Answer:
sodium
Explanation:
(Na) atom is paramagnetic and sodium is a na atom.
When a thin glass tube is put into water, the water rises 1.4 cm. When the same tube is put into hexane, the hexane rises only 0.4 cm. Complete the sentences to best explain the difference. Match the words to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer.
1. The strongest force observed at the surface of glass is:________
2. Water is___________ and interacts, generating adhesive interactions with the only weak dispersion strong hydrogen bonding polar glass
3. Hexane is________ and interacts, generating____________ adhesive interactions with the glass.
a. dipole
b. nonpolar
c. only weak
d. dispersion
e. strong
f. hydrogen bonding
g. polar
Answer:
Explanation:
1 . The strongest force observed at the surface of glass is:__DIPOLE______
2. . Water is__POLAR_________ and interacts, generating adhesive interactions with the only weak dispersion strong hydrogen bonding polar glass.
3 . Hexane is_NON-POLAR _______ and interacts, generating__ONLY WEAK __________ adhesive interactions with the glass.
Calculate the equilibrium constant at 298 K for the reaction of formaldehyde (CH2O) with hydrogen gas using the following information. CH2O(g) + 2H2(g) LaTeX: \longleftrightarrow⟷ CH4(g) + H2O(g) LaTeX: \DeltaΔH°= –94.9 kJ; LaTeX: \DeltaΔS°= –224.2 J/K A. 1.92 B. 9.17 x 10-6 C. 2.07 x 1028 D. 1.10 x 105 E. 8.08 x 104 F. 3.98 x 1011 Group of answer choices
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,
A molecule of aluminum fluoride has one aluminum atom. How many fluorine atoms are present?
Answer:
3 fluorine atoms will be present
Answer:
3
Explanation:
The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.
If you want further tutoring help in chemistry or other subjects for FREE, check out growthinyouth.org.
Part C: complete the third column
Part D: complete the fourth column
Answer:
Part C: P2 = 0.30 atm
Part D: V1 = 16.22 L.
Explanation:
Part C:
Initial pressure (P1) = 2.67 atm
Initial volume (V1) = 5.54 mL
Final pressure (P2) =.?
Final volume (V2) = 49 mL
The final pressure (P2) can be obtained as follow:
P1V1 = P2V2
2.67 x 5.54 = P2 x 49
Divide both side by 49
P2 = (2.67 x 5.54)/49
P2 = 0.30 atm
Therefore, the final pressure (P2) is 0.30 atm
Part D:
Initial pressure (P1) = 348 Torr
Initial volume (V1) =?
Final pressure (P2) = 684 Torr
Final volume (V2) = 8.25 L
The initial volume (V1) can be obtained as follow:
P1V1 = P2V2
348 x V1 = 684 x 8.25
Divide both side by 348
V1 = (684 x 8.25)/348
V1 = 16.22 L
Therefore, the initial volume (V1) is 16.22 L
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.
1. In the addition of HBr to conjugated dienes, is the product which results from 1,2-addition or that which results from 1,4-addition the product of kinetic control?
A. From 1,2-addition
B. From 1,4-addition
2. Which of the following is the strongest acid?
A. CH3CH20H
B. CHзOCH3
C. CH3CH
D. CH3COCH3
E. CH3COH
Answer:
The answer to this question can be defined as follows:
In question 1, the answer is "Option A".
In question 2, the answer is "[tex]\bold{CH_3COOH}[/tex]".
Explanation:
In the second question, there is mistype error in the choices so the correct answer to this question can be defined as follows:
The product From 1,2-addition as its consequence of 1,4-addition is the result of kinetic regulation by HBr in conjugated dienes.The chemical name of the [tex]CH_3COOH[/tex] is the acetic acid, it is one of the carboxylic acids quite basic. It is a major chemical production factor for use as disposable soft drinks, movies or wood glue, polyethylene terephthalate, and many plastics, fibers, and fabrics. It is also used in the storage of the water and soft drinks in the bottles.1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice
Answer:
0.1M solution of NaOH
Explanation:
1 mole of NaOH - 40g
? moles - 1 g = 1/40 = 0.025 moles.
Molarity of 1.00g of NaOH in 0.25L (250 mL) = no. of moles/volume
= 0.025/0.25
= 0.1M.
Given the initial rate data for the reaction being A + B + C --> D determine the rate expression for the reaction and the (k) rate constant. (The units of [A] [B] and [C] are all moles/liter and the units of IRR is moles/liter seconds). If the [A]=[B]=[C]=.30M, what would the IRR be? [A] [B] [C] IRR 0.20 0.10 0.40 .20 0.40 0.20 0.20 1.60 0.20 0.10 0.20 .20 0.20 0.20 0.20 .80
Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
A sample of neon gas at a pressure of 0.609 atm and a temperature of 25.0 °C, occupies a volume of 19.9 liters. If the gas is compressed at constant temperature to a
volume of 12.7 liters, the pressure of the gas sample will be
atm.
Answer:
The pressure of the gas sample will be 0.954 atm.
Explanation:
Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
o P * V = k
To determine the change in pressure or volume during a transformation at constant temperature, the following is true:
P1 · V1 = P2 · V2
That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.
In this case:
P1= 0.609 atmV1= 19.9 LP2=?V2= 12.7 LReplacing:
0.609 atm* 19.9 L= P2* 12.7 L
Solving:
[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]
P2= 0.954 atm
The pressure of the gas sample will be 0.954 atm.
18. All of the following sets of quantum numbers are allowed EXCEPT a. n = 1, = 0, = 0 b. n = 2, = 2, = +1 c. n = 3, = 1, = –1 d. n = 4, = 1, = 0 e. n = 5, = 4, = –3
Answer:
b
Explanation:
it is impossible for n & l to be equal
There are 4 quantum numbers:
Principal Quantum number [tex](n)[/tex] specifies the energy of the electron in a shell.Azimuthal Quantum number [tex](l)[/tex] specifies the shape of an orbital. The value of it lies in the range of 0 to (n-1)Magnetic Quantum number [tex](m)[/tex] specifies the orientation of the orbital in space. The value of it lies in the range of -l to +lSpin Quantum number [tex](s)[/tex] specifies the spin of an electron in an orbital. It can either have a value of [tex]+\frac{1}{2}[/tex] or [tex]-\frac{1}{2}[/tex]To find the forbidden set, we need to know the following facts about quantum numbers:
The value of [tex]l[/tex] will always be less than the value of [tex]n[/tex]The value of [tex]m[/tex] and [/tex]l[/tex[ be be equal to 0 but the value of [tex]n[/tex] can never be equal to 0.Thus, we can say that the option that affirms that n = 2, l = 2, m = +1 is forbidden because the value of [tex]l[/tex] is equal to the value of [tex]n[/tex].
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Draw a Lewis structure for one important resonance form of HBrO4 (HOBrO3). Include all lone pair electrons in your structure. Do not include formal charges in your structure.
Answer:
The Lewis structure is attached with the answer -
Explanation:
Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.
HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.
The Lewis structure is attached in form of image with representation of lone pairs of electrons.
Initial mass of triphenyl methanol in g = 0.220g Initial volume of 33% HBr solution in ml = 0.60 ml Find mas of triphenyl bromide in g = 240 g
Answer:
0.792g of triphenyl bromide are produced.
Explanation:
The reaction of triphenyl methanol with HBr is:
triphenyl methanol + HBr → Triphenyl bromide.
Reaction (1:1), 1 mole of HBr reacts per mole of triphenyl methanol.
To know the mass of triphenyl bromide assuming a theoretical yield (Yield 100%) we need to find first limiting reactant:
Moles triphenyl methanol (Molar mass: 260.33g/mol) =
0.220g × (1mol / 260.33g) = 8.45x10⁻³ moles Triphenyl methanol
Moles HBr (Molar mass: 80.91g/mol; 33%=33g HBr/100mL) =
0.60mL ₓ (33g / 100mL) ₓ (1mol / 80.91g) = 2.45x10⁻³ moles HBr
As amount of moles of HBr is lower than moles of triphenyl methanol, HBr is limiting reactant.
As HBr is limiting reactant, moles produced of triphenyl bromide = moles HBr = 2.45x10⁻³ moles
As molar mass of triphenyl bromide is 323.2g/mol, mass of triphenyl bromide is:
2.45x10⁻³ moles × (323.2g / mol) =
0.792g of triphenyl bromide are produced.bleaching powder reaction, mechanism, use
Answer:
Bleaching Powder's chemical formula is CaOCl2 and is called Calcium Oxychloride. It is prepared on dry slaked lime by chlorine gas. 2. ... It gives calcium chloride, chlorine and water when bleaching powder reacts with hydrochloric acid.
Explanation:
Calculate the mass defect in Mo-96 if the mass of a Mo-96 nucleus is 95.962 amu. The mass of a proton is 1.00728 amu and the mass of a neutron is 1.008665 amu.
Answer:
0.81167 amu
Explanation:
Number of protons=42
Number of neutrons = 54
Mass of all the 42 protons = 42× 1.00728= 42.30576 amu
Mass of all 54 neutrons= 54 × 1.008665 = 54.46791 amu
Calculated mass of protons and neutrons in Mo-96 nucleus= 42.30576 amu + 54.46791 amu= 96.77367 amu
Actual mass of Mo-96 nucleus= 95.962 amu
Therefore mass defect of Mo-96 nucleus= 96.77367 amu - 95.962 amu = 0.81167 amu
Where possible, classify these systems as reactant-favored or product-favored at 298 K. If the direction cannot be determined from the information given, classify the reaction as "Insufficient information."
A. Reactant-favored
B. Product-favored
C. Insufficient information
1. A(s) + B(g) 2C(g) delta H degree = -109 kJ
2. A(s) + 2B(g) C(g) delta H degree=+271 kJ
3. 2A(g) + B(g) 4C(g) delta H degree=+322 kJ
4. A(g) + 2B(g) 2C(g) delta H degree=-89 kJ
Answer:
There is insufficient information to know direction of these systems
Explanation:
Delta H of a reaction is defined as the amount of energy involved when it occurs. The ΔH < 0 represents the reaction will release energy and ΔH > 0 the reaction will absorb energy.
As you can see, ΔH doesn't give information about the direction of a reaction (Spontaneity). In fact, to know spontaneity of a reaction you must know ΔG involved in this reaction.
As the reactions have ΔH but not ΔG,
There is insufficient information to know direction of these systemsDraw the structure of 1,4-hexanediamine.
Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced TemplateTowbars. The single bond is active by default. Include all hydrogen atoms.
View Available Hint(s)
Answer:
1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.
Explanation:
1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.
The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.
Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].
1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.
The structure of 1,4-hexanediamine is shown below.
when air molecules collide with things around us, it produces _______ (2 words), which is measured with a _______.
✔ When air molecules collide with things around us, it produces pressing force , which is measured with a Pressure gauge.
A compound is found to contain 63.65 % nitrogen and 36.35 % oxygen by mass. The molar mass for this compound is 44.02 g/mol. The molecular formula for this compound is
Answer:
THE MOLECULAR FORMULA FOR THE COMPOUND IS N20
Explanation:
To calculate the molecular formula for the compound, we follow the following steps:
Write out the percentage abundance of the individual elements
N = 63.65 %
O = 36.35 %
2. Divide the percentage composition by the atomic masses of the elements
N = 63.65 / 14 = 4.546
O = 36.35 / 16 = 2.272
3. Divide the values by the lowest value
N = 4.546 / 2.272 = 2.00
O = 2.202 / 2.272 = 1
4. The empirical formula of the compound will be:
N2O
5. Calculate the molecular mass
(N2O ) x = 44.02 g/mol.
(14 * 2 + 16) x = 44.02
(28 + 16) x = 44.02
44 x = 44.02
x = 44.02 / 44
x = 1
The molecular formula for the compound is N2O
Calculate the amount of energy absorbed by 45.0 g sample of water to raise its temperature from 18.0C to 48.0 C. The specific heat of water is 4.18 J/g C. 1000 J= 1kj
Answer:
5.643 kJ
Explanation:
The quantity of heat released or absorbed by a substance (Q) is given by the equation:
Q = mcΔT
Where m is the mass of the substance, c is the specific heat of substance and ΔT is the difference between the final temperature and the initial temperature.
Given that:
m = 45 g, Final temperature = 48°C, Initial temperature = 18°C, c = specific heat of water = 4.18 J/g°C
ΔT = Final temperature - Initial temperature = 48°C - 18°C = 30°C
The quantity of heat is:
Q = mcΔT = 45 g × 4.18 J/g°C × 30°C = 5643 J
Q = 5.643 kJ
Answer please and thank you
Answer:
Option B. 30 KJ.
Explanation:
The following data were obtained from the question:
Temperature (T) = 5000 K
Enthalpy change (ΔH) = – 220 kJ/mol
Change in entropy (ΔS) = – 0.05 KJ/mol•K
Gibbs free energy (ΔG) =...?
The Gibbs free energy, ΔG can be obtained by using the following equation as illustrated below:
ΔG = ΔH – TΔS
ΔG = – 220 – (5000 x – 0.05)
ΔG = – 220 – (– 250)
ΔG = – 220 + 250
ΔG = 30 KJ
Therefore, the Gibbs free energy, ΔG is 30 KJ.
A compound containing only C, H, and O, was extracted from the bark of the sassafras tree. The combustion of 32.3 mg produced 87.7 mg of CO2 and 18.0 mg of H2O. The molar mass of the compound was 162 g/mol. Determine its empirical and molecular formulas.
Answer:
Empirical formula: C₅H₅O
Molecular formula: C₁₀H₁₀O₂
Explanation:
When a compound containing C, H and O elements is combusted, the general reaction is:
CₐHₓOₙ + O₂ → a CO₂ + X/2 H₂O
Thus, you can find moles of carbon and hydrogen knowing moles of CO₂ and H₂O that are produced.
Moles CO₂ = Moles C = 0.0877g × (1mol / 44g) =
2.0x10⁻³ moles of CO₂ = moles C
Moles H₂O = 1/2 Moles H = 0.018g × (1mol / 18g) =
1x10⁻³ moles of H₂O; 2.0x10⁻³ moles H
The mass of the moles of C and H are:
2x10⁻³ moles C ₓ (12g / mol) = 0.024g C
2x10⁻³ moles H ₓ (1g / mol) = 0.002g H
Thus, mass of Oxygen is 32.3mg - 24mg C - 2mg O = 6.3mg O
Moles are:
0.0063g O ₓ (1mol / 16g) = 4x10⁻⁴ moles O
Empirical formula is the simplest ratio of atoms in a compound. Dividing each amount of moles for each atom in the 4x10⁻⁴ moles of oxygen (The lower moles), you will obtain:
C: 2.0x10⁻³ / 4x10⁻⁴ = 5
H: 2.0x10⁻³ / 4x10⁻⁴ = 5
O: 4x10⁻⁴ / 4x10⁻⁴ = 1
Thus, empirical formula is:
C₅H₅OThe molar mass of the empirical formula is:
12×5 + 1×5 + 16×1 = 81g/mol
As molar mass of the compound is 162g/mol, molecular formula is twice empirical formula:
C₁₀H₁₀O₂Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.
a) ethylbenzene
b) m-Xylene (1,3- dimethylbenzene)
c) 4-Propyl-3-t-butyltoluene
Answer:
Explanation:
a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .
C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O
O is provided by KMnO₄
b ) In this reaction isophthalic acid is formed .
C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂
c)
4-Propyl-3-t-butyltoluene
In this oxidation , three side chains of ring are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .
The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )
A general chemistry student found a chunk of metal in the basement of a friend's house. To figure out what it was, she tried the following experiment. First she measured the mass of the metal to be 385.8 grams. Then she dropped the metal into a measuring cup and found that it displaced 17.8 mL of water. Calculate the density of the metal. Density = _______ g/mL Use the table below to decide the identity of the metal. This metal is most likely _________.
substances density g/cm3
water 1.00
aluminium 2.72
chromium 7.25
nickel 8.91
silver 10.50
lead 11.34
1. 21.67g/ml
2. aluminium
Explanation:
1. density = mass/volume
385.8/17.8= 21.67ml
2. 1g/ml=0.1g/cm^3
21.67g/ml = 2.167g/cm^3
..... substance is probably aluminium
1. the density of the metal is 21.67g/ml
2. This metal is most likely aluminum
The calculation is as follows;
1.
[tex]density = mass \div volume[/tex]
[tex]385.8\div 17.8= 21.67ml[/tex]
2.
1g/ml=0.1g/cm^3
So,
21.67g/ml = 2.167g/cm^3
Therefore, substance is probably aluminum
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Which element has the largest atomic radius? As N P Sb
Answer:
Sb
Explanation:
The periodic trend for atomic radius is that it decreases from left to right and increases from top to bottom, therefore the elements with the larger atomic radius will be the ones which are closest to the bottom left corner of the periodic table. Since all of these elements are in the same group, the one with the largest atomic radius will be the one at the "bottom", and that is Sb.
If 35.2 g of magnesium react with excess hydrochloric acid, how many grams of magnesium chloride are produced? Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(aq)
Answer:
139.33 g of magnesium chloride, MgCl2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(aq)
Next, we shall determine the mass of Mg that reacted and the mass of MgCl2 from the balanced equation.
This is illustrated below:
Molar mass of Mg = 24 g/mol
Mass of Mg from the balanced equation = 1 x 24 = 24 g
Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol
Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g
From the balanced equation above,
24 g of Mg reacted to produce 95 g of MgCl2.
Finally, we shall determine the mass of MgCl2 produced by reacting 35.2 g of Mg.
This can be obtained as follow:
From the balanced equation above,
24 g of Mg reacted to produce 95 g of MgCl2.
Therefore, 35.2 g of Mg will react to produce = (35.2 x 95)/24 = 139.33 g of MgCl2.
From the calculations made above, 139.33 g of magnesium chloride, MgCl2 were produced.
during the electrolysis of an aqueous solution of sodium nitrate, a gas forms at the anode, what gas is it?
Answer: The answer is B
Explanation:
Answer:
oxygen
Explanation:
Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for this reaction at 298.15K if the pressure of each gas is 22.20 mm Hg.
Answer:
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
ΔG ≅ 199.91 kJ
Explanation:
Consider the reaction:
[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]
temperature = 298.15K
pressure = 22.20 mmHg
From, The standard Thermodynamic Tables; the following data were obtained
[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :
[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]
[tex]K_p = \dfrac{1}{ (22.20)}[/tex]
[tex]K_p[/tex] = 0.045
[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]
where;
R = gas constant = 8.314 × 10⁻³ kJ
[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]
[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]
[tex]\Delta G =207.6+ (-7.687048037)[/tex]
[tex]\Delta G =[/tex] 199.912952 kJ
ΔG ≅ 199.91 kJ
For each of the processes, determine whether the entropy of the system is increasing or decreasing. The system is underlined.
1. a snowman melts on a spring day
2. a document goes through a paper shredder
3. a water bottle cools down in a refrigerator
4. silver tarnishes
5. dissolved sigar precipitates out of water to form rock candy
A. Entropy is increasing
B. Entropy is decreasing
Entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms droplets and water cools down.
Entropy can be defined as the degree of randomness or disorder of a particular system.
Entropy is equal to zero (0) for a perfectly ordered system.
Heat increases the entropy of the system because more energy excites the molecules and it increases the amount of random activity.
Moreover, the cooling decreases the entropy of the system because molecules are more ordered and it decreases the amount of random activity.
In conclusion, entropy is INCREASING when a snowman melts, a document goes through paper shredder, silver tarnishes, while it is DECREASING when dissolved sugar precipitates, water vapor forms water droplets and the water cools down.
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Which of the following is an important intermediate in the mechanism of the reaction
in the box?
Answer:
Explanation:
.
What class of organic product results when 1-heptyne is treated with a mixture of mercuric acetate (HgSO4) in aqueous sulfuric acid (H2O/H2SO4)
Answer:
heptan-2-one
Explanation:
In this case, the final product would be a ketone: heptan-2-one. To understand why this molecule is produced we have to check the reaction mechanism.
The first step is the protonation of the triple bond to produce the more stable carbocation (a secondary one) by the action of sulfuric acid [tex]H_2SO_4[/tex]. The next step is the attack of water to the carbocation to produce a new bond between C and the O, producing a positive charge in the oxygen. Then, a deprotonation step takes place to produce an enol. Finally, we will have a rearrangement (keto-enol tautomerism) to produce the final ketone.
See figure 1
I hope it helps!