/tex]
The energy of combustion of butter is 31.5 kJ/g
To find the energy of combustion per gram of butter, you first compute the total energy using the formula q = C*deltaT, and then divide this total energy by the mass of the butter. The calculated energy of combustion per gram of butter in this case is 31.55 kJ/g.
Explanation:The energy of combustion per gram of butter is calculated by first finding the total energy produced, which is done using the formula: q = C*deltaT where q is the heat energy, C is the heat capacity, and deltaT is the change in temperature. Therefore, q = (2.67 kJ/°C) * (26.1°C - 23.5°C) = 6.94kJ
To find the energy of combustion per gram, you divide the total energy by the mass of the butter. So, Energy of combustion per gram = 6.94 kJ / 0.22 g = 31.55 kJ/g.
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Standard reduction potentials are 1.455 V for the PbO2(s)/Pb(s) couple, 1.82 V for Co3 (aq)/Co2 (aq), 3.06 V for F2(g)/HF(aq), 1.07 V for Br2(l)/Br-(aq), and 1.77 V for H2O2(aq)/H2O(l). Under standard-state conditions, arrange the oxidizing agents in order of decreasing strength.
Answer:
F2(g)/HF(aq)>Co3 (aq)/Co2 (aq)> H2O2(aq)/H2O(l)> PbO2(s)/Pb(s)>Br2(l)/Br-(aq)
Explanation:
The tendency of any specie to function as oxidizing agent is a highly dependent on the reduction potential of the couple. The more positive the value of the reduction potential of the couple, the better it does as an oxidizing agent.
This implies that we could know a good oxidizing agent by looking at their respective reduction potentials. The couple having the greatest (most positive) reduction potential is selected as the best oxidizing agent. If there are a number of couples at having different reduction potentials, the order of oxidizing ability can be obtained by arranging the species in order of decreasing positive reduction potentials just as we have done in the answer above.
Decreasing strength will be:
F₂(g)/HF(aq)>Co₃ (aq)/Co₂ (aq)> H₂O₂(aq)/H₂O(l)>PbO₂(s)/Pb(s)>Br₂(l)/Br-(aq)
Trend for Oxidizing agents:The tendency of any specie to function as an oxidizing agent is highly dependent on the reduction potential of the couple. The more positive the value of the reduction potential of the couple, the better it does as an oxidizing agent.
The couple having the greatest (most positive) reduction potential is selected as the best oxidizing agent.
Thus, the decreasing strength for oxidizing agents will be:
F₂(g)/HF(aq)>Co₃ (aq)/Co₂ (aq)> H₂O₂(aq)/H₂O(l)>PbO₂(s)/Pb(s)>Br₂(l)/Br-(aq)
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Identify the reactant that is a Brønsted−Lowry acid in the following reaction: HI(aq)+H2O(l)→I−(aq)+H3O+(aq) Express your answer as a chemical formula. nothing Request Answer Part B Identify the reactant that is a Brønsted−Lowry base in the following reaction: HI(aq)+H2O(l)→I−(aq)+H3O+(aq) Express your answer as a chemical formula. nothing Request Answer Part C Identify the reactant that is a Brønsted−Lowry acid in the following reaction: F−(aq)+H2O(l)⇌HF(aq)+OH−(aq) Express your answer as a chemical formula. nothing Request Answer Part D Identify the reactant that is a Brønsted−Lowry base in the following reaction: F−(aq)+H2O(l)⇌HF(aq)+OH−(aq) Express your answer as a chemical formula.
Answer:
(1) HI is Bronsted-Lowry acid and [tex]H_{2}O[/tex] is Bronsted-Lowry base.
(2) [tex]H_{2}O[/tex] is Bronsted-Lowry acid and [tex]F^{-}[/tex] is Bronsted-Lowry base.
Explanation:
A Bronsted-Lowry acid is a species which donates proton ([tex]H^{+}[/tex]).
A Bronsted-Lowry base is a species which accepts proton ([tex]H^{+}[/tex]).
(1) [tex]HI(aq)+H_{2}O(l)\rightarrow I^{-}(aq)+H_{3}O^{+}(aq)[/tex]
Reactants: HI and [tex]H_{2}O[/tex]
Here HI donates proton and [tex]H_{2}O[/tex] accepts proton.
Hence HI is Bronsted-Lowry acid and [tex]H_{2}O[/tex] is Bronsted-Lowry base.
(2) [tex]F^{-}(aq)+H_{2}O(l)\rightleftharpoons HF(aq)+OH^{-}(aq)[/tex]
Reactants: [tex]F^{-}[/tex] and [tex]H_{2}O[/tex]
Here [tex]H_{2}O[/tex] donates proton and [tex]F^{-}[/tex] accepts proton.
Hence [tex]H_{2}O[/tex] is Bronsted-Lowry acid and [tex]F^{-}[/tex] is Bronsted-Lowry base.
Let's consider Brønsted−Lowry acid-base theory:
An acid is a species that donates H⁺.A base is a species that accepts H⁺.We have 4 equations and we want to identify which reactant is the acid or the base.
What are the reactants?The substance(s) to the left of the arrow in a chemical equation are called reactants.
Part A. Identify the reactant that is a Brønsted−Lowry acid in the following reaction:HI(aq) + H₂O(l) → I⁻(aq) + H₃O⁺(aq)
HI donates H⁺ to H₂O, so it is an acid.
Part B. Identify the reactant that is a Brønsted−Lowry base in the following reaction:HI(aq) + H₂O(l) → I−(aq) + H₃O⁺(aq)
H₂O accepts H⁺ from HI, so it is a base.
Part C. Identify the reactant that is a Brønsted−Lowry acid in the following reaction:F⁻(aq) + H₂O(l) ⇌ HF(aq) + OH⁻(aq)
H₂O donates H⁺ to F⁻, so it is an acid.
Part D. Identify the reactant that is a Brønsted−Lowry base in the following reaction:F⁻(aq) + H₂O(l) ⇌ HF(aq) + OH⁻(aq)
F⁻ accepts H⁺ from H₂O, so it is a base.
In the reaction HI(aq) + H₂O(l) → I⁻(aq) + H₃O⁺(aq), HI is a Brønsted−Lowry acid and H₂O a Brønsted−Lowry base.In the reaction F⁻(aq) + H₂O(l) ⇌ HF(aq) + OH⁻(aq) , H₂O is a Brønsted−Lowry acid and F⁻ a Brønsted−Lowry base.Learn more about Brønsted−Lowry acid-base theory here: https://brainly.com/question/7256753
4. To how much water should 50 mL of 12 M hydrochloric acid be added to
produce o 40 M solution?
Answer : The volume of water added are, 15 mL
Explanation :
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the initial molarity and volume of HCl.
[tex]M_2\text{ and }V_2[/tex] are the final molarity and volume of water.
We are given:
[tex]M_1=12M\\V_1=50mL\\M_2=40M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]12M\times 50mL=40M\times V_2\\\\V_2=15mL[/tex]
Hence, the volume of water added are, 15 mL
The balanced equation below shows the products that are formed when pentane (C5H12) is combusted.
C5H12 + 802 → 10CO2 + 6H20
What is the mole ratio of oxygen to pentane?
Answer:
8 : 1
Explanation:
The balanced equation for the reaction is given below:
C5H12 + 8O2 → 5CO2 + 6H2O
From the balanced equation above,
1 mole of C5H12 reacted with 8 moles of O2.
Thus the mole ratio of O2 to C5H12 is:
8 : 1
In an Honors organic chemistry lab, a student devised an experiment in which she would treat benzoic acid with t-butanol in an acid-catalyzed esterification reaction using concentrated sulfuric acid. Regretfully, the synthetic yield of the expected ester was exceedingly low. Please explain this outcome in terms of the chemistry that actually occurred in the reaction flask. [This attempted esterification reaction would have been better suited to dicyclohexylcarbodiimide (DCC/pyridine) esterification conditions.]
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
1. Take the reaction: NH3 + O2 + NO + H2O. In an experiment, 3.25g of NH3 are allowed
to react with 8.12g of O.
c. How many grams of NO are formed?
Answer:
5.74g of NO
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
4NH3 + 5O2 —> 4NO + 6H2O
Step 2:
Determination of the masses of NH3 and O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:
Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol
Mass of NH3 from the balanced equation = 4 x 17 = 68g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 5 x 32 = 160g
Molar Mass of NO = 14 + 16 = 30g/mol
Mass of NO from the balanced equation = 4 x 30 = 120g
From the balanced equation above,
68g of NH3 reacted with 160g of O2 to produce 120g of NO.
Step 3:
Determination of the limiting reactant.
We need to determine the limiting because it will be used to calculate the maximum yield of the reaction. This is illustrated below:
From the balanced equation above,
68g of NH3 reacted with 160g of O2.
Therefore, 3.25g of NH3 will react with = (3.25 x 160)/68 = 7.65g of O2.
From the simple illustration above, we can see that lesser mass of O2 is needed to react with 3.25g of NH3. Therefore, NH3 is the limiting reactant while O2 is the excess reactant.
Step 4:
Determination of the mass of NO produced from the reaction.
In this case the limiting reactant will be used because all of it were used in the reaction.
The limiting reactant is NH3.
From the balanced equation above,
68g of NH3 reacted to produce 120g of NO.
Therefore, 3.25g of NH3 will react to produce = (3.25 x 120)/68 = 5.74g of NO.
From the calculations made above, 5.74g of NO is produced.