Answer:
a) 40 %
b) [tex]4.04~g~CO_2[/tex]
c) [tex]5.53x10^2^3~molecules~of~O_2[/tex]
Explanation:
For a) we will have to calculate the molar mass of [tex]C_6H_1_2O_6[/tex], so the first step is to find the atomic mass of each atom and multiply by the amount of atoms in the molecule.
C => 12*(6) = 72
H => 1*(12) = 12
O => 6*(16) = 96
Molar mass = 180 g/mol
Then we can calculate the percentage by mass:
[tex]Percentage~=~\frac{72}{180}*100=40[/tex]
For b) we have to start with the reaction of glucose:
[tex]C_6H_1_2O_6~+~6O_2~->~6CO_2~+~6H_2O[/tex]
Then we have to convert the grams of glucose to moles, the moles of glucose to moles of carbon dioxide and finally the moles of carbon dioxide to grams. To do this we have to take into account the following conversion ratios:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol carbon dioxide
-) 1 mol carbon dioxide = 44 g carbon dioxide
[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~CO_2}{1~mol~C_6H_1_2O_6}\frac{44~g~CO_2}{1~mol~CO_2}=4.04~g~CO_2[/tex]
For C, we have to start with the conversion from grams of glucose to moles, the moles of glucose to moles of oxygen and finally the moles of oxygen to molecules. To do this we have to take into account the following conversion ratios:
-) 180 g of glucose = 1 mol glucose
-) 1 mol glucose = 6 mol oxygen
-) 1 mol oxygen = 6.023x10^23 molecules of O2
[tex]16.55~g~C_6H_1_2O_6\frac{1~mol~C_6H_1_2O_6}{180~g~C_6H_1_2O_6}\frac{6~mol~O_2}{1~mol~C_6H_1_2O_6}\frac{6.023x10^2^3~molecules~O_2}{1~mol~O_2}=~5.53x10^2^3~molecules~of~O_2[/tex]
This detailed answer explains the mass percent of carbon in glucose, the mass of CO2 produced by its combustion, and the number of oxygen molecules needed for complete combustion.
a) Mass Percent of Carbon in Glucose: Glucose has a molar mass of 180.16 g/mol. The molar mass of the carbon in one mole of glucose is 72.06 g (6 carbons in a molecule of glucose). Therefore, the mass percent of carbon in glucose is 40%.
b) Mass of CO2 Produced: The combustion of 16.55 g of glucose will produce 44.01 g of CO2.
c) Oxygen Molecules Needed: For the complete combustion of 16.55 g of glucose, 23 molecules of O2 are required.
Which statement is false?
O For the complete conversion of glucose to CO2, approximately 32 ATP molecules can be synthesized.
O The PDH complex is considered to be a reaction of the citric acid cycle.
O Succinate dehydrogenase is an integral membrane protein.
O Succinyl–CoA synthetase and succinic thiokinase are two names for the same enzyme.
Answer:
"The PDH complex is considered to be a reaction of the citric acid cycle" is the false statement
Explanation:
The PDH complex is involved in the oxidation of piruvate, producing acetate and free CO2 (besides ATP and NADH). Concomitantly, in presence of coenzyme A, it binds the acetate molecule to form Acetyl-CoA.
Acetyl-CoA, later, will enter the Krebs cycle reacting with oxalacetate to form citric acid.
The other statements options are true
Marianne gets dressed and puts on some of her favorite perfume. While applying the perfume, she accidentally gets some in her mouth and notices that it has abitter taste. After getting dressed, she heads to the kitchen to prepare some breakfast. She fills a small bowl with plain yogurt and takes a taste. The yogurt is too sour for her liking, so she adds honey to it for some sweetness. When she is done eating, she washes her bowl with soap and water. The soap is so slippery that she accidentally drops the bowl while cleaning it, causing it to break on the kitchen floor. Frustrated, Marianne sweeps up the broken glass and puts it in the trash, and then heads to work. Which substance displays a characteristic of an acid? A:soap B:yogurt C:perfume D:honey
Answer:
Yogurt displays characteristics of an acid.
Explanation:
The characteristics of an Acid are as follows:
It has an pH of less than 7It has a sour aftertaste(though one shouldn't try it)It increases the [tex]H^{+}[/tex] concentration of water.It reacts with Metals to liberate Hydrogen Gas.It gives out [tex]H^{+}[/tex] ions in aqueous solution.It turns blue litmus to red.Only Yogurt displays one of these properties ie. sour taste, therefore it is the only material having characteristic of acid.
Bittery taste and slippery touch are characteristics of bases.
Answer:
its c
Explanation:
2 A(g) + B(g) \longrightarrow⟶ 2 C(g)
Rate = k [A][B]
At the beginning of one trial of this reaction, [A] = 3.0 M and [B] = 1.0 M. The observed rate for the formation of C is 0.36 mol L-1 sec-1.
The numerical value of k, the rate constant, is closest to
108
0.040
0.12
6.0
Answer:
The correct answer is 0.12.
Explanation:
According the reaction kinetics we all know that
Rate=k[A][B]
K= rate constant
[A] = Concentration of A
[B] =Concentration of B
R=0.36mol/litre/sec
K=?
K= rate÷[A][B]
= 0.36÷3
=0.12
A chemistry student is given 3.00 L of a clear aqueous solution at 17.° C . He is told an unknown amount of a certain compound X is dissolved in the solution. The student allows the solution to cool to 17.° C . The solution remains clear. He then evaporates all of the water under vacuum. A precipitate remains. The student washes, dries and weighs the precipitate. It weighs 0.15 kg . Using the above information can you calculate the solubility, X, in water at 17 degrees C
Answer:
Solubility of X in water at 17°C is > 0,05 kg/L
Explanation:
When the solution is at 17°C the solution remains clear. That means that every X molecule is dissolved in water. The student obtained 0,15 kg of X in 3,00L of water. That means that under this conditions the concentration is:
[tex]\frac{0,15 kg X}{3,00L}[/tex] = 0,05kg/L
The solubility is defined as the concentration of a saturated solution. A saturated solution is a solution in which the solute (X) is in the maximum possible concentration. That means that solubility of X in water at 17°C is > 0,05 kg/L because X was completely dissolved in water and there is not enough data to obtain the exact solubility.
You need to add more X, heat the solution and then, cool it at 17°C obtaining a precipitate. The concentration of X substracting the mass of X that precipitate is the solubility of X in water.
I hope it helps!
Sodium metal reacts with water to produce hydrogen gas according to the following equation: 2Na(s) + 2H2O(l)2NaOH(aq) + H2(g) The product gas, H2, is collected over water at a temperature of 20 °C and a pressure of 754 mm Hg. If the wet H2 gas formed occupies a volume of 8.77 L, the number of moles of Na reacted was mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.
Answer:
Number of moles of sodium reacted = 0.707 moles
Explanation:
P(H₂) = P(T) – P(H₂O)
P(H₂) = 754 – 17.5 = 736.5 mm Hg
Use the ideal gas equation which
PV= nRT, where P is the pressure V is the volume, n is the number of moles R is the Gas Constant and T is temperature
Re- arrange to calculate the number of moles and using the data provided
n = P x V/R x T
n =736.5 x 8.77/62.36367 x (mmHg/mol K) x (20 + 273)
n = 0.35348668
n = 0.353 moles H₂
from the equation we know that
0.353 mole H₂ x 2mole Na/1mole H₂, So
0.353 x 2 = 0.707 mole Na
The number of moles of Sodium metal reacted were 0.707 moles.
0.708 moles of Na react with water to produce 8.77 L of H₂ collected over water at 20 °C and 754 mmHg.
Sodium metal reacts with water to produce hydrogen gas according to the following equation:
2 Na(s) + 2 H₂O(l) ⇒ 2 NaOH(aq) + H₂(g)
Hydrogen is collected over water. The total pressure of the gaseous mixture is 754 mmHg. If the vapor pressure of water is 17.5 mmHg, the partial pressure of hydrogen is:
[tex]P = pH_2 + pH_2O\\\\pH_2 = P - pH_2O = 754 mmHg - 17.5 mmHg = 737 mmHg[/tex]
Then, we will convert 20 °C to Kelvin using the following expression.
[tex]K = \° C + 273.15 = 20\° C + 273.15 = 293 K[/tex]
Hydrogen occupies 8.77 L at 293 K and 737 mmHg. We can calculate the moles of hydrogen using the ideal gas equation.
[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{ R \times T} = \frac{737mmHg \times 8.77L}{ (62.4mmHg/mol.K) \times 293K} = 0.354 mol[/tex]
The molar ratio of Na to H₂ is 2:1. The moles of Na that produced 0.354 moles of H₂ are:
[tex]0.354 mol H_2 \times \frac{2molNa}{1molH_2} = 0.708 mol Na[/tex]
0.708 moles of Na react with water to produce 8.77 L of H₂ collected over water at 20 °C and 754 mmHg.
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A compound decomposes by a first-order process. If 26 of the compound decomposes in 60 minutes, the half-life of the compound is ________minutes.
(A) 26
(B) 31
(C) 18
(D) 5
(E) 138
Answer:
(E) 138
Explanation:
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given:
26 % is decomposed which means that 0.26 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.26 = 0.74
t = 60 min
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.74=e^{-k\times 60}[/tex]
Taking natural log both sides, we get that:-
[tex]ln\ 0.74=-k\times 60[/tex]
k = 0.005018 min⁻¹
Also, Half life expression for first order:-
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]t_{1/2}=\frac {ln\ 2}{0.005018}\ min=138\ min[/tex]
The correct option is:- (E) 138
The half-life for a first-order reaction is constant and independent of the concentration of the reactant. Since only 26% decomposes in 60 minutes, the half-life must be greater than 60 minutes. Therefore, the correct answer is (E) 138 minutes.
Explanation:The student's question involves determining the half-life of a compound that decomposes by a first-order process. Given that 26% of the compound decomposes in 60 minutes, we understand that this is not a complete half-life since a half-life would be the time it takes for 50% of the compound to decompose.
In the case of a first-order reaction, the half-life is constant and does not depend on the concentration of the reactant. Therefore, even without the exact rate constant, knowing that after 60 minutes only 26% has decomposed tells us that the half-life must be greater than 60 minutes. Since none of the provided answers (A, B, C, D) are greater than 60, the correct answer must be (E) 138 minutes, assuming that the half-life is indeed not dependent on the concentration for this reaction.
The unit cell in a crystal of diamond belongs to a crystal system different from any we have discussed. The volume of a unit cell of diamond is 0.0454 nm3 and the density of diamond is 3.52 g/cm3. Find the number of carbon atoms in a unit cell of diamond.
Answer:
8.03 atoms of carbon
Given data;
Volume of unit cell diamond = 0.0454 nm³ (4.54×10⁻²³cm³)
Density = 3.52 g/cm³
Number of atoms = ?
Solution;
d = m/v
m = d × v
m = 3.52 g/cm³ × 4.54×10⁻²³cm³
m = 16 ×10⁻²³ g
Now we will determine the moles:
Number of moles of carbon = mass/ molar mass
Number of moles of carbon = 16 ×10⁻²³ g/ 12 g/mol
Number of moles of carbon = 1.3 ×10⁻²³ mol
one mole = 6.022 ×10²³ atoms
1.3 ×10⁻²³ mol × 6.022 ×10²³ atoms / 1mol
8.03 atoms of carbon
There are eight carbon atoms in the unit cell of a diamond, calculated using the provided unit cell volume and the density of diamond. The unit cell volume was first converted from nm³ to cm³, and then the mass of the unit cell was calculated using the formula for density. Finally, the number of atoms was found by dividing this mass by the molar mass of carbon and multiplying by Avogadro's number.
Explanation:The number of carbon atoms in a unit cell of a diamond can be calculated using the provided information of the diamond's crystal system, unit cell volume, and density. When pure, every carbon atom in a diamond forms four single bonds to four other atoms at the corners making the diamond a giant molecule. Consequently, diamond crystals are very hard and have high melting points.
First, convert the unit cell volume from nm³ to cm³. In the case of diamond, the unit cell volume is 0.0454 nm³, which is equivalent to 4.54 × 10-²³ cm³. Then you can use the formula for density (d=mass/volume) to calculate the mass of the unit cell, which is approximately 1.6 × 10-²² g. To find the number of atoms in the unit cell, we divide this mass by the molar mass of carbon, 12.01 g/mol, which gives us the number of moles, and then multiply by Avogadro's number (6.022 x 10²³ atoms/mole) to get the number of atoms. Therefore, there are eight carbon atoms in a unit cell of diamond.
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Arrange the following alkyl halides in order from most reactive to least reactive in an SN1 reaction: 2-bromo-2-methylpentane, 2-chloro-2-methylpentane, 3-chloropentane, and 2-iodo-2-methylpentane.
Answer:
2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-metylpentane > 3-chloropentane.
Explanation:
At a substitution reaction by SN1, the alkyl halide must lose its halide, and then an intermediary will be formed: a carbonium, which is an alkyl group with a positive charge in the carbon. The halide lost will be formed the halide ion, which is also an intermediary of the reaction.
The reactivity depends on the stability of the intermediaries (first of the carbonium, and second of the halide ion). As more bonded with carbons is the carbonium, more stable it is. The order of stability of the halides ions is from their electronegativity: as lower is it, as stable is the ion. The order is then: I⁻ > Br⁻ > Cl⁻ > F⁻.
2-bromo-2-methylpentane, 2-chloro-2-metylpentane, and 2-iodo-2-methylpentane, will form a 3-degree intermediary, so they will be more reactive than 3-chloropentane, which form a 2-degree intermediary. So, for the order of the stability of the halide ions, the order of reactivity is:
2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-metylpentane > 3-chloropentane.
The carbocation stability and quality of leaving group determine reactivity in SN1 reactions. The more stable the carbocation and better the leaving group, the higher the reactivity. Therefore, the alkyl halides are listed in the order of 2-iodo-2-methylpentane, 2-bromo-2-methylpentane, 2-chloro-2-methylpentane, and 3-chloropentane, from most reactive to least reactive.
Explanation:In SN1 reactions, stability of the carbocation intermediate is key to reactivity. As the carbon-halogen bond breaks, a carbocation is formed, and the more stable the cation, the faster it will form. Tetra-substituted carbocations are the most stable, followed by tri-, di-, and mono-substituted. So, in terms of reactivity, it goes as following:
2-iodo-2-methylpentane2-bromo-2-methylpentane2-chloro-2-methylpentane3-chloropentaneThe first three substances generate a tertiary carbocation which is more stable than the secondary carbocation generated by 3-chloropentane. Iodine is the most reactive because it's the best leaving group, then bromine, and chlorine is the least reactive among the three.
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Match each term with its definition. 1) Reaction coordinate A) represents the additional free energy that molecules must have to attain the transition state 2) ΔG° B) a generalized measure of the progress of the reaction through intermediate states 3) ΔG°‡ C) free energy of the reaction
Answer:
Reaction coordinate ⇒ a generalized measure of the progress of the reaction through intermediate states.
ΔG° ⇒ free energy of the reaction.
ΔG°‡ ⇒ represents the additional free energy that molecules must have to attain the transition state.
Explanation:
Reaction coordinate is a diagrammatic chart expressing a generalized measure of the progress of the reaction through intermediate states.
The free energy, G , of a chemical system is the free energy which is available for doing work. It is the driving force that brings about a chemical change.
In the Arrhenius equation, the term activation energy (Ea) is used to describe the energy required to reach the transition state. A superficially similar mathematical relationship, the Eyring equation, is used to describe the rate of a reaction.THUS, Instead of also using Ea, the Eyring equation uses the concept of Gibbs energy and the symbol ΔG‡ to denote the Gibbs energy of activation to achieve the transition state.
The slowest step of a reaction mechanism is called the
A.) Elementary Step
B.) Inhibitor
C.) Rate Law
D.) Rate-Determining Step
Answer:
D.) Rate-Determining Step
Explanation:
Rate - Determining Step -
The slowest step of any of the chemical reaction helps to find the overall rate of the chemical reaction , and , hence is known as the rate - determining step .
There are two type of reaction possible , one is elementary reaction and a complex reaction .
Elementary reaction is a single step reaction , and hence , that very step determined the rate of the reaction and hence is known as the rate determining step .
And ,
In case of a complex reaction , the reaction is preceded by many steps , and hence , the slowest step among other steps is known as the rate determining step of the reaction .
A 250 mL container of CO2 exerting a pressure of 1.00 atm is connected through a valve to a 500 mL container of O2 exerting a pressure of 2.00 atm.
When the valve is opened, the gases mix, forming a 750 mL mixture of CO2 and O2.
What is the total pressure of this mixture?
Explanation:
The given data is as follows.
[tex]P_{1}[/tex] = 1 atm, [tex]P_{2}[/tex] = 2 atm
[tex]V_{1}[/tex] = 250 ml, [tex]V_{2}[/tex] = 500 ml
Total volume = [tex]V_{1} + V_{2}[/tex] = 750 ml
Therefore, total pressure will be as follows.
[tex]P_{total} V_{total} = P_{1}V_{1} + P_{2}V_{2}[/tex]
[tex]P_{total} = \frac{P_{1}V_{1} + P_{2}V_{2}}{V_{total}}[/tex]
= [tex]\frac{1 atm \times 250 ml + 2 atm \times 500 ml}{750 ml}[/tex]
= 1.66 atm
Thus, we can conclude that total pressure of the given mixture is 1.66 atm.
Given the electronegativities below, which covalent single bond is most polar?
Element: H C N O
Electronegativity: 2.1 2.5 3.0 3.5
A) N-H B) O-C C) O-N D) C-H E) O-H
Answer:
So the correct answer is e) O-H
Explanation:
To determine the most polar bond we calculate the electronegativity difference in each bond and determine the highest value:
a) N-H = 3.0 - 2.1 = 0.9
b) O-C = 3.5 - 2.5 = 1
c) O-N = 3.5 - 3 = 0.5
d) C-H = 2.5 - 2.1 = 0.4
e) O-H = 3.5 - 2.1 = 1.4
So the correct answer is e) O-H
A piece of unknown metal weighs 217 g . When the metal absorbs 1.43 kJ of heat, its temperature increases from 24.5°C to 39.1°C. Determine the specific heat of the metal.
Answer:
The specific heat of the metal is 0.451 J/g°C (This could be iron).
Explanation:
Step 1: Data given
Mass of the unknown metal = 217 grams
The metal absorbs 1.43 kJ of heat
The temperature increases from 24.5°C to 39.1 °C
Step 2: Calculate the specific heat of the metal
q = m*c*ΔT
⇒ q = the heat transfer = 1.43 kJ
⇒ m = the mass of the unknown metal = 217 grams
⇒ c = the specific heat of the metal = TO BE DETERMINED
⇒ ΔT = the change of temperature = T2 - T1 = 39.1 - 24.5 = 14.6 °C
q = 1430 J = 217g * C * 14.6 °C
C = 0.451 J/ g°C
The specific heat of the metal is 0.451 J/g°C (This could be iron).
Hydrazine (N2H4) is used as a rocket fuel. The combustion products are nitrogen (N2) and water (H2O). Use the following information about bond energies to estimate the change in enthalpy associated with the combustion of 1 mol of hydrazine
Bond Bond Energy (kJ/mol)
N–N 163
N=N 418
N N 941 <- triple bond
N–H 388
O–O 146
O=O 495
O–H 463
a. −583 kJ/mol d. −358 kJ/mol b. +2,450 kJ/mol e. −1,970 kJ/mol c. −1,078 kJ/mol
Answer:
a. −583 kJ/mol
Explanation:
First we start with the balanced chemical equation
N2H4 + O2 ------> N2 + 2 H2O
Breaking bonds is endothermic, it requires energy whereas bond forming is exothermic, it expels energy. So the net energy change during this reaction can be calculated this way
ΔH = bond breaking energy - bond forming energy
= (4*388+163 + 495) - (941+4*463)
= -583 kJ/mol
Answer:
N–N 163
N=N 418
N N 941 <- triple bond
N–H 388
O–O 146
O=O 495
O–H 463
a. −583 kJ/mol d. −358 kJ/mol b. +2,450 kJ/mol e. −1,970 kJ/mol c. −1,078 kJ/mol
Explanation:
. If the following mixtures of two compounds are being separated by gas chromatography, which one of the compounds would be expected to elute from the column first (have the shortest retention time)? Consider boiling points that are only I or 2°C apart to be essentially identical. Consider compounds with the same functional groups to be identical in polarity if they differ by no more than a few carbons. Look up any necessary structures. Mixture Compound with the shortest retention time Polarity of the column packing a. ethyl acetate (bp 77°C) and ethyl butyrate (bp 120°C) polar nonpolar b, ethyl acetate (bp 77"C) and ethyl butyrate (bp 120°C) c. acetic acid (bp l 18°C) and n-butyl alcohol (bp 117c) polar d. sec-butyl alcohol (bp 100"C) nonpolar and ethyl propionate (bp 99c)
Answer:
a. Ethyl acetate.
b. Ethyl acetate.
c. N-butyl alcohol.
d. Sec-butyl alcohol.
Explanation:
In gas chromatography, the elution order depends of boiling point and polarity of compounds. The lower boiling point will elute first and the elution depends of kind of column you are using; if column is polar, non-polar compounds will elute first and vice versa.
a. ethyl acetate (bp 77°C) and ethyl butyrate (bp 120°C) polar:
As the structures of these two molecules just change in 2 carbons it is possible to consider that are the same polarity. Thus, we will see just boiling point, as ethyl acetate has lower boiling point will elute first.
b. ethyl acetate (bp 77°C) and ethyl butyrate (bp 120°C) nonpolar
Again, we will consider the same polarity for both compunds and ethyl acetate will elute first because its boiling point
c. acetic acid (bp 118°C) and n-butyl alcohol (bp 117c) polar.
As boiling point are similar we will see polarity: Acetic acid is a carboxilic acid and n-butyl alcohol has just a OH group in tis structure, that means acetic acid is more polar. As column is polar, acetic acid will be more retained doing n-butyl alcohol elutes first.
d. sec-butyl alcohol (bp 100"C) and ethyl propionate (bp 99c) nonpolar
As boiling point are similar we will see polarity: Sec-butyl alcohol has just a OH group but ethyl propionate is an esther, that means sec-butyl alcohol is more polar. As column is nonpolar, the polar compound will elute first, that is sec-butyl alcohol.
i hope it helps!
In gas chromatography, a compound with a lower boiling point tends to elute first from the column. Ethyl acetate, due to its lower boiling point, would elute first in both nonpolar and polar column conditions for mixtures a and b. For mixtures c and d, acetic acid and sec-butyl alcohol may elute first, though factors such as polarity may play a role in the precise order.
Explanation:Order of Elution in Gas Chromatography
In gas chromatography, the compound that elutes first is generally the one with the lower boiling point and weaker intermolecular forces, since it interacts less strongly with the stationary phase and hence moves faster through the column.
For mixture a with a nonpolar column packing, ethyl acetate (bp 77°C) should elute first due to its lower boiling point compared to ethyl butyrate (bp 120°C).For mixture b with a polar column packing, the compound with the shortest retention time would still be ethyl acetate because the boiling points are more decisive in this context.For mixture c, acetic acid and n-butyl alcohol have nearly identical boiling points, so other factors such as the difference in polarity might influence their elution order. However, if their polarities are similar here, the order may still depend on subtle differences in boiling points or interactions with the polar column.For mixture d, with a nonpolar column, sec-butyl alcohol (bp 100°C) would be expected to elute slightly before ethyl propionate (bp 99°C) due to the slightly higher boiling point suggesting stronger interactions with the stationary phase.The conclusion for each mixture takes into account boiling points as a primary factor, with polarity as a secondary factor when boiling points are comparable.
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In which of the following compounds does the transition-metal have the highest oxidation number? In which of the following compounds does the transition-metal have the highest oxidation number? Rb3[MoO3F3] K4[Mn(CN)6] Na[Ag(CN)2] K2[PtCl6] [Co(NH3)4Cl2]
Answer:
The transition-metal have the highest oxidation number is Mo in Rb₃[MoO₃F₃]
Explanation:
Transition metals are metals that are found in the middle of the periodic table.
For the compounds:
Rb₃[MoO₃F₃]: Rb is an alkali metal that has a charge of +1. that means [MoO₃F₃] is -3. The oxyfluoride O₃F₃ has a charge of -9. That means the Mo is +6
K₄[Mn(CN)₆]: Potassium is +1, thus, the [Mn(CN)₆] is -4, As the CN is -1 and there are 6, the Mn is +2
Na[Ag(CN)₂]: As Na is +1, the [Ag(CN)₂] is -1. As CN are -1, the Ag is +1
K₂[PtCl₆]: Potassium is +1, the [PtCl₆] is -2, As the Cl are -1, the Pt is +4
[Co(NH₃)₄Cl₂]: The NH₃ are neutral, The Cl are -1, that means the Co is +2.
The transition-metal have the highest oxidation number is Mo +6
I hope it helps!
In the given examples, the compound with the transition metal in the highest oxidation state is Rb3[MoO3F3] where molybdenum (Mo) is in the +6 oxidation state.
Explanation:The question you've asked is about transition metals and their oxidation states in various compounds. To find the highest oxidation state, we must identify the compound wherein the transition metal loses the maximum number of electrons. In the given compounds,
the oxidation state can be calculated for each metal respectively: Mo in Rb3[MoO3F3] has +6, Mn in K4[Mn(CN)6] has +4, Ag in Na[Ag(CN)2] has +1, Pt in K2[PtCl6] has +4, and Co in [Co(NH3)4Cl2] has +3. Thus, the compound where the transition metal has the highest oxidation number is Rb3[MoO3F3] where Mo has an oxidation state of +6.
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Given only the following data, what can be said about the following reaction?
3H2(g) + N2(g)---> 2NH3(g) ΔH=-92kJ
A.) The enthalpy of products is greater than the enthalpy of reactants
B.) The total bond energies of products are greater than the total bond energies of reactants
C.) The reaction is very fast
D.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia
Answer:
The correct answer is B the total bond energies of the product is greater than the total bond energies of the reactants.
Explanation:
3H2+N2=2NH3
This reaction is an endothermic reaction that means that reaction consume heat energy.
Due the above reaction is endothermic the enthalpy change of the reaction is negative(-ve).
During endothermic reaction the total bond energies of the product is greater than the total bond energies of the reactants.
Fe(II) can be precipitated from a slightly basic (aq) solution by bubbling oxygen through the solution, which converts Fe(II) to insoluble Fe(III):4Fe(OH)+(aq)+ 4OH-(aq)+O2(g)+2H2O(l)=4Fe(OH)3(s)how many grams of O2 are consumed to precipitate all of the iron in 85ml of 0.075 M Fe(II).
Answer:
0,051g of O₂ are consumed
Explanation:
The reaction of precipitation of Fe(II) is:
4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)
The moles of Fe(II) you have in 85ml of 0.075 M Fe(II) are:
0,085L×0,075M = 6,4x10⁻³ moles of Fe(II)
For the reaction, you require 4 moles of Fe(II) and 1 mol of O₂(g) to precipitate the iron. Thus, moles of O₂(g) you require are:
6,4x10⁻³ moles of Fe(II)×[tex]\frac{1molO_2}{4molesFe(II)}[/tex] = 1,6x10⁻³ moles of O₂(g)
These moles are:
1,6x10⁻³ moles of O₂(g)×[tex]\frac{32g}{1mol}[/tex] = 0,051g of O₂ are consumed
I hope it helps!
Which of these statements is true?
(a) All spontaneous reactions have a negative enthalpy change
(b) All spontaneous reactions have a positive entropy change
(c) All spontaneous reactions have a positive free-energy change
(d) All spontaneous reactions have a negative free-energy change
(e) All spontaneous reactions have a negative entropy change
Answer:
(d) All spontaneous reactions have a negative free-energy change
Explanation:
All spontaneous reactions releases free energy which can be used later for the work to be done. A reaction with a negative value for ΔG releases free energy and is thus spontaneous
Among the options listed, the correct statement is that all spontaneous reactions result in a negative free-energy change, reflecting the energy released during the process. So the correct option is d.
Explanation:The true statement among the options provided is:
(d) All spontaneous reactions have a negative free-energy change.
This is based on the principles of thermodynamics. The second law of thermodynamics states that the entropy of the universe increases in the course of spontaneous reactions. Nevertheless, a characteristic of spontaneous processes is that they release free energy, leading to a negative change in Gibbs free energy (ΔG). Therefore, not all spontaneous reactions have a negative enthalpy change (ΔH), nor do they always have a positive entropy change (ΔS), but they must result in a decrease in Gibbs free energy (negative free-energy change).
Ammonium iodide dissociates reversibly to ammonia and hydrogen iodide. NH4I (s) ⇌ NH3 (g) + HI (g) At 400 ºC, Kp = 0.215. Calculate the partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400 ºC. Complete the ICE box below as part of your answer.
A. 0.103 atmB. 0.215 atmC. 0.232 atmD. 0.464 atmE. 2.00 atm
Answer: The partial pressure of ammonia is 0.464 atm
Explanation:
For the given chemical equation:
[tex]NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)[/tex]
Initial: 1
At eqllm: 1-x x x
The expression for [tex]K_p[/tex] for the following equation is:
[tex]K_p=p_{NH_3}\times p_{HI}[/tex]
The partial pressures of solids and liquids are taken as 1 in the equilibrium expression.
We are given:
[tex]K_p=0.215[/tex]
Putting values in above equation, we get:
[tex]0.215=x\times x\\\\x=0.464atm[/tex]
Hence, the partial pressure of ammonia is 0.464 atm
Determine the enthalpy of neutralization in Joules/mmol for a solution resulting from 19 mL of 0.9 M NaOH solution and 19 mL of a HCl with the same molarity. If separately, each had a temperature of 28.8 degrees Celsius, and upon addition, the highest temperature reached by the solution was graphically determined to be 37.8 degrees Celsius. Round to the nearest whole number.
Answer:
3.00 J/mmol
Explanation:
The temperature of the solution is increasing, so it's an endothermic reaction. The heat that is being absorbed can be calculated by:
Q = m*c*ΔT
Where m is the total mass (HCl + NaOH), c is the specific heat, and ΔT is the variation of the temperature (final - initial). Because the molarity is small, the solutions are basically water, so c = 4.184 J/g°C.
The mass is the molar mass multiplied by the number of moles (n), which is the volume multiplied by the molarity:
nHCl = 0.9 mol/L *0.019 = 0.0171 mol
nNaOH = 0.9 mol/L *0.019 = 0.0171 mol
Molar masses: HCl = 36.5 g/mol; NaOH = 40 g/mol
mHCl = 36.5*0.0171 = 0.62415 g
mNaOH = 40 * 0.0171 = 0.6840 g
m = 1.30815 g
Q = 1.30815*4.184*(37.8 - 28.8)
Q = 49.2597 J
Because the number of moles of NaOH is equal to the number of moles of HCl, and the stoichiometry of the neutralization is 1:1, they both react completely, so the enthalpy can be calculated based in any of them. The enthalpy is the heat divided by the number of moles. In mmol, n is 17.1 mmol.
ΔH = 49.2597/17.1
ΔH = 2.88 J/mmol
ΔH = 3.00 J/mmol
To determine the enthalpy of neutralization, we need to calculate the moles of NaOH and HCl used, the change in temperature, and use the formula q = mcΔT. The enthalpy of neutralization is the heat energy released or absorbed during the neutralization reaction between an acid and a base to form 1 mole of water.
Explanation:The enthalpy of neutralization is the heat energy released or absorbed during the neutralization reaction between an acid and a base to form 1 mole of water. To determine the enthalpy of neutralization, we use the formula q = mcΔT, where q is the heat energy, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature. In this case, we are given the volumes and molarities of NaOH and HCl solutions, as well as the temperatures. We can use these values to calculate the enthalpy of neutralization.
First, we need to calculate the moles of NaOH and HCl used:
Convert the volume of NaOH solution to moles using the formula:Moles of NaOH = volume (L) × molarity (mol/L)
Calculate the moles of HCl using the same formula as above.Next, we calculate the change in temperature:
ΔT = final temperature - initial temperature
Finally, we can calculate the enthalpy of neutralization using the formula:
Enthalpy of neutralization = q / moles of acid
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The spectrum of a hydrogen atom contains a total of 26 lines, and that of lithium possess approximately eight times as many. This provides evidence for the quantum mechanical model of the atom due to
larger mass of the lithium atom.
greater number of lithium isotopes.
greater attraction between the electrons and the lithium nucleus.
greater number of occupied sublevels in the lithium atom
Answer:
greater number of occupied sublevels in the lithium atom
Explanation:
Although Hydrogen has only one electron, the atom has several different energy levels. When the electron transitions from a higher energy level to a lower one, photons are released. The photons absorb light at different wavelengths and colors.then they appear as lines on the spectrum of the hydrogen atom. Lithium has a greater number of occupied sub-levels which means more electrons will transition through the energy levels leading to more lines appearing on its spectrum .
A 9.96 mL sample of nitric acid required 13.25 mL of 0.085 M strontium hydroxide for titration. Calculate the molarity of the acid solution. (Hint: It's stoichiometry, you need the balanced equation)
The molarity of the nitric acid solution can be calculated using the principles of stoichiometry and titration. By knowing the molarity and volume of strontium hydroxide, we can determine the moles of nitric acid due to a 1:1 stoichiometry in the balanced equation. The molarity of nitric acid is found to be 1.13 M.
Explanation:The subject of the question is related to a concept in chemistry known as titration which is used to determine the concentration of an unknown solution (in this case, nitric acid) using a known solution (strontium hydroxide). Given that the volume of nitric acid is 9.96 mL and the volume and molarity of strontium hydroxide are 13.25 mL and 0.085 M respectively, we can determine the molarity of the nitric acid.
The balanced chemical equation for this acid-base reaction is HNO3(aq) + Sr(OH)2(aq) -> Sr(NO3)2(aq) + 2H2O(l). From this, we can see that one molecule of nitric acid reacts with one molecule of strontium hydroxide. Hence, we can apply the principle of stoichiometry and say that the moles of strontium hydroxide used in the titration are equal to the moles of nitric acid present in the solution.
The moles of strontium hydroxide can be calculated as Molarity x Volume (in L), which gives 0.085 mol/L * 13.25 mL = 0.01126 mol. Since the stoichiometry between strontium hydroxide and nitric acid is 1:1, this is also the moles of nitric acid in our sample. The molarity of nitric acid is then calculated as moles of solute/volume of solution in L, providing us with the molarity. Molarity = 0.01126 mol / 0.00996 L = 1.13 M.
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Calculate the molar solubility of Cd(OH)2 when buffered at a pH = 12.30. The Ksp for Cd(OH)2 is 2.5 x 10-14. Calculate the molar solubility of Cd(OH)2 when buffered at a pH = 12.30. The Ksp for Cd(OH)2 is
a.2.5 x 10-14 M.
b. 8.5 x 10-6 M
c. 6.3 x 10-11 M
d. 1.3 x 10-12 M
e. 5.0 x 10-2 M
f. 1.8 x 10-5 M
Answer:
c. 6,3x10⁻¹¹M
Explanation:
The solubility of a buffer is defined as the concentration of the dissolved solid in a saturated solution. For the Cd(OH)₂, solubility is:
[Cd²⁺] = S
The dissolution of Cd(OH)₂ is:
Cd(OH)₂ ⇄ Cd²⁺ + 2OH⁻
And the ksp is defined as:
ksp = [Cd²⁺][OH⁻]²
As ksp = 2,5x10⁻¹⁴ and [OH⁻] at pH=12,30 = 10^-(14-12,30) = 0,01995M
2,5x10⁻¹⁴ = [Cd²⁺]×(0,01995M)²
[Cd²⁺] = 6,3x10⁻¹¹M
That means solubility is c. 6,3x10⁻¹¹M
I hope it helps!
The molar solubility of Cd(OH)2 when buffered at a pH of 12.30 can be calculated using the concept of hydrolysis. The correct answer is 6.3 x 10^(-11) M.
Explanation:To calculate the molar solubility of Cd(OH)2 when buffered at a pH of 12.30, we need to use the concept of hydrolysis. Cd(OH)2 is a slightly soluble salt that undergoes hydrolysis in aqueous solution. At a high pH value, OH- ions react with water to form more OH- ions, shifting the equilibrium towards the hydrolysis reaction.
First, we write the balanced equation for the hydrolysis reaction: Cd(OH)2(s) ⇌ Cd2+(aq) + 2OH-(aq)Since OH- is being produced, we can assume that the concentration of OH- is much greater than that of Cd2+. Therefore, we can ignore the concentration of Cd2+ when calculating the solubility product (Ksp).Next, we use the equation for the hydrolysis reaction to write the expression for the solubility product constant (Ksp): Ksp = [Cd2+][OH-]^2The concentration of OH- ions in a basic solution is related to the pH by the equation: pOH = 14 - pHUsing this equation, we can calculate the pOH of the buffered solution: pOH = 14 - 12.30 = 1.70Then, we convert the pOH back to OH- concentration: [OH-] = 10^(-pOH) = 10^(-1.70)Finally, we substitute the calculated [OH-] into the expression for Ksp to solve for the molar solubility of Cd(OH)2: [Cd(OH)2] = sqrt(Ksp / [OH-]^2)After performing the calculations, the molar solubility of Cd(OH)2 when buffered at a pH of 12.30 is approximately 6.3 x 10^(-11) M. Therefore, the correct answer is option c. 6.3 x 10^(-11) M.
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Calculate the lattice energy for CaCl2 from the following information: Energy needed to vaporize one mole of Ca(s) is 192kJ. For calcium the first ionization energy is 589.5kJ/mol and the second ionization energy is 1146kJ/mol. The electron affinity of Cl is-348kJ/mol. The bond energy of Cl2 is 242.6kJ/mol of CI-Cl bonds. The standard heat of formation of CaCl2 is -795kJ/mol. (include the sign and your numerical answer, do not include units, do not use scientific notation, and round your answer to 1 decimal point)
Calculate the total energy for individual process, including energy to vaporize Ca(s), first and second ionization energies of Ca, bond energy of Cl2 and electron affinity of Cl. The lattice energy is the result of subtracting standard heat of formation of CaCl2 from the total energy calculated, which equals to 2635.9 kJ.
Explanation:The lattice energy for CaCl2 can be calculated using the given information and by understanding that the process of forming a salt like CaCl2 involves several different energies. First, there is the energy required to vaporize Ca(s), then the energy for first and second ionization of the calcium atom, followed by the electron affinity for Cl and the bond energy of Cl2.
To start calculating, first add the energy to vaporize Ca(s), the first and second ionization energies for Ca: 192kJ + 589.5kJ + 1146kJ = 1927.5kJ. When 2 Cl react with Ca, this process includes the bond energy for Cl2 and twice the electron energy for Cl: 242.6kJ + 2 * -348kJ = -453.4kJ. To calculate the energy required to form lattice, subtract the standard heat of formation of CaCl2 from the total energy calculated: 1927.5kJ - (-453.4kJ + -795kJ) = 2635.9 kJ
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Toluene and pyridine form ideal solutions. Consider a solution of toluene and pyridine prepared at 25°C. Assuming that the mole fractions of toluene and pyridine in the vapor phase are equal, calculate the composition of the solution. At 25°C the vapor pressures of toluene and pyridine are 28 and 21 torr, respectively.
A.χpyridine:B.χtoluene:
Answer:
X pyridine = 0.57
X toluene = 0.43
Explanation:
We will solve this question utilizing Raoult´s law for ideal solution of two volatile componens.
When we are told that the mole fractions of toluene and pyridine in the vapor phase are equal that implies that their partial pressures are also equal since partial pressure = χ Ptotal .
We also know that according to Raoult´s law the partial pressure of a compound A above a solution is given by
PA = XA x PºA where PA is the patial P of A, Xa is the mol fraction of A in solution and PºA is its pure vapor pressure, therefore we have for the two components in this question:
Ppyridine = Xpyridine Pºpyridine
Ptoluene = Xtoluene Pºtoluene
Ppyridine = Ptoluene ∴ XpyridinePºpyridine = Xtoluene Pºtoluene
but for a 2 component solution Xtoluene = (1 - Xpyridine)
so we have an equation with one unknown and we can solve for the mole fraction of pyridine:
XpyridinePºpyridine = (1-Xpyridine)Pºtoluene
substituting with our values:
Xpyridine x (21 torr) = (1-Xpyridine) x (28torr) (We can cancel the torr units)
Xpyridine x (21 ) = 28 - Xpyridine (28)
Xpyridine x ( 21 + 28 ) = 28
Xpyridine = 28/ (28+21) = 0.57
Xtoluene = 1 - 0.43 = 0.43
Phosphorus will react with oxygen to form diphosphorus pentoxide, P2O5 4P(s)+5O2(g) → 2P2O5(s) How many grams of P2O5 are formed when 4.89 g of P react with excess oxygen?
Answer:
11.2 gExplanation:
Given,
Mass of P = 4.89 g
Molar mass of [tex]P[/tex] = 30.9738 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{4.89\ g}{30.9738\ g/mol}[/tex]
[tex]Moles_{P}= 0.158\ mol[/tex]
According to the reaction shown below as:-
[tex]4P_{(s)}+5O_2_{(g)}\rightarrow 2P_2O_5_{(s)}[/tex]
4 moles of P on reaction forms 2 moles of diphosphorus pentoxide
Also,
1 mole of P on reaction forms 2/4 mole of diphosphorus pentoxide
0.158 moles of P on reaction forms [tex]\frac{2}{4}\times 0.158[/tex] moles of diphosphorus pentoxide
Moles of diphosphorus pentoxide = 0.079 moles
Molar mass of diphosphorus pentoxide = 141.9445 g/mol
Mass = Moles * Molar mass = 0.079 moles * 141.9445 g/mol = 11.2 g
A student is asked to determine the molarity of a strong base by titrating it with 0.250 M solution of H2SO4. The students is instructed to pipet a 20.0 mL portion of the strong base solution into a conical flask, to add two drops of an indicator that changes at pH = 7, and to dispense the standard H2SO4 solution from a buret until the solution undergoes a permanent color change. The initial buret reading is 5.00 mL and the final reading is 30.00 mL at the equivalence point.Which procedural error will result in a strong base molarity that is too high?Using 4 drops of indicator rather than the recommended 2 dropsUsing a buret with a tip filled with air rather than the H2SO4 solutionUsing an indicator that changes at pH = 5 rather than at pH = 7Using a conical flask that contains several drops of H2O
Answer:
The molarity of the strong base is 0.625 M
Which procedural error will result in a strong base molarity that is too high?
⇒ Using a buret with a tip filled with air rather than the H2SO4 solution
Explanation:
Step 1: Data given
Molarity of H2SO4 = 0.250 M
The initial buret reading is 5.00 mL
The final reading is 30.00 mL
Step 2: Calculate volume of H2SO4 used
30.00 mL - 5.00 mL = 25.00 mL
Step 3: Calculate moles of H2SO4
0.250 M = 0.250 mol/L
Since there are 2 H+ ions per H2SO4
0.250 mol/L * 2 = 0.500 mol/L
The number of moles H2SO4 = 0.500 mol/L * 0.025 L
Number of moles H2SO4 = 0.0125 mol
Step 4: Calculate moles of OH-
For 1 mol H2SO4, we need 1 mol of OH-
For 0.0125 mol of H2SO4, we have 0.0125 mol of OH-
Step 5: Calculate the molarity of the strong base
Molarity = moles / volume
Molarity OH- = 0.0125 mol / 0.02 L
Molarity OH - = 0.625 M
Which procedural error will result in a strong base molarity that is too high?
⇒ Using a buret with a tip filled with air rather than the H2SO4 solution
One mole of an ideal gas is compressed isothermally but irreversibly at 130oC from 2.5 bar to 6.5bar in a piston/cylinder device. The work required is 30% greater than the work of reversible, isothermal compression. The heat transferred from the gas during compression flows to a heat reservoir at 25oC. Claculate the entropy changes of the gas, the heat reservoir, and ΔStotal.
Answer:
Entropy change of gas = -7.944 J/mol*K
Entropy change of the reservoir = 13.96 J/mol*K
Total entropy change = 6.02 J/mol*K
Explanation:
Step 1: Data given
1 mol of ideal gas is compressed isothermally ( = constant temperature)
Temperature : 130 °C
Pressure = 2.6 bar to 6.5 bar
The work required is 30% greater than the work of reversible
The heat transferred from the gas during compression flows to a heat reservoir at 25°C
Step 2: Calculate entropy change for gas
ΔSgas = Cp*ln(T2/T1) - R*ln (P2/P1)
⇒ Since the temperature is constant, T2 = T1 so ln(T2/T1) = 0
ΔSgas = - R*ln (P2/P1)
⇒ with R = gas constant = 8.314
⇒ with P1 = initial pressure = 2.5 barr
⇒ with P2 = final pressure = 6.5 bar
ΔSgas = -8.314 *ln(6.5/2.5)
ΔSgas = -7.944 J/mol*K
Step 3: Calculate reversible work done
Reversible work done = R*T*ln (P2/P1)
Reversible work done = 8.314 * 403.15 * ln (6.5/2.5)
Reversible work done = = 3202. 67 J/mol
Step 4: Actual work done
Actual work done = 1.3 * 3202.67 = 4163.47 J/mol
For isothermal compression:
Q = -W
Q = -4163.47 J/mol
Step 5: Calculate entropy change for reservoir
Entropy change for reservoir ΔSres = -Q/Tres = 4163.47 /298.15
ΔSres = 13.96 J/mol*K
Step 6: Calculate total entropy change
ΔStotal = ΔSgas + ΔSres = -7.944 J/mol*K +13.96 J/mol*K
ΔStotal = 6.02 J/mol*K
The entropy change of the gas is 6.02 J/mol.
How to calculate entropy?Firstly, the entropy change for gas will be:
= 1 × 8.314 × In(2.5/6.5)
= -7.944 J/K
The reversible work done will be:
= 8.314 × 403.15 × In(6.5/2.5)
= 3202.67 J/mol
The actual work done will be:
= 1.3 × 3202.67
= 4163.47 J/mol
The entropy change for reservoir will be:
= -7.944 + 13.96
= 6.02 J/mol
In conclusion, the entropy change is 6.02 J/mol
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The pressure in car tires is often measured in pounds per square inch (lb/in.2), with the recommended pressure being in the range of 25 to 45 lb/in.2. Suppose a tire has a pressure of 27.0 lb/in.2 . Convert 27.0 lb/in.2 to its equivalent in atmospheres.
Answer:
1.36 atm
Explanation:
First, we have to transform lb and in² into the International System of Units.
The international unit for force is the Newton. The equivalence (based on Earth's gravity) is 1 lb = 4.448 N.The international unit for area is the m². The equivalence is 1 m² = 1550 in².The resulting unit is known as Pascal. 1 Pa = 1 N/m².[tex]\frac{27.0lb}{in^{2}} .\frac{4.448N}{1lb} .\frac{1550in^{2}}{1m^{2} } =1.38 \times 10^{5} Pa[/tex]
The equivalence between Pascal and atm is 101,325 Pa = 1 atm. Then,
[tex]1.38 \times 10^{5} Pa.\frac{1atm}{101,325Pa} =1.36atm[/tex]