Answer:
(d) 2
Explanation:
Lets say that there are 2 apples or 1+1 if you count them you would do 1,2 so 2 would be the final answer
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.
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
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.
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 patientWhen 3-methylpent-2-ene is treated with mercury(II) acetate in methanol and the resulting product isreacted with NaBH4, what is the primary organic compound which results
Answer:
3-methylpentan-3-ol
Explanation:
In this case, we have an "Oxymercuration reaction". With this in mind, we will have to add an "OH" to the most substituted carbon of the double bond and we will obtain 3-methylpentan-3-ol. To understand how this molecule is produced we have to check the mechanism:
The mercury(II) acetate ([tex]Hg(OAC)_2[/tex]) is an ionic substance. So, this substance can be ionized into his ions and we will have the cation [tex]HgOAc^+[/tex] and the anion [tex]AcO^-[/tex]. The cation will attack the double bond and vice-versa to produce a "cyclic intermediate". Then a water molecule will attack the most substituted carbon and the cyclic compound would be broken producing a new bond C-O with a positive charge in the oxygen. Then a deprotonation step takes place and finally, the [tex]NaBH_4[/tex] would reduce the compound to produce the final alcohol.
See figure 1
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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]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.
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]
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
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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]
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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.
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The nutrition label on the back of a package of hotdogs (purchased within the US) indicates that one hotdog contains 100 calories. How many calories does a hotdog actually have?
A. 1,000
B. It depends on how many hotdogs you eat
C. 100
D. 10
E. 100,000
Answer:
C. 100
Explanation:
Biochemical researches and studies have found out that an average health hotdog has a calorie of between 100 and 150 which is usually dependent on the additives.
Since the nutrition label on the back of a package of hotdogs (purchased within the US) indicates that one hotdog contains 100 calories then it truly contains such amount of calories. The standard number of calories present in a hotdog is independent of the amount eaten by individuals.
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%
If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of B is 0.82 g/ml; the concentration of C is 0.64 M; and the concentration of D is 0.38 M.
A(s) + 3 B(l) _____ 2(aq) + D(aq)
Pick the correct statement about this system.
A. Q < K and reaction shifts left
B. Q > K and reaction shifts left
C. Q > K and reaction shifts right
D. Q = K and reaction does not shift
E. Q < K and reaction shifts right
Answer:
E. Q < K and reaction shifts right
Explanation:
Step 1: Write the balanced equation
A(s) + 3 B(l) ⇄ 2(aq) + D(aq)
Step 2: Calculate the reaction quotient (Q)
The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.
Q = [C]² × [D]
Q = 0.64² × 0.38
Q = 0.15
Step 3: Compare Q with K and determine in which direction will shift the reaction
Since Q < K, the reaction will shift to the right to attain the equilibrium.
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.
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.
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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=searchResultsWhat 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.
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
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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
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.
A 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.
If D+2 would react with E-1, what do you predict to be the formula?
Answer:
DE2
Explanation: for every one D+2 you need two E-1 because +2=-2
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.
Briefly in your own words, describe the face-centered cubic unit cell in such a way that someone reading only your description, would be able to make a reasonable sketch of the structure.
Answer:
Face-centered Cubic Unit Cell (FCC) is consists of 8 atoms at the corners (one atom at each corner) and 6 atoms are present at the face-center (one atom at each face). Each eight atoms at corners contribute 1/8 and six atoms at faces contribute 1/2 in each unit cell.
How 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
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
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.
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.
Given these data in a study on how the rate of a reaction was affected by the concentration of the reactants,
Experiment [A] [B] [C] Rate (mol L‑1 hr‑1 )
1 0.200 0.100 0.600 5.0
2 0.200 0.400 0.400 80.0
3 0.600 0.100 0.200 15.0
4 0.200 0.100 0.200 5.0
5 0.200 0.200 0.400 20.0
From this data, what is the numerical value of the rate constant, (k), for this reaction (value that would be found using the same units used in the data above)?
a. 2083
b. 694
c. 417
d. 2500
e. 83.3
Answer:
d. 2500
Explanation:
In a kinetic study with 3 different reactants, you change concentrations of the reactants to see how this concentration affects rate of reaction. General law is:
v = k [A]ᵃ [B]ᵇ [C]ⁿ
If you see 1 and 3 experiments, the concentration of C change from 0.600M to 0.200M but reaction rate doesn't change, thus n=0:
v = k [A]ᵃ [B]ᵇ [C]⁰
v = k [A]ᵃ [B]ᵇ×1
Now, reaction 2 and reaction 4 change B from 0.400M to 0.200M having the other reactants constant. When B is duplicated, rate increase 4 times. That means b = 2:
v = k [A]ᵃ [B]ᵇ
v = k [A]ᵃ [B]²
Finally, if you see 3 and 4 reactions, A change from 0.200M to 0.600M and the reaction rate change from 15.0 to 5.0, That means if the concentration of A is triplicated, reaction rate will be triplicated to. Thus a=1:
v = k [A]ᵃ [B]²
v = k [A] [B]²
Relpacing this equation in any experiment (Experiment 5, for example):
20.0 = k [0.200] [0.200]²
2500 = k
That means right answer is:
d. 2500What 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
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