Answer:
E. 8.08 x 10⁴.
Explanation:
Hello,
In this case, for the reaction:
[tex]CH_2O(g) + 2H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)[/tex]
We can compute the Gibbs free energy of reaction via:
[tex]\Delta G\°=\Delta H\°-T\Delta S\°[/tex]
Since both the entropy and enthalpy of reaction are given at 298 K (standard temperature), therefore:
[tex]\Delta G\°=-94.9kJ-(298K)(-224.2\frac{J}{K}*\frac{1kJ}{1000kJ} )\\\\\Delta G\°=-28.1kJ[/tex]
Then, as the equilibrium constant is computed as:
[tex]K=exp(-\frac{\Delta G\°}{RT} )[/tex]
We obtain:
[tex]K=exp(-\frac{-28.1kJ/mol}{8.314x10^{-3}\frac{kJ}{mol* K}}*298K )\\\\K=8.08 x10^4[/tex]
For which the answer is E. 8.08 x 10⁴.
Best regards,
Select the correct answer. A certain reaction has this form: aA bB. At a particular temperature and [A]0 = 2.00 x 10-2 Molar, concentration versus time data were collected for this reaction and a plot of ln[A]t versus time resulted in a straight line with a slope value of -2.97 x 10-2 min-1. What is the half-life of this reaction? A. 23.33 seconds B. 0.043 minutes C. 0.0043 seconds D. 23.33 minutes E. 1680 minutes
Answer:
[tex]\large \boxed{\text{D. 23.34 min}}[/tex]
Explanation:
1. Find the order of reaction
Use information from the graph to find the order.
If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.
2. Find the half-life
[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]
The half life of the reaction is 23.33 minutes.
We know that for a first order reaction;
ln[A]t = ln[A]o - kt
A plot of ln[A]t against time (t) will yield a straight line graph with a slope of -k.
From the question, the slope is -2.97 x 10-2 min-1.
So, -2.97 x 10-2 min-1 = - k
k = 2.97 x 10-2 min-1
The half life of a first order reaction is obtained from;
t1/2 = 0.693/k
t1/2 = 0.693/2.97 x 10-2 min-1
t1/2 = 23.33 minutes
Learn more: https://brainly.com/question/3902440
An oxide has a chemical formula with the form X2O3. Which group is element X more likely to be a member of? Select the correct answer below: group 12 group 13 group 14 group 2
Answer:
Group 13
Explanation:
You know X has 3 valence electrons, as oxygen has a subscript of 3. This means X has an ionic charge of +3. Group 13 consists mainly of metalloids but it also has metals such as aluminum, which has a +3 charge. If you use aluminum as an example, you know that when combined with oxygen, it forms Al2O3. Group 12 has transition metals that don't have +3 ionic charges, group 14 has metalloids, metals that don't have ionic charges of +3, and nonmetals, and group 2 has metals with ionic charges of +2. Group 13 is the answer.
4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?
Answer: 9.08 L
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles[/tex]
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
According to stoichiometry :
4 moles of [tex]Al[/tex] require = 3 moles of [tex]O_2[/tex]
Thus 0.54 moles of [tex]Al[/tex] will require=[tex]\frac{3}{4}\times 0.54=0.405moles[/tex] of [tex]O_2[/tex]
Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.
According to the ideal gas equation:
[tex]PV=nRT[/tex]
P = Pressure of the gas = 1 atm
V= Volume of the gas = ?
T= Temperature of the gas = 273 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 0.405
[tex]V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L[/tex]
Thus 9.08 L of [tex]O_2[/tex] at STP would be required
Considering the reaction stoichiometry and STP conditions, 9.072 L of O₂ at STP would be required.
The balanced reaction is:
4 Al + 3 O₂ → 2 Al₂O₃
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Al: 4 moles O₂: 3 moles Al₂O₃: 2 moles
Being 27 g/mole the molar mass of Al, this is the amount of mass that a substance contains in one mole, then if 14.6 grams Al are reacted, the number of moles of Al that react is calculated as:
[tex]14.6 gramsx\frac{1 mole}{27 grams}= 0.54 moles[/tex]
Then you can apply the following rule of three: if by stoichiometry 4 moles of Al react with 3 moles of O₂, 0.54 moles of Al react with how many moles of O₂?
[tex]amount of moles of O_{2} =\frac{0.54 moles of Alx3 moles of O_{2} }{4 moles of Al}[/tex]
amount of moles of O₂= 0.405 moles
On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.
Then you can apply the following rule of three: if by definition of STP 1 mole of O₂ occupies 22.4 L, 0.405 moles of O₂, how much volume does it occupy?
[tex]volume=\frac{0.405 moles of O_{2}x22.4 L }{1 mole of O_{2} }[/tex]
volume= 9.072 L
Finally, 9.072 L of O₂ at STP would be required.
Learn more:
brainly.com/question/16487206?referrer=searchResults brainly.com/question/14446695?referrer=searchResults brainly.com/question/11564309?referrer=searchResults brainly.com/question/4025026?referrer=searchResults brainly.com/question/18650135?referrer=searchResultsA certain reaction has the following general form. aA → bB At a particular temperature and [A]0 = 2.80 ✕ 10−3 M, concentration versus time data were collected for this reaction, and a plot of 1/[A] versus time resulted in a straight line with a slope value of +3.40 ✕ 10−2 L mol−1 s−1. (a) Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. (Rate expressions take the general form: rate = k . [A]a . [B]b.) rate law:
Answer:
Rate law: [tex]r=k[A]^2[/tex]
Integrated rate law: [tex]\frac{1}{[A]}=kt+ \frac{1}{[A]_0}[/tex]
Rate constant: [tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]
Explanation:
Hello,
In this case, since the slope is obtained by plotting 1/[A] and it has the units L/(mol*s) or 1/(M*s), we can infer the reaction is second-order, therefore, its rate law is:
[tex]r=k[A]^2[/tex]
The integrated rate law:
[tex]\frac{1}{[A]}=kt+ \frac{1}{[A]_0}[/tex]
That is obtained from the integration of:
[tex]\frac{d[A]}{dt}=-k[A]^2[/tex]
And of course, since the slope equals the rate constant, its value is:
[tex]k=3.40x10^{-2}\frac{L}{mol*s}[/tex]
Regards.
When hydrogenation of two alkenes produce the same alkane, the more stable alkene has the___________ smaller heat of hydrogenation.
Explanation:
Heat of hydrogenation of alkenes is a measure of the stability of carbon-carbon double bonds.
In general, the lower the value of the heat of hydrogenation the more stable the double bond of the alkene.
Also, heat of hydrogenation of alkenes always have a negative value.
In a mixture of argon and hydrogen, occupying a volume of 1.66 L at 910.0 mmHg and 54.9oC, it is found that the total mass of the sample is 1.13 g. What is the partial pressure of argon
Answer:
Partial pressure (Ar) = 316.1mmHg
Explanation:
In the mixture of Ar and H₂ you can find the total moles of both gases using general gas law and with the mass of the sample and molar weight of each gas find the mole fraction of Argon and thus, its partial pressure.
Moles of gases:
PV = nRT
P = 910.0mmHg ₓ (1atm / 760mmHg) = 1.1974atm
V = 1.66L
n = Moles gases
R = 0.082atmL/molK
T = 54.9°C + 273.15K = 328.05K
PV = nRT
1.1974atm*1.66L = n*0.082atmL/molK*328.05K
0.0739 moles = total moles of the sampleKnowing H₂ = 2.016g/mol and Ar = 39.948g/mol you can write:
1.13g = 2.016X + 39.948Y (1)
Where X = moles of hydrogen and Y = moles of Argon.
Also we can write:
0.0739moles = X + Y (2)
Total moles of the sample are moles of hydrogen + moles Argon
Replacing 2 in 1:
1.13g = 2.016(0.0739-Y) + 39.948Y
1.13 = 0.1564 - 2.016Y + 39.948Y
0.9736 = 37.932Y
0.02567 = Y = moles of Argon
As total moles are 0.0739moles, mole fraction of Ar in the sample are:
XAr = 0.02567mol / 0.0739mol
X Ar = 0.347
Last, partial pressure of Ar = X Ar * total pressure.
Partial pressure (Ar) = 0.347*910.0mmHg
Partial pressure (Ar) = 316.1mmHg
Which of the following is an endothermic process? Question 1 options: work is done by the system on the surroundings heat energy flows from the system to the surroundings work is done on the system by the surroundings heat energy is evolved by the system none of the abo g
Answer:
work is done on the system by the surroundings
Explanation:
An endothermic process is any process in which energy is absorbed by the system from the surrounding, usually in the form of heat energy. Energy intake is usually associated with bond braking, and the energy from the surrounding goes into breaking the chemical bond between atoms int the reaction. The quantity of heat absorbed in any endothermic process can be calculated for by analyzing the Gibbs free energy of the reaction system. Endothermic process is mostly associated with heat energy intake by the system from the surrounding, but can also be used to refer to any system that undergoes any form of energy input into the system, from the surrounding even in the form of work.
A reaction mixture at 175 K initially contains 522 torr of NO and 421 torr of O2. At equilibrium, the total pressure in the reaction mixture is 748 torr. Calculate Kp at this temperature. Express your answer to three significant figures.
Answer:
[tex]Kp=0.0386[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]
For which the equilibrium expression is:
[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]
Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:
[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]
And the total pressure:
[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]
Yet it is 748 torr, for which the extent is:
[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]
Therefore, Kp turns out:
[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]
Best regards.
A chemist prepares a solution of barium acetate by measuring out of barium acetate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.
The given question is incomplete. The complete question is :
A chemist prepares a solution of barium acetate by measuring out 32 g of barium acetate into a 350 ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's barium acetate solution. Round your answer to significant digits.
Answer: The concentration of barium acetate solution is 0.375 mol/L
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in ml
moles of [tex]Ba(CH_3COO)_2[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{32g}{255g/mol}=0.125mol[/tex]
Now put all the given values in the formula of molality, we get
[tex]Molarity=\frac{0.125\times 1000}{350ml}[/tex]
[tex]Molarity=0.357M[/tex]
Therefore, the concentration of solution is 0.375 mol/L
For the carbonate ion, CO3 2− 1- Draw the electron orbital diagram for the valence electrons of the central carbon before and after hybridization. 2- identify which carbon and oxygen electron orbitals overlap to create each single and double C-O bond in the structure
Answer:
See explanation below
Explanation:
Carbon has four electrons in its outermost shell. The CO3^2- anion is found to be in the trigonal planar geometry. For a carbon atom in the trigonal planar geometry, the carbon is sp2 hybridized. This implies that an s orbital mixes with two p orbitals to yield the hybrid orbitals in the ion.
Carbon forms three double bonds to three oxygen atoms using these hybrid sp2 orbitals. Recall that the actual bonding in each C-O linkage lies between that of a pure C-O single bond and C-O double bonds.
Note that there are two p orbitals and one s orbital participating in this hybridization hence three hybrid orbitals are expected to be formed.
Draw the Lewis structure of acetaldehyde (CH₃CHO) and then choose the appropriate pair of molecular geometries of the two central atoms.
a) tetrahedral/trigonal pyramidal.
b) trigonal planar / linear.
c) tetrahedral / trigonal planar.
d) trigonal/tetrahedral.
e) planar /trigonal planar.
Given that,
Draw the Lewis structure of acetaldehyde (CH₃CHO).
We know that,
The Lewis structure shows the number of electrons around an atom.
According to structure,
We need to find the molecular geometries of the two central atoms
Using molecular geometries
For first central atom,
Number of bond pair = 2
Here, double bond to O count as single bond
The number of lone pair is zero.
The geometry is Trigonal planar.
For second central atom,
Number of bond pair = 4
The number of lone pair is zero.
The geometry is tetrahedral
Hence, The molecular geometries of the two central atoms are trigonal planar and tetrahedral.
(d) is correct option.
The central carbon atoms in acetaldehyde have a tetrahedral geometry and a trigonal planar geometry respectively.
Acetaldehyde has two central carbon atoms. The Lewis structure of acetaldehyde shows the arrangement of electrons around the atoms in the compound. The lone pairs are shown as dots while the bond pairs are represented using a single dash.
The first central carbon atom in acetaldehyde has a tetrahedral geometry while the second central carbon atom in acetaldehyde has a trigonal planar geometry.
Learn more: https://brainly.com/question/7558603
Which of the following is a property of salts? Undergo combustion Do not make ionic bonds easily Do not conduct electricity as solids Formed due to reaction of acid with water
Answer:
Do not conduct electricity as solids.
Explanation:
Hello,
In this case, we should remember that salts are formed when an acid and base react in order to yield the salt and water due to the ions exchange during neutralization chemical reactions. For instance, when hydrochloric acid (acid) reacts with potassium hydroxide (base), sodium chloride (salt) and water are yielded via:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
Moreover, it is widely known that salts are formed by electrovalent/ionic bonds which involves electron transfer so the metallic atom becomes positively charged (cation) whereas the non-metallic atom becomes negatively charged (anion) once the electrons are received so it can conduct electricity when dissolved in water yet not when solid since electron transfer is facilitated by the aqueous media, otherwise, ions remain together. Thereby, answer is do not conduct electricity as solids.
Regards.
Answer:
c
Explanation:
Predict the product of the following Wittig reaction. Be sure your answer accounts for stereochemistry, where appropriate. If multiple stereoisomers form, be sure to draw all products using appropriate wedges and dashes.
1. PPh3
5-iodo-1-phenyl-1-pentanone →
2. n- BuLi
Answer:
Final product: cyclopent-1-en-1-ylbenzene
Explanation:
In this case, we have a Wittig reaction. The addition of [tex]PPh_3[/tex] and n-Buli will produce the "Ylide compound". First, we will have an Sn2 reaction in which the iodide is replaced by triphenylphosphine. Then the base n-Buli will remove a hydrogen atom to form a double bond (Ylide compound). Then the double bond will be delocalized to produce a carbanion. This carbanion, will attack the carbon in the carbonyl group generating a negative charge in the oxygen. Then the negative charge will attack the phosphorous atom to produce a cyclic structure. Finally, the cyclic structure is broken to produce the alkene (cyclopent-1-en-1-ylbenzene).
See figure 1
I hope it helps!
When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4. The balanced equation is:
Answer:
2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3
Explanation:
The general schemefor a reaction is given as;
Reactants --> Products
In this question, the reactants are AgNO3 and Na2SO4. The product is Ag2SO4.
The equation is given as;
AgNO3 + Na2SO4 --> Ag2SO4
The other poduct formed in this reaction is NaNO3.
The full reaction is given as;
AgNO3 + Na2SO4 --> Ag2SO4 + NaNO3
The above reaction is not balanced because there are unequal number of atoms of the elements on both sides of the reaction.
The balanced equation is given as;
2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3
In this equation, there are equal number of moles of the atoms on both sides.
What compound is formed when 2,2-dimethyloxirane (2-methyl-1,2-epoxypropane) is treated with ethanol containing sulfuric acid
Answer:
2-ethoxy-2-methylpropan-1-ol
Explanation:
On this reaction, we have an "epoxide" (2-methyl-1,2-epoxypropane). Additionally, we have acid medium (due to the sulfuric acid [tex]H_2SO_4[/tex]). The acid medium will produce the hydronium ion ([tex]H^+[/tex]). This ion would be attacked by the oxygen of the epoxide. Then a carbocation would be produced, in this case, the most stable carbocation is the tertiary one. Then an ethanol molecule acts as a nucleophile and will attack the carbocation. Finally, a deprotonation step takes place to produce 2-ethoxy-2-methylpropan-1-ol.
See figure 1
I hope it helps!
please tell me the formula for these types of questions
Answer:
The half life of the radio - active isotope is 8 hours
Explanation:
We can tell that half - life of this radio - active isotope will be the time span with which 100 cpm of the substance remains, as half of 100 cpm is 200 cpm. When 25 cm remains it takes the duration of 24 hours / 1 day.
25 cm / 200 cm = 1 / 8
Therefore 1 / 8 of the substance remains after 24 hours. We want to calculate the time it takes for 1 / 2 of the substance to remains, which should clearly be less than 24 hours,
1 / 2 [tex]*[/tex] 1 / 2
24 / 3 = 8 hours - three half lives fit into 1 / 8, and hence 24 / 3 = 8 hours. We can check this solution by considering this 8 hours. After 8 hours one - half of the substance remains, or 100 cpm. After another 8 hours one - half of 100 cpm remains, or 50 cpm. And after another 8 hours one - half of 50 cpm remains, or 25 cpm. 3, 8 hours is a duration of 24 hours - the remaining amount being 25 cpm.
Write the molecular equation and net ionic equation for the reaction of hydroiodic acid and potassium hydroxide. Include phases (states). Enter the formula for water as H2O .
The molecular equation is :[tex]HI(aq)+KOH(aq) -- > KI (aq) + H_2O (l)[/tex]
The net ionic equation is: [tex]H^+(aq) + OH^-(aq) -- > H_2O(l)[/tex]
Reaction between hydroiodic acid and potassium hydroxide:When hydroiodic acid reacts with potassium hydroxide, this will result in the formation of a salt i.e. potassium iodide, and water is obtained as a by-product.
The molecular equation can be represented as:
[tex]HI(aq)+KOH(aq) -- > KI (aq) + H_2O (l)[/tex]
The net-ionic equation can be represented as:
[tex]H^+(aq) + OH^-(aq) -- > H_2O(l)[/tex]
Find more information about molecular equation here:
brainly.com/question/1603500
A molecular equation is a balanced chemical equation that represents a chemical reaction by showing the complete chemical formulas of all reactants and products involved. The net ionic equation for the reaction is as follows:
[tex]H+(aq) + OH-(aq) = H_2O(l)[/tex]
The molecular equation for the reaction between hydroiodic acid (HI) and potassium hydroxide (KOH) can be written as follows:
[tex]HI(aq) + KOH(aq) = KI(aq) + H_2O(l)[/tex]
In this equation, (aq) represents aqueous solutions, indicating that the substances are dissolved in water, and (l) represents a liquid phase for water.
To write the net ionic equation, we need to remove the spectator ions that do not participate in the actual chemical reaction. In this case, potassium ion (K+) and iodide ion (I-) are spectator ions, meaning they appear on both sides of the equation without undergoing any change.
Therefore, the net ionic equation for the reaction is as follows:
[tex]H+(aq) + OH-(aq) = H_2O(l)[/tex]
In the net ionic equation, H+ represents the hydrogen ion from hydroiodic acid, and OH- represents the hydroxide ion from potassium hydroxide. These ions combine to form water ([tex]H_2O[/tex]) as the only product.
For more details regarding molecular equations, visit:
https://brainly.com/question/14286552
#SPJ6
What is the name of this molecule?
Answer:
[tex]\boxed{Butyne}[/tex]
Explanation:
Triple Bonds => So it is an alkyne
The suffix used will be "-yne"
4 Carbons => The prefix used will be "But-"
Combining the prefix and suffix, we get:
=> Butyne
Answer:
[tex]\boxed{\mathrm{Butyne}}[/tex]
Explanation:
Alkynes have triple bonds ≡. The molecule has one triple bond.
Suffix ⇒ yne
The molecule has 4 carbon atoms and 6 hydrogen atoms.
Prefix ⇒ But (4 carbons)
The molecule is Butyne.
[tex]\mathrm{C_4H_6}[/tex]
A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the percent mass of the tin and percent by mass of oxygen in the sample
Answer:
Percentage mass of Tin = 96.3%
Percentage mass of oxygen = 6.40%
Explanation:
The product of the reaction is an oxide of tin.
Assuming all of the 0.500 g sample of tin reacted with oxygen to produce the oxide:
Mass of oxide = 0.534 g
Mass of tin present in the oxide = 0.500 g
Mass of oxygen in the oxide = 0.534 g of oxide - 0.500 g Sn = 0.034 g O
Percentage composition = mass of element/mass of compound × 100%
Percentage composition of Sn = 0.500 g/0.534 g × 100 = 93.6% Sn
Percentage composition of oxygen = 0.034 g/0.534 g × 100 = 6.40%
The following initial rate data apply to the raction
F2(g) + 2Cl2O(g) ---> 2FClO2(g) +Cl2(g)
Expt. [F2] (M) [Cl2O] (M) Intitial rate (M/s)
1 0.05 0.010 5 x 10^-4
2 0.05 0.040 2.0 x 10^-3
3 0.10 0.010 1.0 x 10^-3
Which of the following is the rate law (rate equation) for this reaction?
A. rate= k[F2]^2 [Cl2O]^4
B. rate= k[F2]^2 [Cl2O]
C. rate= k[F2] [Cl2O]
D. rate= k[F2] [Cl2O]^2
E. rate= k[F2]^2 [Cl2O]^2
Answer:
C. rate = k[F₂] [Cl₂O]
Explanation:
Based on the reaction, rate law can be obtained from the initial concentration of reactants thus:
rate = k[F₂]ᵃ [Cl₂O]ᵇ
Where the exponents a and b can be finded doing a experiment changing initial concentrations and seeing how a variation contribute in rate law.
If you analize experiments 1 and 2, the only change is [Cl₂O] (From 0.010 to 0.040, four times more) that changes its concentration in four times. This change produce rate law change from 5x10⁻⁴ to 2.0x10⁻³, also four times. That means the exponent b of [Cl₂O] is 1.
rate = k[F₂]ᵃ [Cl₂O]ᵇ
rate = k[F₂]ᵃ [Cl₂O]¹
Now, comparing experiments 1 and 3, the [F₂] change from 0.05 to 0.10, (Twice), and initial rate change from 5x10⁻⁴ to 1x10⁻³ (Also, twice). That means a = 1 and rate law is:
rate = k[F₂]¹ [Cl₂O]
rate = k[F₂] [Cl₂O]
Thus, right answer is:
C. rate = k[F₂] [Cl₂O]What mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A? Express your answer using two significant figures.
Answer
mass of aluminum metal= 7 .0497g of Al
Explanation:
current = 21 A
time = 1 hour = 60 X 60 = 3600 s
quantity of electricity passed = current X time = 21X 3600 = 75600 C
Following the electrolysis the below reaction will occur :
Al3+ + 3e- --------> Al
therefore, 3F i.e. 3 X 96500 C = 289500 C gives 1 mole of Al
so 1 C will produce 1/289500 moles of Al
so 108000 C will produce 1/289500 X 75600 = 0.2611 moles of Al
now 1 mole of aluminium weighs = 27 g/mole
so 0.2611 moles of Al = 0.2611 X 27 = 7 .0497 g
mass of aluminum metal= 7 .0497 g of Al
The mass of aluminum metal can be produced per hour in the electrolysis of a molten aluminum salt by a current of 21 A is 7.05 g
We'll begin by calculating the the quantity of electricity used. This can be obtained as follow:
Current (I) = 21 A
Time(t) = 1 h = 60 × 60 = 3600 s
Quantity of electricity (Q) =?Q = it
Q = 21 × 3600
Q = 75600 CFinally, we shall determine the mass of the aluminum metal produced. Al³⁺ + 3e —> AlRecall:
1 mole of Al = 27 g
1 electron (e) = 96500 C
Thus,
3 electrons = 3 × 96500 = 289500 C
From the balanced equation above,
289500 C of electricity produced 27 g of Al.
Therefore,
75600 C of electricity will produce = (75600 × 27) / 289500 = 7.05 g of Al
Thus, the mass of the aluminum metal obtained is 7.05 g
Learn more: https://brainly.com/question/25626152
Provide the reagents necessary to carry out the following conversion. Group of answer choices KMnO4, NaOH,H2O KMnO4, H3O , 75oC H2SO4, heat 1. mCPBA 2. H3O none of these
Answer:
KMnO4,H3O^+,75°C
Explanation:
The conversion of cyclohexene to trans-1,2-cylohexanediol is an oxidation reaction. Alkenes are oxidized in the presence of potassium permanganate and acids to yield the corresponding diols.
These diols may also be called glycols. They are molecules that contain two -OH(hydroxyl) groups per molecule. The reaction closely resembles the addition of the two -OH groups of hydrogen peroxide to an alkene.
The bright color of potassium permanganate disappears in this reaction so it can be used as a test for alkenes.
The displacement of a bromine atom by an amine is a substituion reaction. Write out the mechanism of this reaction (2-->3) Why might you expect that the reaction you have performed, using t-BuNH2, to be much slower than the same reaction using methylamine
Answer:
An alkyl halide can undergo SN2 reaction with an amine
Explanation:
The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.
The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).
This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.
The detailed mechanism of this reaction has been attached to this answer.
For the product of the reaction below, which proton is removed irreversibly by NaNH2 base, thus preventing any isomerization of the alkyne bond in the product?
Answer:
The Highly acidic proton joined to one of the carbon in the ALKYNE bond.
(Kindly Check the attachment for the drawing because the solution will need us to draw).
Explanation:
So, let us start by defining some major key terms in this particular Question given above;
(1). ISOMERIZATION: isomerization can simply be defined as the kind is of chemical rearrangement whichay lead to the breaking and the formation of new bonds.
(2). NaNH2 BASE: Sodium amide is a Chemical compound which has a Molar mass of 39.01 g/mol and Heat capacity (C) of 66.15 J/mol K. It is also known as sodamide. It is a good nucleophile.
(3). ALKYNE BOND: it is a C-C joined together by three bonds.
The chemical reaction given in the Question is given in the attachment too.
Therefore, The Highly acidic proton joined to one of the carbon in the ALKYNE bond is removed irreversibly by NaNH2 base.
Which of the following functional groups is formed from the condensation of carboxylic acids???
a. acid anhydride
b. acid halide
c. amide
d. ester
e. ether
Answer:
a
Explanation:
its made up of carbon and hydrogen
What is an example of a molecular compound
Answer:
Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).
Which phase change is an example of an exothermic process?
A.
solid to liquid
B.
solid to gas
C.
liquid to solid
D.
liquid to gas
E.
solid to plasma
Reset
Answer:
C
Explanation:
Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.
Calculate the pH of the 1L buffer composed of 500 mL 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of NaOH is added (Ka HC2H3O2 = 1.75 x 10-5). Report your answer to the hundredths place.
Answer:
pH = 4.79
Explanation:
The pH of the acetic buffer can be determined using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is -logKa = 4.76
pH = 4.76 + log [sodium Acetate] / [Acetic Acid]
Where [] can be taken as moles of each specie.
Thus, to find pH of the buffer we need to calculate molesof acetic acid and sodium acetate.
Initial moles:
Initial moles of acetic acid and sodium acetate are:
500mL = 0.500L ₓ (0.60moles / L) = 0.30 moles of both acetic acid and sodium acetate
Moles after reaction:
Now, 0.010 moles of NaOH are added to the buffer reacting with acetic acid, CH₃COOH, producing more acetate ion, as follows:
NaOH + CH₃COOH → CH₃COO⁻ + H₂O
That means after reaction moles of both species are:
Acetic acid: 0.30mol - 0.010mol (Moles that react) = 0.29 moles
Acetate: 0.30mol + 0.010mol (Moles produced) = 0.31 moles
Replacing in H-H equation:
pH = 4.76 + log [0.31] / [0.29]
pH = 4.79
1)The average lethal dose of Valium is 1.52 mg/kg of body weight. Estimate how many grams of Valium would be lethal for a 200.-lb woman. Show all your calculations. (1lb = 453.6 g)
2) A patient in hospital is receiving the antibiotic amoxcillin IV at the rate of 50. mL/h. The IV contains 1.5 g of the antibiotic in 1000. mL. (IV stands for intravenous). Calculate the mg/min of the drip. Show all your calculations
Answer:
1. 0.138g of valium would be lethel in the woman
2. 125mg/min is the drip of the patient
Explanation:
1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.
A person that weighs 200lb requires:
200lb × (453.6g / 1lb) × (1kg / 1000g) = 90.72kg (Weight of the woman in kg)
90.72kg × (1.52mg / kg) =
137.9mg ≡
0.138g of valium would be lethel in the woman2. The IV contains 1.5g = 1500mg/mL.
If the patient is receiving 5.0mL/h, its rate in mg/h is:
5.0mL/h × (1500mg/mL) = 7500mg/h
Now as 1h = 60min:
7500mg/h × (1h / 60min) =
125mg/min is the drip of the patientHow many moles of aqueous sodium ions and sulfide ions are formed when 2.05 mol of sodium sulfide dissolved in water
Explanation:
We have to put out the balanced chemical equation before proceeding. The equation for the dissociation of sodium sulfide in water is given as;
Na₂S --> 2Na⁺ + S²⁻
From the stoichiometry of the reaction we can tell that;
1 mol of Na₂S produces 2 moles of Na⁺ and 1 mol of S²⁻.
Sodium ion:
1 mol of Na₂S = 2 mol of Na⁺
2,05 = x
upon solving for x;
x = 2 * 2.05 = 4.10 moles
Sulfide ion:
1 mol of Na₂S = 1 mol of S²⁻
2.05 = x
upon solving for x;
x = 1 * 2.05 = 2.05 moles
Answer:
HEY ITS RIGHT ABOVE!!!
Explanation:
i did 1.85 instead of 2.05 and it was right... im so sorry