Answer:
0,
Explanation:
if n was 2, then 1,0,-1
13C NMR is a technique in which the total number of signals represents the number of unique carbon atoms in a molecule. Propose a structure that is consistent with the following data.
a. The IR includes peaks at 1603 and 1495 cm^-1
b. The 13c NMR has a total of 7 signals
c. The compound has one acidic proton.
Answer:
D. Poop Butt.
Explanation:
Based on the given data, we can propose a possible structure that fits the criteria: a. carbonyl group (C=O) and an aromatic ring b. there are seven unique carbon environments. c. Presence of a functional group like a carboxylic acid or phenol .
a. The IR peaks at 1603 [tex]cm^{-1}[/tex] and 1495[tex]cm^{-1}[/tex]suggest the presence of both a carbonyl group (C=O) and an aromatic ring.
b. The 13C NMR having a total of 7 signals indicates that there are seven unique carbon environments in the molecule.
c. Considering the presence of an acidic proton, it suggests the presence of a functional group like a carboxylic acid (COOH) or phenol ([tex]C_6H_5OH[/tex]).
Putting all this information together, a possible structure that fits the data could be benzoic acid ([tex]C_6H_5COOH[/tex]). It contains a benzene ring (giving 6 unique carbon environments), a carbonyl group (giving 1 unique carbon environment), and an acidic proton in the carboxylic acid group. This structure satisfies all the given data.
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A reaction is performed in a lab whereby two solutions are mixed together. The products are a liquid and a solid precipitate. What procedures would facilitate measurement of actual yield of the solid
Answer:
filtration, drying, and weighing
Explanation:
The procedures that would facilitate the measurement of the actual yield of the solid would be filtration of the precipitate, drying of the precipitate, and weighing of the precipitate respectively.
The liquid/solid mixture resulting from the reaction can be separated by the process of filtration using a filter paper. The residue in the filter paper would be the solid while the filtrate would be the liquid portion of the reaction's product.
The residue can then be allowed to dry, and then weighed using a laboratory-grade weighing balance. The weight of the solid represents the actual yield of the solid.
A chemist adds of a M barium chlorate solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.
The given question is incomplete, the complete question is:
A chemist adds 200.0 ml of a 0.52M barium chlorate (Ba(CIO3)2) solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.
Answer:
The correct answer is 32 grams.
Explanation:
Based on the given solution, the molarity of barium chlorate solution given is 0.52 M, this shows that the solution will comprise 0.52 moles in 1 L or 1000 ml of the solution.
Therefore, in 200 ml, it will comprise 0.52/1000 × 200 moles of Ba(ClO₃)₂,
= 0.52/1000 × 200 = 0.104 moles
The molecular mass of Ba(ClO₃)₂ is 304.23 gram per mole
So, the mass of Ba(ClO₃)₂ in 0.104 moles will be,
= 304.23 g/mol × 0.104
= 31.639 grams or 32 grams.
The equilibrium constant is given for two of the reactions below. Determine the value of the missing equilibrium constant. A(g) + 2B(g) ↔ AB2(g) Kc = 59 AB2(g) + B(g) ↔ AB3(g) Kc = ? A(g) + 3B(g) ↔ AB3(g) Kc = 478
Answer:
The correct answer is 8.10
Explanation:
Given:
A(g) + 2B(g) ↔ AB₂(g) Kc = 59 ---- Eq. 1
A(g) + 3B(g) ↔ AB₃(g) Kc = 478 ----- Eq. 2
We have to rearrange the chemical equations in order to obtain:
AB₂(g) + B(g) ↔ AB₃(g) Kc = ?
AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478. The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:
AB₂(g) ↔ A(g) + 2B(g) Kc₁= 1/59
A(g) + 3B(g) ↔ AB₃(g) Kc₂= 478
-----------------------------------------
AB₂(g) + B(g) ↔ AB₃(g)
If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:
Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10
The value of the missing equilibrium constant is 8.10.
The value of the missing equilibrium constant is 8.10
Chemical Equations:Since
A(g) + 2B(g) ↔ AB₂(g) Kc = 59 ---- Eq. 1
A(g) + 3B(g) ↔ AB₃(g) Kc = 478 ----- Eq. 2
Now we have to rearrange the chemical equations in order to obtain:
AB₂(g) + B(g) ↔ AB₃(g) Kc = ?
Here AB₂(g) represents a reactant, so we have to applied the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant should be the same: Kc= 478.
The following is the sum of rearranged chemical equations, and the compounds in bold and italic should be canceled:
AB₂(g) ↔ A(g) + 2B(g) Kc₁= 1/59
A(g) + 3B(g) ↔ AB₃(g) Kc₂= 478
-----------------------------------------
AB₂(g) + B(g) ↔ AB₃(g)
In the case when we add reactions at equilibrium, the equilibrium constants Kc are multiplied as follows:
Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10
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Charcoal from the dwelling level of the Lascaux Cave in France gives an average count of 0.97 disintegrations of ^14 C per minute per gram of sample. Living wood gives 6.68 disintegrations per minute per gram. Estimate the date of occupation and hence the probable date of the wall painting in the Lascaux Cave. Hint: Disintegrations per minute per gram" has the same units as the time-derivative of concentration for a radioactive decay model. (You may use the fact that the half-life of ^14C is 5568 years.)
Answer:
Explanation:
count given by old sample = .97 disintegrations per minute per gram
count given by fresh sample = 6.68 disintegrations per minute per gram
Half life of radioactive carbon = 5568 years
rate of disintegration
dN / dt = λ N
In other words rate of disintegration is proportional to no of radioactive atoms present . As number reduces rate also reduces .
Let initial no of radioactive be N₀ and after time t , number reduces to N
N₀ / N = 6.68 / .97
Now
[tex]N=N_0e^{-\lambda t}[/tex]
[tex]\frac{N}{N_0} =e^{-\lambda t}[/tex]
[tex]\frac{6.68}{.97} = e^{\lambda t}[/tex]
λ is disintegration constant
λ = .693 / half life
= .693 / 5568
= .00012446 year⁻¹
Putting the values in the equation above
[tex]\frac{6.68}{.97} = e^{.00012446\times t}[/tex]
[tex]6.8866 = e^{.00012446\times t}[/tex]
1.929577 = .00012446 t
t = 15503.6 years .
Hydrogen iodide decomposes according to the equation: 2HI(g) H 2(g) + I 2(g), K c = 0.0156 at 400ºC A 0.660 mol sample of HI was injected into a 2.00 L reaction vessel held at 400ºC. Calculate the concentration of HI at equilibrium.
Answer:
[HI] = 0.264M
Explanation:
Based on the equilibrium:
2HI(g) ⇄ H₂(g) + I₂(g)
It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:
Kc = 0.0156 = [H₂] [I₂] / [HI]²
As initial concentration of HI is 0.660mol / 2.00L = 0.330M, the equlibrium concentrations will be:
[HI] = 0.330M - 2X
[H₂] = X
[I₂] = X
Where X is reaction coefficient.
Replacing in Kc:
0.0156 = [X] [X] / [0.330M - 2X]²
0.0156 = X² / [0.1089 - 1.32X + 4X² ]
0.00169884 - 0.020592 X + 0.0624 X² = X²
0.00169884 - 0.020592 X - 0.9376 X² = 0
Solving for X:
X = - 0.055 → False solution, there is no negative concentrations
X = 0.0330 → Right solution.
Replacing in HI formula:
[HI] = 0.330M - 2×0.033M
[HI] = 0.264MUsing the bond energy data from your text (or the internet), determine (show calculations for) the approximate enthalpy change , ∆H, for each of the following reactions: (a) Cl2 (g) + 3F2 (g) ⟶ 2ClF3 (g)
Answer:
∆H= 438 KJ/mol
Explanation:
First, we have to find the energy bond values for each compound:
-) Cl-Cl = 243 KJ/mol
-) F-F = 159 KJ/mol
-) F-Cl = 193 KJ/mol
If we check the reaction we can calculate the number of bonds:
[tex]Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)[/tex]
In total we will have:
-) Cl-Cl = 1
-) F-F = 3
-) F-Cl = 6
With this in mind. we can calculate the total energy for each bond:
-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol
-) F-F = (3*159 KJ/mol) = 477 KJ/mol
-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol
Now, we can calculate the total energy of the products and the reagents:
Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol
Products = 1158 KJ/mol
Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:
∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol
I hope it helps!
The approximate enthalpy change is:
∆H= 438 KJ/mol
Calculation for enthalpy change:First, we have to find the energy bond values for each compound:
Cl-Cl = 243 KJ/mol
F-F = 159 KJ/mol
F-Cl = 193 KJ/mol
Balanced chemical reaction:
Cl₂ (g) + 3F₂ (g) ⟶ 2ClF₃ (g)
Total number of bond for each:
Cl-Cl = 1
F-F = 3
F-Cl = 6
Total bond energy will be:
Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol
F-F = (3*159 KJ/mol) = 477 KJ/mol
F-Cl = (6*193 KJ/mol) = 1158 KJ/mol
Now, we can calculate the total energy of the products and the reactants:
Reactants = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol
Products = 1158 KJ/mol
Finally, to calculate the total enthalpy change we have to do a subtraction between products and reagents:
∆H= 1158 KJ/mol-720 KJ/mol = 438 KJ/mol
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What was one idea Dalton taught about atoms?
A. Atoms contained negatively charged particles scattered inside.
B. Atoms of one type would not react with atoms of another type.
C. All atoms of one type were identical in mass and properties.
D. Atoms changed into new elements when they formed compounds.
Answer:
C
Explanation:
I had this question and C is the right answer
One idea that Dalton taught about atoms was that all atoms of one type were identical in mass and properties.
What is an atom?
An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.
The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.
Atoms of the same element are similar as they have number of sub- atomic particles which on combination do not alter the chemical properties of the substances.
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Applied Exercises (40 points) Answer the following questions in complete sentences. 1) In molecules with the same number of electron groups but different molecular geometries, discuss what happens to the bond angle? 2) What happens to the bond angle as you increase the number of bonding groups? 3) In 5 electron group molecules, what is the difference between axial and equatorial positions? Which groups are removed as lone pairs are added? 4) What is the difference between tetrahedral bent and
Answer:
See explanation
Explanation:
In molecules with the same number of electron groups but different molecular geometries, the bond angles differ markedly owing to the presence of lone pairs on the central atom. Recall that lone pairs of electrons take up more space around the central atom and causes more repulsion thus squeezing the bond angle and making it less than the value expected on the basis of the Valence Shell Electron Pair Repulsion Theory.
As the number of bonding groups increases, the bond angle increases since the repulsion due to lone pairs of electrons is being progressively removed by increase in the number of bonding groups.
For 5 electron group molecules, the axial groups are oriented at a bond angle of 90° while the equatorial groups are oriented at a bond angle of 120°. In the presence of lone pairs, the equatorial bonds are removed because the equatorial bonds often have a greater bond length than the axial bonds.
In the tetrahedral geometry, four groups are bonded to the central atom while in a bent molecular geometry, only two groups are bonded to the central atom with two lone pairs present in the molecule.
A study of the system, 4NH3(g) + 7O2(g) <--> 2N2O4(g) + 6H2O(g), was carried out. A system was prepared with [NH3] = [O2] = 3.60 M as the only components initially. At equilibrium, [N2O4] is 0.600 M. Calculate the equilibrium concentration of NH3(g).
Answer:
The equilbrium concentration of NH₃(g) is 2.4 M
Explanation:
The balanced reaction is:
4 NH₃(g) + 7 O₂(g) ⇔ 2 N₂O₄(g) + 6 H₂O(g)
By stoichiometry of the reaction, 2 moles of N₂O₄ are formed from 4 moles of NH₃.
Considering that the concentration is [tex]concentration=\frac{number of moles}{volume}[/tex] and with a volume of 1 liter, it is possible to apply the following rule of three: if 2 M of N₂O₄ are formed from 4 M of NH₃, 0.6 M of N₂O₄ from what concentration of NH₃ are formed?
[tex]concentration of NH_{3}=\frac{0.6 M of N_{2}O_{3} *4MofNH_{3} }{2 M of N_{2}O_{3} }[/tex]
concentration of NH₃= 1.2 M
By subtracting the moles of NH3 in equilibrium from the moles of NH₃ initially, you will see how many moles of NH₃ were converted and remain in equilibrium: 3.6 M - 1.2 M= 2.4 M
The equilbrium concentration of NH₃(g) is 2.4 M
Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use the method of successive approximations in your calculations or the quadratic formula.
Answer:
[tex][H^+]=0.000285[/tex]
[tex]pH=3.55[/tex]
Explanation:
In this, we can with the ionization equation for the hydrazoic acid ([tex]HN_3[/tex]). So:
[tex]HN_3~<->~H^+~+~N_3^-[/tex]
Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:
[tex]Ka=\frac{[H^+][N_3^-]}{[HN_3]}[/tex]
For each mol of [tex]H^+[/tex] produced we will have 1 mol of [tex]N_3^-[/tex]. So, we can use "X" for the unknown values and replace in the Ka equation:
[tex]Ka=\frac{X*X}{[HN_3]}[/tex]
Additionally, we have to keep in mind that [tex]HN_3[/tex] is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:
[tex]Ka=\frac{X*X}{0.004-X}[/tex]
Finally, we can put the ka value and solve for "X":
[tex]2.2X10^-^5=\frac{X*X}{0.004-X}[/tex]
[tex]2.2X10^-^5=\frac{X^2}{0.004-X}[/tex]
[tex]X= 0.000285[/tex]
So, we have a concentration of 0.000285 for [tex]H^+[/tex]. With this in mind, we can calculate the pH value:
[tex]pH=-Log[H^+]=-Log[0.000285]=3.55[/tex]
I hope it helps!
The [H+] and pH of a 0.0040 M hydrazoic acid solution is 0.000296648 and 3.527759
pH based problem:What information do we have?
Hydrazoic acid solution = 0.0040 M
Ka of hydrazoic acid = 2.20 × 10⁻⁵
We know that weak acids
[H+] = √( Ka × C)
[H+] = √( 2.2 × 10⁻⁵ × 0.0040)
[H+] = 0.000296648
So,
pH = -log [H+]
pH = -log [0.000296648]
Using log calculator
pH = 3.527759
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2) 2.5 mol of an ideal gas at 20 oC under 20 atm pressure, was expanded up to 5 atm pressure via; (a) adiabatic reversible and (b) adiabatic irreversible process. Calculate the values of w, q, ΔU, ΔH for each process. (Cv = 5 cal / mol.K ≈ 5/2 R; R ≈ 2 cal / mol.K) (Please find the desired values by making the corresponding derivations
Answer:
a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)
workdone (w) = -8442.6 J ≈ -8.443 KJ
heat transferred (q) of the ideal gas = - w
q = 8.443 KJ
b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0
the workdone(w) in the ideal gas= - 4567.5 J ≈ - 4.57 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Explanation:
given
mole of an ideal gas(n) = 2.5 mol
Temperature (T) = 20°C
= (20°C + 273) K = 293 K
Initial pressure of the ideal gas(P₁) = 20 atm
Final pressure of the ideal gas(P₂) = 5 atm.
2) (a)for adiabatic reversible process,
note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.
Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]
= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]
= 6090.01 J × [-1.3863]
= -8442.6 J ≈ -8.443 KJ
So, the work done (w) of ideal gas = -8.443 KJ
For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-8.443 KJ)
q = 8.443 KJ
heat transfer (q) of the ideal gas = 8.443 KJ
(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.
Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )
= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]
= - 6090.01 J × 0.75
= - 4567.5 J ≈ - 4.57 KJ
∴work done(w) of an ideal gas = - 4.57 KJ
For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-4.5675 KJ)
q = 4.5675 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Calculate the volume in liters of a M mercury(II) iodide solution that contains of mercury(II) iodide . Round your answer to significant digits.
Answer:
41L
Explanation:
Of a 4.8x10⁻⁵M mercury (II) iodide that contains 900mg of mercury (II) iodide. 2 significant digits
Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (Mercury (II) iodide in this case) per Liter of solution.
A 4.8x10⁻⁵M solution contains 4.8x10⁻⁵ moles of solute per liter.
Now, 900mg = 0.900g of mercury (II) iodide (Molar mass: 454.4g/mol) are:
0.900g × (1mol / 454.4g) = 1.98x10⁻³moles of HgI₂
If in 1L there are 4.8x10⁻⁵ moles of HgI₂, There are 1.98x10⁻³moles of HgI₂ in:
1.98x10⁻³moles of HgI₂ ₓ (1L / 4.8x10⁻⁵moles) =
41L30. What is the Bronsted base of H2PO4- + OH- ⟶HPO42- + H2O?
Answer:
OH⁻ is the Bronsted-Lowry base.
Explanation:
A Bronsted-Lowry base is a substance that accepts protons. In the chemical equation, OH⁻ accepts a proton from H₂PO₄⁻ to become H₂O. H₂PO₄⁻ would be a Bronsted-Lowry acid because it donates a proton to OH⁻ and becomes HPO₄²⁻.
Hope that helps.
A student obtained a clean flask. She weighed the flask and stopper on an analytical balance and found the total mass to be 34.232 g. She then filled the flask with water and found the new mass to be 60.167 g. The temperature of the water was measured to be
Answer:
25.99mL is the volume internal volume of the flask
Explanation:
To complete the question:
The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask
The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.
To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:
Mass water = Mass filled flask - Mass of clean flask
Mass water = 60.167g - 34.232g
Mass water = 25.935g of water.
To convert this mass to volume:
25.935g × (1mL / 0.997992g) =
25.99mL is the volume internal volume of the flaskCalculate the pH of a buffer solution that contains 0.25 M benzoic acid (C 6H 5CO 2H) and 0.15M sodium benzoate (C 6H 5COONa). [K a = 6.5 × 10 –5 for benzoic acid]
Answer:
3.97
Explanation:
pH of buffer solution = pKa+Log(Cb/Ca)
pH of buffer solution = -log(Ka)+log(Cb/Ca)............... Equation 1
Where Ca = concentration of acid, Cb = concentration of base.
Given: Ka = 6.5×10⁻⁵, Ca = 0.25 M, Cb = 0.15 M
Substitute into equation 1
pH of buffer solution = -log(6.5×10⁻⁵)+log(0.15/0.25)
pH of buffer solution = 4.19+(0.22)
pH of buffer solution = 3.97.
what is the IUPAC name of KNO3
Give the formulas for all of the elements that exist as diatomic molecules under normal conditions. See if you can do this without looking anything up.
Answer:
They are:
H2, N2, O2, F2, Cl2, Br2, and I2.
Note: whether the element At molecule is monoatomic or diatomic is incredibly arguable. While some say it exists as diatomic because it is a halogen like bromine, iodine etc, At is in fact extremely unstable and no one has ever really studied the molecules on it, so, when others say it is monoatomic, this is also based on calculations. But the other 7 elements listen above is for sure diatomic.
Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2).
Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2) are the formulas of the elements that is present as diatomic molecules under normal environmental conditions. Diatomic molecules refers to those molecules that is composed of only two atoms of the same or different elements. There are large number of diatomic molecules which is made up of two similar elements or different elements.
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Zinc bromide is considered which of the following?
A) molecular compound
B) atomic element
C) molecular element
D) ionic compound
Answer:
D
Explanation:
soluble in water and acidic5.00 mol of ammonia are introduced into a 5.00 L reactor vessel in which it partially dissociates at high temperatures. 2NH 3(g) 3H 2(g) + N 2(g) At equilibrium and a particular temperature, 1.00 mole of ammonia remains. Calculate K c for the reaction.
Explanation:
system at equilibrium, will the reaction shift towards reactants ~
--?'
2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an
exothermic reaction. Will heating the equilibrium system increase o~e amount of
ammonia produced? . .co:(
3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will
the reaction shift? ':'\
.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,
4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)
Temperature (K) Kc
300 1.5x104
600 55 k ' pr, cl l<..J~
e- ~ r fee, ct o. ~ 1<
900 3.4 X 10-3
Is the reaction endothermic or exothermic (explain your answer)?
t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\
exothe-rnh't.-- ,.. ..,. (/.., ,~.
5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.
What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)
N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..
~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb
J. [,v 1+3] ~
I
4,:i.~ = 0,05
5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.
Initially, there are 5.00 mol of ammonia in a 5.00 L reactor vessel. The initial concentration of ammonia is:
[tex][NH_3]_i = \frac{5.00mol}{5.00L} = 1.00 M[/tex]
At equilibrium, there is 1.00 mole of ammonia in the 5.00 L vessel. The concentration of ammonia at equilibrium is:
[tex][NH_3]eq = \frac{1.00mol}{5.00L} = 0.200 M[/tex]
We can calculate the concentrations of all the species at equilibrium using an ICE chart.
2 NH₃(g) ⇄ 3 H₂(g) + N₂(g)
I 1.00 0 0
C -2x +3x +x
E 1.00-2x 3x x
Since the concentration of ammonia at equilibrium is 0.200 M,
[tex]1.00-2x = 0.200\\\\x = 0.400 M[/tex]
The concentrations of all the species at equilibrium are:
[tex][NH_3] = 0.200 M\\[H_2] = 3x = 1.20 M\\[N_2] = x = 0.400 M[/tex]
The concentration equilibrium constant (Kc) is:
[tex]Kc = \frac{[H_2]^{2} [N_2]}{[NH_3]^{2} } = \frac{(1.20^{3})(0.400) }{0.200^{2} } = 17.3[/tex]
5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.
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Water (2190 g ) is heated until it just begins to boil. If the water absorbs 5.83×105 J of heat in the process, what was the initial temperature of the water?
Answer:
The initial temperature was [tex]36.4^\circ \:C[/tex]
Explanation:
[tex]\Delta t=\frac{q}{m\cdot C_s}=\frac{5.83\times10^5}{2190\times 4.184}\\\\=63.6^\circ\:C[/tex]
The temperature difference [tex]=100-63.6=36.4^\circ\:C[/tex]
Best Regards!
A chemistry student weighs out of chloroacetic acid into a volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to significant digits.
Answer:
11.6mL of the 0.1400M NaOH solution
Explanation:
0.154g of chloroacetic acid diluted to 250mL. Titrated wit 0.1400M NaOH solution.
The reaction of chloroacetic acid, ClCH₂COOH (Molar mass: 94.5g/mol) with NaOH is:
ClCH₂COOH + NaOH → ClCH₂COO⁻ + Na⁺ + H₂O
Where 1 mole of the acid reacts per mole of the base.
That means the student will reach equivalence point when moles of chloroacetic acid = moles NaOH.
You will titrate the 0.154g of ClCH₂COOH. In moles (Using molar mass) are:
0.154g ₓ (1mol / 94.5g) = 1.63x10⁻³ moles of ClCH₂COOH
To reach equivalence point, the student must add 1.63x10⁻³ moles of NaOH. These moles comes from:
1.63x10⁻³ moles of NaOH ₓ (1L / 0.1400moles NaOH) = 0.0116L of the 0.1400M NaOH =
11.6mL of the 0.1400M NaOH solutionWhy was it important to establish the Clean Air Act?
Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.
Explanation:
Answer:
Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.
A student is given an antacid tablet that weighs 5.8400 g. The tablet is crushed and 4.2800 g of the antacid is added to 200. mL of simulated stomach acid. It is allowed to react and then filtered. It is found that 25.00 mL of this partially neutralized stomach acid required 11.6 mL of a NaOH solution to titrate it to a methyl red end point. It takes 29.0 mL of this NaOH solution to neutralize 25.00 mL of the original stomach acid. How much of the original 200. mL of stomach acid (in mL) is neutralized by the 4.2800 g crushed sample of the tablet
Answer:
Explanation:
Given that:
mass of the antacid tablet = 5.8400 g
required mass of the antacid tablet = 4.2800 g was added to 200. mL of simulated stomach acid.
The amount of the original 200. mL of stomach acid (in mL) needed to neutralize the 4.2800 g crushed sample of the tablet can be calculated as:
= 11.6 mL of NaOH × 25.00 mL /29.0 mL NaOH
= 10.00 mL original stomach acid
Now; since it requires 11.6 mL of NaOH o neutralize 10.00 mL of original acid , then:
the antacid neutralized = 200 mL - 10.00 mL
the antacid neutralized = 190.00 mL
Which element would have the most valence electrons and also be able to react with hydrogen?
Answer:
Fluorine, Chlorine, Bromine, or Iodine
Explanation:
These all have an ALMOST full valence shell. And they need one more electron so they'd react with hydrogen
Answer:
its chlorine
Explanation:
just trust me do i look like i would lie too you ;-)
btw i just took the test :-)
A study of the following system, 4 NH3(g) + 7 O2(g) <--> 2 N2O4(g) + 6 H2O(g), was carried out. A system was prepared with [N2O4] = [H2O] = 3.60 M as the only components initially. At equilibrium, [H2O] is 0.600 M. Calculate the equilibrium concentration of O2(g).
Answer:
3.50 M
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)
Step 2: Make an ICE chart
4 NH₃(g) + 7 O₂(g) ⇄ 2 N₂O₄(g) + 6 H₂O(g)
I 0 0 3.60 3.60
C +4x +7x -2x -6x
E 4x 7x 3.60-2x 3.60-6x
Step 3: Calculate the value of x
The concentration of water at equilibrium is 0.600 M. Then,
3.60-6x = 0.600 M
x = 0.500 M
Step 4: Calculate the concentration of O₂ at equilibrium
The concentration of O₂ at equilibrium is 7x = 7(0.500M) = 3.50 M
An acetic acid buffer solution is required to have a pH of 5.27. You have a solution that contains 0.010 mol of Acetic acid. What molarity of sodium acetate will you need to add to the solution
Answer:
Molarity of sodium acetate you will need to add is 0.0324M
Explanation:
Assuming volume of the buffer is 1L.
The pH of a buffer can be determined using Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is pKa of the weak acid, [A⁻] molar concentration of conjugate base and [HA] molar concentration of weak acid
Replacing for the acetic buffer (pKa = 4.76):
pH = 4.76 + log [Sodium Acetate] / [Acetic Acid]
As you have 0.010 moles of acetic acid in 1L:
[Acetic Acid] = 0.010mol / 1L = 0.010M
And you require a pH of 5.27:
5.27 = 4.76 + log [Sodium Acetate] / [0.010M]
0.51 = log [Sodium Acetate] / [0.010M]
10^0.51 = [Sodium Acetate] / [0.010M]
3.236 = [Sodium Acetate] / [0.010M]
3.236 [0.010M] = [Sodium Acetate]
0.0324M = [Sodium Acetate]
Molarity of sodium acetate you will need to add is 0.0324M
The combustion of propane (C 3H 8) in the presence of excess oxygen yields CO 2 and H 2O: C 3H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2O (g) When 2.5 mol of O 2 are consumed in their reaction, ________ mol of CO 2 are produced.
Answer:
1.5 mol of CO₂
Explanation:
Use the mole ratio to find how many moles of CO₂ are produced from the reaction.
For every 5 moles of O₂, three moles of CO₂ is produced.
2.5 mol O₂ × 3 mol CO₂ ÷ 5 mol O₂
= 2.5 mol O₂ × 0.6
= 1.5 mol CO₂
When 2.5 mol of O₂ is consumed in the reaction, 1.5 mol of CO₂ is produced.
Hope that helps.
Write the equilibrium constant expression for the experiment you will be studying this week. 2. If the equilibrium values of Fe3+, SCN- and FeSCN2+ are 9.5 x 10-4 M, 3.6 x 10-4 M and 5.7 x 10-5 M respectively, what is the value of Kc? 3. Write the general form of the dilution equation. 4. A solution is prepared by adding 18 mL of 0.200 M Fe(NO3)3 and 2 mL of 0.0020 M KSCN. Calculate the initial concentrations of Fe3+ and SCNin the solution.
Answer:
Kc = 166.7
[Fe³⁺] = 0.18 M
[SCN⁻] = 2×10⁻⁴ M
Explanation:
In the reaction of Fe³⁺ and SCN⁻, it is formed a complex:
Fe³⁺ + SCN⁻ ⇄ FeSCN²⁺ Kc
Let's make the expression for Kc → [FeSCN²⁺] / [Fe³⁺] . [SCN⁻]
5.7×10⁻⁵ / 9.5×10⁻⁴. 3.6×10⁻⁴ = 166.7
We determine the mmoles, we add from each reactant:
18 ml . 0.2M = 3.6 mmoles of Fe³⁺
2 ml . 0.002M = 4×10⁻³ mmoles of SCN⁻
General form of the dilution equation is:
Concentrated [C] . Concentrated Volume = Diluted [C] . Diluted Volume
Total volume = 20mL
[Fe³⁺]: 3.6 mmoles /20mL = 0.18 M
[SCN⁻]: 4×10⁻³ mmoles /20mL = 2×10⁻⁴ M
The value should be 1.67 x 10^2
The initial concentration should be 0.18 M and 2.0 x 10^(-4) M
Calculation of the value and initial concentration:The value is
= 5.7 x 10^(-5)/(9.5 x 10^(-4) x 3.6 x 10^(-4))
= 167
= 1.67 x 10^2
we know that
Initial moles = volume x concentration
So,
= 18/1000 x 0.200
= 0.0036 mol
Now
Initial moles = volume x concentration
= 2/1000 x 0.0020
= 4.0 x 10^(-6) mol
So,
Total volume should be
= 18 + 2
= 20 mL
= 0.02 L
Now
Initial concentration
= moles /total volume
= 0.0036/0.02
= 0.18 M
Now
Initial concentration
= moles /total volume
= 4.0 x 10^(-6)/0.02
= 2.0 x 10^(-4) M
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Which of the following statements about metal elements is correct?
A. Metals tend to easily gain more valence electrons.
B. Metal elements are always heavier than non-metal elements.
C. Metals tend to easily lose their valence electrons.
D. A metal atom can take an electron from a non-metal atom.
Answer: C. Metals tend to easily lose their valence electrons.
Explanation:
Metals are those substances which have tendency to loose their valence electrons to attain noble gas configuration and forms positive ions called as cations.
Example: Gold, potassium etc
[tex]M\rightarrow M^++e^-[/tex]
Non metals are those substances which have tendency to gain valence electrons to attain noble gas configuration and form negative ions called as anions.
Example: Sulphur, Chlorine
[tex]N+e^-\rightarrow N^-[/tex]