Answer:
The correct answer is - option B - lowest b.p.... isobutane < acetone < isopropanol ...highest b.p.
Explanation:
Isobutane has lowest boiling point due to no hydrogen bonding and no diole to dipole interaction found in them. Isobutane only shows weak dispersion force.
Acetone has dipole dipole interaction but due to lack of Hydrogen bonding they have low boiling point than isopropanol but higher than isobutanol.
Isopropanol is the compound that has ability to form hydrogen bonding with other molecule its boiling point is maximum among all three.
Thus, the correct answer is - option B - lowest b.p.... isobutane < acetone < isopropanol ...highest b.p.
Given the data: Ag2O(s), = ‑31.1 kJ mol-1, S° = +121.3 J mol-1 K-1 Ag(s), = 0.00 kJ mol-1, S° = +42.55 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = +205.0 J mol-1 K-1 Calculate the temperature at which = 0 for the reaction, Ag2O(s) → 2 Ag(s) + ½ O2(g). Assume that, since the physical states do not change, and are independent of temperature between ‑50.0 °C and 950.0 °C.
Answer:
[tex]T=469.1K\\\\T=195.9\°C[/tex]
Explanation:
Hello,
In this case, for the given decomposition reaction, we can compute the enthalpy of reaction considering the enthalpy of formation of each involved species (products minus reactants):
[tex]\Delta _rH=2\Delta _fH_{Ag}+\frac{1}{2} \Delta _fH_{O_2}-\Delta _fH_{Ag_2O}\\\\\Delta _rH=2*0.00+\frac{1}{2} *0.00-(-31.1)=31.1kJ/mol[/tex]
Next, the entropy of reaction considering the standard entropy for each involved species (products minus reactants):
[tex]\Delta _rS=2S_{Ag}+\frac{1}{2} S_{O_2}-S_{Ag_2O}\\\\\Delta _rS=2(42.55)+\frac{1}{2} (205.0)-(121.3)=66.3J/(mol*K)[/tex]
Next, since the Gibbs free energy of reaction is computed in terms of both the enthalpy and entropy of reaction at the unknown temperature, for such Gibbs energy equaling 0, the temperature (in K and °C) turns out:
[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\0=31.1kJ/mol-T(66.3\frac{J}{mol*K}*\frac{1kJ}{1000J} )\\\\T=\frac{31.1kJ/mol}{0.0663kJ/(mol*K)} =469.1K\\\\T=195.9\°C[/tex]
Which is within the given rank.
Best regards.
For the following reaction, 6.99 grams of oxygen gas are mixed with excess nitrogen gas . The reaction yields 10.5 grams of nitrogen monoxide . nitrogen ( g ) oxygen ( g ) nitrogen monoxide ( g ) What is the theoretical yield of nitrogen monoxide
Answer:
13.11 g.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below :
N2 + O2 —> 2NO
Next, we shall determine the mass of O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 1 x 32 = 32 g.
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO from the balanced equation = 2 x 30 = 60 g.
From the balanced equation above,
32 g of O2 reacted to produce 60 g of NO.
Finally, we shall determine the theoretical yield of NO as follow:
From the balanced equation above,
32 g of O2 reacted to produce 60 g of NO.
Therefore, 6.99 g of O2 will react to produce = (6.99 x 60)/32 = 13.11 g of NO.
Therefore, the theoretical yield of nitrogen monoxide, NO is 13.11 g.
The decomposition of nitramide in aqueous solution at 25 °C NH2NO2(aq)N2O(g) + H2O(l) is first order in NH2NO2 with a rate constant of 4.70×10-5 s-1. If an experiment is performed in which the initial concentration of NH2NO2 is 0.384 M, what is the concentration of NH2NO2 after 31642.0 s have passed? M
Answer:
[tex][NH_2NO_2]=0.0868M[/tex]
Explanation:
Hello,
In this case, for the given chemical reaction, the first-order rate law is:
[tex]r=\frac{d[NH_2NO_2]}{dt} =-k[NH_2NO_2][/tex]
Which integrated is:
[tex][NH_2NO_2]=[NH_2NO_2]_0exp(-kt)[/tex]
Thus, the concentration after 31642.0 s for a 0.384-M solution is:
[tex][NH_2NO_2]=0.384M*exp(-4.70x10^{-5}s^{-1}*31642.0s)\\[/tex]
[tex][NH_2NO_2]=0.0868M[/tex]
Best regards.
Answer:
[A] = 0.0868 M
Explanation:
Rate constant = 4.70×10-5 s-1
First order reaction
Initial concentration, [A]o = 0.384 M
Final concentration, [A] = ?
Time, t = 31642.0 s
All these variables are related by the following equation;
[A] = [A]o e^(-kt)
[A] = 0.384 e^(-4.70×10-5 x 31642.0)
[A] = 0.384 e^(-1.4872)
[A] = 0.384 * 0.2260
[A] = 0.0868 M
Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answerβ-1,4- and α-1,6-glycosidicβ-1,4-glycosidicgalactosean unbranchedglucosea branchedfructoseα-1,6-glycosidicAmylose is ......... polymer of ....... units joined by ........ bonds. Amylopectin is ....... polymer of .......units joined by ........ bonds.
The words given are not clear, so the clear question is as follows:
Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:
A. β-1,4- and α-1,6-glycosidic
B. α-1,4-glycosidic
C. α-1,4-galactose
D. an unbranched glucose
E. a branched fructose
F. α-1,6-glycosidic
Amylose is ......... polymer of ....... units joined by ........ bonds.
Amylopectin is ....... polymer of .......units joined by ........ bonds.
Answer:
D. an unbranched glucose
C. α-1,4-galactose
B. α-1,4-glycosidic
E. a branched fructose
A. β-1,4- and α-1,6-glycosidic
F. α-1,6-glycosidic
Explanation:
Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.
Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.
Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.
Hence, the correct answers in the sequential order are:
Amylose:
D. an unbranched glucose
C. α-1,4-galactose
B. α-1,4-glycosidic
Amylopectin:
E. a branched fructose
A. β-1,4- and α-1,6-glycosidic
F. α-1,6-glycosidic
A small amount of solid calcium hydroxide is shaken vigorously in a test tube almost full of water until no further change occurs and most of the solid settles out. The resulting solution is:______.
Answer:
Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.
Explanation:
Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.
Small amounts of calcium hydroxide salt, [tex]Ca(OH)_{2}_(s)[/tex] is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.
I hope this explanation is helpful.
Write the equation for the reaction described: A solid metal oxide, , and hydrogen are the products of the reaction between metal and steam. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
Pb + 2H2O --> PbO2 + 2H2
Explanation:
Products:
Solid metal; PbO2
Hydrogen; H
Reactants:
Metal; Pb
Steam; H2O
Reactants --> Products
Pb + H2O --> PbO2 + H2
Upon balancing we have;
Pb + 2H2O --> PbO2 + 2H2
A student is using a coffee-cup calorimeter to determine the enthalpy change of the endothermic reaction of two aqueous solutions. After both solutions are added to the cup, the student neglects to put the lid on the cup. This would cause the magnitude of the calculated ΔH° value to be: the answer is: too small, since the solution will absorb heat from the room. But why? Wouldn't depend on if the reaction releases or absorbs heat. Wouldn't it be too large because heat escapes the cup? I'm so confused
Answer:
Explanation:
In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .
The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .
Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.
Sulfur dioxide reacts with oxygen to form sulfur trioxide. What change in hybridization of the sulfur occurs in this reaction ? g
Answer:
PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.
Explanation:
0.25 L of aqueous solution contains 0.025g of HCLO4 (strong acid) what will be the Ph of the solution g
Answer:
The pH of the solution will be 3
Explanation:
The strength of acids is determined by their ability to dissociate into ions in aqueous solution. A strong acid is any compound capable of completely and irreversibly releasing protons or hydrogen ions, H⁺. That is, an acid is said to be strong if it is fully dissociated into hydrogen ions and anions in solution.
Being pH=- log [H⁺] or pH= - log [H₃O⁺] and being a strong acid, all the HClO₃ dissociates:
HClO₄ + H₂O → H₃O⁺ + ClO₄-
So: [HCLO₄]= [H₃O⁺]
The molar concentration is:
[tex]molar concentration=\frac{number of moles of solute}{volume solution}[/tex]
The molar mass of HClO₄ being 100 g / mole, then if 100 grams of the compound are present in 1 mole, 0.025 grams in how many moles are present?
[tex]moles of HClO_{4} =\frac{0.025 grams*1 mole}{100 grams}[/tex]
moles of HClO₄= 0.00025
Then:
[tex][HClO_{4}]=\frac{0.00025 moles}{0.25 L}[/tex]
[tex][HClO_{4}]=0.001 \frac{ moles}{ L}[/tex]
Being [HCLO₄]= [H₃O⁺]:
pH= - log 0.001
pH= 3
The pH of the solution will be 3
A sample of radioactive silver contains two isotopes, 108Ag (denoted A) and 110Ag (denoted B). The second of these (B) has a half life of 24 seconds, whereas the first (A) has a half life of 2.3 minutes. If a sample contains equal numbers of each of these isotopes at the beginning of an experiment that runs for an hour, which of the following statements is correct?
A. At the end of the hour, isotope B has a greater decay constant λ than isotope A
B. At the end of the hour, isotope A has the same decay constant λ as isotope B
C. At the end of the hour, isotope A has a greater decay constant λ than isotope B
Answer:
A : At the end of the hour, isotope B has a greater decay constant λ than isotope A
Explanation:
Firstly, we need to understand that radioactive decay follows a first order rate law.
What this means is that we can calculate the radioactive decay constant using the following formula from the half-life
Mathematically;
[tex]t_{1/2}[/tex] = 0.693/λ
where λ represents the radioactive decay constant.
Rearranging the equation, we can have
λ = 0.693/[tex]t_{1/2}[/tex]
Now, to have a fair level playing ground, it is best that the half-life of both isotopes are in the same unit of time(seconds)
For A, the half-life = 2.3 minutes which is same as 2.3 × 60 = 138 seconds
For B, the half-life is 24 seconds
Thus, at the end of the hour, the decay constant for isotope A will be;
λ = 0.693/138 = 0.0050 [tex]s^{-1}[/tex]
For isotope B, the decay constant will be;
λ = 0.693/24 = 0.028875 [tex]s^{-1}[/tex]
We can see that the decay constant of isotope B is higher than that of A at the end of the experiment
A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
A 1.362 g sample of an iron ore that contained Fe3O4 was dissolved in acid with all of the iron being reduced to iron (II). The solution was acidified with sulfuric acid and titrated with 39.42 mL of 0.0281 M KMnO4, which oxidized the iron (II) to iron (III) while reducing the permanganate to manganese (II). Generate the balanced net ionic equation for the reaction. What is the mass percent of iron in this iron ore sample?
Answer:
a. MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
b. 18.17% of Fe in the sample
Explanation:
a. In the reaction, Fe²⁺ is oxidized to Fe³⁺ and permanganate, MnO₄⁺ reduced to Mn²⁺, thus:
Fe²⁺ → Fe³⁺ + 1e⁻
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O
5 times the iron and suming the manganese reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂Ob. Moles of permanganate in the titration are:
0.03942L × (0.0281 moles / L) = 1.108x10⁻³ moles of MnO₄⁻
Based on the reaction, 1 mole of permanganate reacts with 5 moles of iron, if 1.108x10⁻³ moles of MnO₄⁻ reacts, moles of iron are:
1.108x10⁻³ moles of MnO₄⁻ × (5 moles Fe²⁺ / 1 mole MnO₄⁻) =
4.431x10⁻³ moles of Fe²⁺. Molar mass of Fe is 55.845g/mol. 4.431x10⁻³ moles of Fe²⁺ are:
4.431x10⁻³ moles of Fe²⁺ ₓ (55.845g / mol) =
0.2474g of Fe you have in your sample.Percent mass is:
0.2474g Fe / 1.362g sample ₓ 100 =
18.17% of Fe in the sampleThe mass percent of iron in the sample is 22.6%.
The net ionic equation of the reaction is;
5Fe^2+(aq) + 8H^+(aq) + MnO4^- -----> 5Fe^3+(aq) + Mn^2+(aq) + 4H2O(l)
Number of moles of MnO4^- = 39.42/1000 L × 0.0281 M = 0.0011 moles
If 5 moles of Fe^2+ reacts with 1 mole of MnO4^-
x moles of Fe^2+ reacts with 0.0011 moles
x = 5 moles × 0.0011 moles/1 mole
x = 0.0055 moles
Mass of Fe^2+ = 0.0055 moles × 56 g/mol = 0.308 g
Mass percent of iron = 0.308 g/ 1.362 g × 100/1
= 22.6%
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Given the following reaction and data, A + B → Products
Experiment A (M) B (M) Rate (M/s)
1 1.50 1.50 0.320
2 1.50 2.50 0.320
3 3.00 1.50 0.640
Required:
a. What is the rate law of the reaction?
b. What is the rate constant?
Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
Rate = k×[A]b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
k = 0.213s⁻¹Calculate the energy required to heat 1.30kg of water from 22.4°C to 34.2°C . Assume the specific heat capacity of water under these conditions is 4.18·J·g−1K−1 . Round your answer to 3 significant digits.
Answer:
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
Explanation:
Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.
The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:
Q = c * m * ΔT
Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).
In this case:
[tex]c=4.18 \frac{J}{g*K}[/tex]m= 1.30 kg= 1,300 g (1 kg=1,000 g)ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.Replacing:
[tex]Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K[/tex]
Q= 64,121.2 J
The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J
11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. How many grams per kilogram of body weight is a lethal dose for 50% of domestic chickens?
Answer:
[tex]0.033g[/tex]
Explanation:
Hello,
In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:
[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]
Thus, we obtain:
[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]
That in grams is:
[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]
Regards.
What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 4.0 g of urea in 32.0 g of methanol, CH3OH
Answer:
0.0630
Explanation:
The molar mass of urea = 60 g/mol
we all know that:
[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
Then; the number of moles of urea
= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]
= 0.0667 mol
Similarly; the number of moles of methanol
= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]
= 0.9988 mol
The total number of moles = (0.0667 + 0.9988) mol
= 1.0655 mol
Finally,the mole fraction of urea [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]
[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]
= 0.0630
You wish to construct a galvanic cell with the anode consisting of a Ni electrode in a 1.0 M Ni(NO3)2 solution. What would be the highest standard cell potential if used as the cathode in this galvanic cell?
Answer:
Au^3+(aq) +3e ------> Au(s). 1.50 V
Explanation:
When we construct the galvanic cell, our intention is to produce energy by spontaneous electrochemical reactions. In order to have a spontaneous electrochemical reaction, E°cell must be positive. The more positive the value of E°cell, the more spontaneous the reaction is.
E°cell= E°cathode - E°anode
If E°cathode= 1.50 V
E°anode= -0.25 V
E°cell= 1.50 -(-0.25)
E°cell= 1.75 V
Hence the process; Au^3+(aq) +3e ------> Au(s) yields the highest standard cell potential
What is the concentration of A after 50.7 minutes for the second order reaction A → Products when the initial concentration of A is 0.250 M? (k = 0.117 M⁻¹min⁻¹)
Answer:
0.101 M
Explanation:
Step 1: Given data
Initial concentration of A ([A]₀): 0.250 MFinal concentration of A ([A]): ?Time (t): 50.7 minRate constant (k): 0.117 M⁻¹.min⁻¹Step 2: Calculate [A]
For a second-order reaction, we can calculate [A] using the following expression.
1/[A] = 1/[A]₀ + k × t
1/[A] = 1/0.250 M + 0.117 M⁻¹.min⁻¹ × 50.7 min
[A] = 0.101 M
g The electronic structure of which ONE of the following species cannot be adequately described by a single Lewis formula? (In other words, the electronic structure of which one can only be described by drawing two or more resonance structures?) A) C2H4 B) SO3 2– C) SO3 D) C3H8 E) HCN
Answer:
C) SO3
Explanation:
Lewis formula shows the bonding between atoms of a molecule and expresses the lone pair present in the atoms.
SO3 or Sulfur trioxide cannot be adequately described by a single Lewis formula because it has majorly 3 resonance structures because Sulfur does not follow the octet rule and can expand electrons in its outer shell.
Hence, the correct answer is C) SO3
What happens if we put raw eggs in a pot full of hot oil?
a sample of gas occupies a volume of 2.62 liters at 25 C and 1.00 atm. what will be the volume at 50 C and 2 atm
Answer:2.62 L
Explanation:
A sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.
What is ideal gas law ?The equation of state for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it is a decent approximation of the behavior of many gases under various situations.
An ideal gas is one in which there are no intermolecular attraction forces and all collisions between atoms or molecules are entirely elastic. It may be seen as a group of perfectly hard spheres that collide but do not else interact with one another.
By using ideal gas equation,
P₁ V ₁ ÷ T = P₂V₂ ÷ T
1 × 2.62 ÷ 25 = 2 × V₂ ÷ 50
V₂ = 1 × 2.62 × 50 ÷ 25 × 2
V₂ = 2.62 liters.
Thus, a sample of gas occupies a volume of 2.62 liters at 25° C and 1.00 atm. and the volume at 50° C and 2 atm then volume is 2.62 liters.
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Which of these species would you expect to have the lowest standard entropy (S°)?
a. CH4(g)
b. H2O(g)
c. NH3(g)
d. HF(g)
Answer:
d. HF(g)
Explanation:
Hello,
In this case, the standard entropy S° could be predicted by looking at the amount of bonds the compound has, thus, the fewer the number bonds, the lower the standard entropy, it means that d. HF(g) has lowest value as it has one bond only whereas methane has four bonds, water two bonds and ammonia three bonds.
Best regards.
Which of the following is a salt that could be generated by combining a weak acid and a weak base? Select the correct answer below:
a) NaCl
b) Na2SO4
c) NH4NO3
d) NH4F
Answer:
d) NH4F
Explanation:
Hello,
In this case, the base resulting from mixing a weak acid and a weak base is d) NH4F since ammonium hydroxide is a wear base and hydrofluoric acid is a weak acid.
Ammonium hydroxide is a weak base since it is not completely ionized in ammonium and hydroxyl ions:
[tex]NH_4OH\rightarrow NH_4^++OH^-[/tex]
Moreover, hydrofluoric acid is a weak acid since it is not completely ionized in hydrogen and fluoride ions:
[tex]HF\rightleftharpoons H^++F^-[/tex]
For the both of the substances, the limit is established by the basic and the acid dissociation constant respectively.
Regards.
A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia
Answer:
0.327 M
Explanation:
Step 1: Write the balanced equation
2 NH₃(g) ⇄ N₂(g) + 3H₂(g)
Step 2: Make an ICE chart
2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)
I x 0 0
C -2y +y +3y
E x-2y y 3y
Step 3: Find the value of y
The concentration of N₂ at equilibrium is 0.0161 M. Then,
y = 0.0161
Step 4: Find the value of x
The concentration of NH₃ at equilibrium is 0.295 M. Then,
x-2y = 0.295
x-2(0.0161) = 0.295
x = 0.327
Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the data and mechanism. a. the mechanism for this reaction is E2 b. an increase in 1-butene was observed when t-butoxide was used c. an increase in 1-butene was observed when methoxide was used d. the mechanism for this reaction is E1 e. no significant difference was observed
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
An object has a mass of 4.9g and a volume of 14.ml.what is the density of the object?
Answer:
0.35 g/mL
Explanation:
Use the formula D = [tex]\frac{m}{v}[/tex], where D is density, m is mass, and v is volume.
D = 4.9/14
D = 0.35
D = 0.35 g/mL
Calculate the [H+] and pH of a 0.0010 M acetic acid solution. The Ka of acetic acid is 1.76×10−5. Use the method of successive approximations in your calculations.
Answer:
[tex][H^+]=0.000123M[/tex]
[tex]pH=3.91[/tex]
Explanation:
Hello,
In this case, dissociation reaction for acetic acid is:
[tex]CH_3COOH\rightleftharpoons CH_3COO^-+H^+[/tex]
For which the equilibrium expression is:
[tex]Ka=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}[/tex]
Which in terms of the reaction extent [tex]x[/tex] could be written as:
[tex]1.74x10^{-5}=\frac{x*x}{[CH_3COOH]_0-x}=\frac{x*x}{0.0010M-x}[/tex]
Thus, solving by using a solver or quadratic equation we obtain:
[tex]x_1=0.000123M\\\\x_2=-0.000141M[/tex]
And clearly the result is 0.000123M, which also equals the concentration of hydronium ion in solution:
[tex][H^+]=0.000123M[/tex]
Now, the pH is computed as follows:
[tex]pH=-log([H^+])=-log(0.000123)\\\\pH=3.91[/tex]
Best regards.
If the Ksp for Li3PO4 is 5.9×10−17, and the lithium ion concentration in solution is 0.0020 M, what does the phosphate concentration need to be for a precipitate to occur?
Answer:
7.4 × 10⁻⁹ M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷
Concentration of lithium ion: 0.0020 M
Step 2: Write the reaction for the solution of Li₃PO₄
Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
Step 3: Calculate the phosphate concentration required for a precipitate to occur
The solubility product constant is:
Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³
[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³
[PO₄³⁻] = 7.4 × 10⁻⁹ M
The standard entropy of a substance refers to its entropy at:__________.
a. absolute zero and 1 bar
b. 0°C and 1 bar
c. 25 °C and 1 bar
d. 25 °C and 0 bar
Answer:
b. 0°C and 1 bar
Explanation:
Hello,
In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.
Best regards.
A student ran the following reaction in the laboratory at 242 K: 2NOBr(g) 2NO(g) Br2(g) When she introduced 0.143 moles of NOBr(g) into a 1.00 liter container, she found the equilibrium concentration of NOBr(g) to be 0.108 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc
Answer:
1.84 × 10⁻³
Explanation:
Step 1: Write the balanced equation
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
Step 2: Calculate the initial concentration of NOBr
0.143 moles of NOBr(g) are introduced into a 1.00 liter container. The molarity is:
M = 0.143 mol / 1.00 L = 0.143 M
Step 3: Make an ICE chart
2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
I 0.143 0 0
C -2x +2x +x
E 0.143-2x 2x x
Step 4: Find the value of x
The equilibrium concentration of NOBr(g) was 0.108 M. Then,
0.143-2x = 0.108
x = 0.0175
Step 5: Calculate the concentrations at equilibrium
[NOBr] = 0.108 M
[NO] = 2x = 0.0350 M
[Br₂] = x = 0.0175 M
Step 6: Calculate the equilibrium constant (Kc)
Kc = [0.0350]² × [0.0175] / [0.108]²
Kc = 1.84 × 10⁻³
