A cylinder contains 3.1 L of oxygen at 300 K and 2.7 atm. The gas is heated, causing a piston in the cylinder to move outward. The heating causes the temperature to rise to 610 K and the volume of the cylinder to increase to 9.4 L.
How many moles of gas are in the cylinder?
Express your answer using two significant figures.
Answer: The moles of gas present in the cylinder is 0.34 moles.
Explanation:
Given: [tex]P_{1}[/tex] = 2.7 atm, [tex]V_{1}[/tex] = 3.1 L, [tex]T_{1}[/tex] = 300 K
[tex]P_{2}[/tex] = ?, [tex]V_{2}[/tex] = 9.4 L, [tex]T_{2}[/tex] = 610 K
Formula used to calculate the final temperature is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm[/tex]
Now, moles present upon heating the cylinder are as follows.
[tex]P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol[/tex]
Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.
How much energy is required to melt 2 kg of gold? Use the table below and this equation.
a. 125.6 kJ
b. 1729 kJ
c. 10.4 kJ
d. 3440kJ
The equation for the energy required to melt 2 kg of gold is 3440 kJ.
What is energy?Energy is the ability to do work or cause change. It is an essential part of everyday life and is present in many forms, such as thermal energy, electrical energy, chemical energy, and mechanical energy. Energy can be converted from one form to another in order to do work.
The equation for calculating the energy required to melt a certain mass of material is Q = m x Lf, where Q is the energy required (in joules), m is the mass of the material (in kilograms), and Lf is the latent heat of fusion (in joules per kilogram).
Using the table below, we can see that the latent heat of fusion for gold is 1760 kJ/kg. Therefore, the equation for the energy required to melt 2 kg of gold is: Q = 2 kg x 1760 kJ/kg = 3440 kJ.
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A buffer is prepared containing 0.75 M NH3 and 0.20 M NH4 . Calculate the pH of the buffer using the Kb for NH3. g
Answer:
pH=8.676
Explanation:
Given:
0.75 M [tex]NH_{3}[/tex]
0.20 M [tex]NH_{4}[/tex]
The objective is to calculate the pH of the buffer using the kb for [tex]NH_3[/tex]
Formula used:
[tex]pOH=pka+log\frac{[salt]}{[base]}\\[/tex]
pH=14-pOH
Solution:
On substituting salt=0.75 and base=0.20 in the formula
[tex]pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\ =4.75+0.5740\\ =5.324[/tex]
pH=14-pOH
On substituting the pOH value in the above expression,
pH=14-5.324
Therefore,
pH=8.676
Calculate the solubility (in mol/L) of Fe(OH)3 (Ksp = 4.0 x 10^-38) in each of the following situations:
(A) Pure water (assume that the pH is 7.0 and remains constant).
(B) A solution buffered at pH = 5.0.
(C) A solution buffered at pH = 11.0.
Answer:
(A) 1.962x10^-10 M solubility in pure water
(B) 4.0 x 10^-33 M solubility
(C) 4.0 x 10^-27 M solubility
Explanation:
(A) Fe(OH)3 would give (Fe3+) and (3OH-)
Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
Let y = [Fe^3+]
Let 3y = [OH-]
4x10^-38 = (y)(3y)^3
4x10^-38 = 27y^4
y^4 = 4x10^-38 ÷ 27
y^4 = 1.481 x 10^-39
y = 1.962x10^-10 M solubility in pure water
(B) pH = 5.0
5.0 = - log [OH-]
-5.0 = log [OH-]
[OH-] = 10^-5.0 = 1.0 x 10^-5 M
So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
[Fe^3+][1.0 x 10^-5] = 4.0 x 10^-38
[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-5
= 4.0 x 10^-33 M solubility
(C) pH = 11.0
11.0 = - log [OH-]
-11.0 = log [OH-]
[OH-] = 10^-11.0 = 1.0 x 10^-11 M
So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
[Fe^3+][1.0 x 10^-11] = 4.0 x 10^-38
[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-11
= 4.0 x 10^-27 M solubility
Linoleic acid is a polyunsaturated fatty acid found, in ester form, in many fats and oils. Its doubly allylic hydrogens are particularly susceptible to abstraction by radicals, a process that can lead to the oxidative degradation of the fat or oil.
a. True
b. Flase
Answer:
True.
Explanation:
The information presented in the question above regarding linoleic acid is true. Linoleic acid is, in fact, found in many oils and fats in the ester form. In addition, linoleic acid is considered a polyunsaturated fatty acid, due to the presence of two unsaturations in its composition. Its chemical formula is CH3-(CH2)4-CH=CH-CH2-CH=CH-(CH2)7COOH and it is an essential fatty acid for the human body, as it is essential in the composition of arachidonic acid that is responsible for building muscle, managing body fat thermogenesis, and regulating core protein synthesis.
What is the law of multiple proportions?
A. The proportion of elements to compounds is constant.
B. All elements are found in equal proportions in nature.
C. Different compounds may contain the same elements but may have different ratios of those elements.
D. All compounds contain the same elements in the same proportions.
Answer:
I think
(d) All compounds contain the same elements in the same properties
How many grams of h2o are needed to produce 45g of NO
Which of the following is the correct way to balance the following chemical question:
2SnO2 + 4H2 -> 2Sn + 4H2O
SnO2 + 2H2 -> Sn + 2H2O
a. Both equation I and II are balanced, but equation I is the correct way to write the balanced equation.
b. Can you divide equation II by another factor and still have it be correct? Why or why not?
c. In a complete sentence, write down a method you could use to determine if an equation is written in the correct way.
Answer:
i have no answer for part A
part B
the one that has a 4 can be divided by 2 because reducing
part c
you can determine if an equation is written in the correct way by balancing the equation as if it had not been done already.
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Which statement is true about the experiments? (5 points)
The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
The same amount of heat is absorbed in both the experiments because the heat absorbed depends only on the final temperature.
The heat absorbed in Trial 2 is about 3,674 J greater than the heat absorbed in Trial 1.
The heat absorbed in Trial 2 is about 5,021 J greater than the heat absorbed in Trial 1.
Answer:
Explanation:
Using the formula below to calculate the heat absorbed in each trial:
Q = m × c × ∆T
Where;
Q = amount of heat absorbed (J)
m = mass of substance (g)
c = specific heat of water (4.184J/g°C)
∆T = change in temperature (°C)
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Q = 30 × 4.184 × (40 - 0)
Q = 30 × 4.184 × 40
Q = 5,020.8J
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Answer:
The same amount of heat is absorbed in both the experiments because the product of mass, specific heat capacity, and change in temperature are equal for both.
Explanation:
Explanation:
Using the formula below to calculate the heat absorbed in each trial:
Q = m × c × ∆T
Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0 °C.
Q = 30 × 4.184 × (40 - 0)
Q = 30 × 4.184 × 40
Q = 5,020.8J
Trial 2: Heat 40.0 grams of water at 10.0 °C to a final temperature of 40.0 °C.
Q=40*4.184*30
Q=5020.8J
Using the following reaction:
H2SO4 (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H2O (l)
Calculate the molarity of the H2SO4 solution if 14.92 mL of NaOH was necessary to reach the endpoint of a titration. The molarity of the NaOH solution was 0.83 M and 25.18 mL of H2SO4 was added to the Erlenmeyer flask.
Answer:
The molar concentration of the H₂SO₄ solution is 0.28 M
Explanation:
Molar concentration = number of moles / volume in litres
Number of moles = molar concentration × volume
From the equation of reaction, molar ratio of acid to base = 1 : 2
Using the formula; Na/Nb = CaVa/CbVb
Where Na is the number of moles of acid; Nb = number of moles of base; Ca = concentration of acid; Va = volume of acid; Cb = concentration of base; Vb = volume of base; Na/Nb = mole ratio of acid to base
Substituting the given values in the equation:
1/2 = Ca × 25.18 / 0.93 × 14.92
Ca = 0.93 × 14.92/ 25.18 × 2
Ca = 0.28M
Therefore, the molar concentration of the H₂SO₄ solution is 0.28 M
Which equation expresses the solubility product of Zn3(PO4)2?
a. Ksp = [Zn2+][PO43]
b. Ksp = [Zn2+]3 [PO43]2
c. Ksp = 6[Zn2+][PO43]2
d. Ksp = 108[Zn2+][PO43]2
⇒b. Ksp = [Zn2+]3 [PO43]2
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define saturated and unsaturated fats
Answer:
unsaturated fats, which are liquid at room temperature,are different from saturated fat because they contain one or more double bonds and fewer hydrogen atoms on their carbon chain.
Predict the products from reaction of 2-hexyne with the following reagents: (a) 2 equiv Br2 (b) 1 equiv HBr (c) Excess HBr (d) Li in NH3 (e) H2O, H2SO4, HgSO4
Answer:
See explanation and image attached
Explanation:
The reactions of the alkynes involved are shown in the image attached to this answer.
First of all, the reaction of two equivalents of bromine with the alkyne converts it to a saturated compound as shown. One equivalent of HBr converts the alkyne to alkene while excess HBr completely reduces the compound to a saturated compound.
Li/NH3 reduces the alkyne to an alkeneby anti addition to the triple bond.
Reaction of the alkyne with H2O, H2SO4, HgSO4 converts it to an aldehyde as shown.
Which of the following elements is the largest in size
a. O
b. He
c. K
d. H
Answer:
C. K
i took this class before
Answer: The largest element is K
Explanation: As K has the largest radius among O,He and H
Give the following reaction: ammonium nitrate—> dinitrogen monoxide + water.
a.) Write a complete balanced chemical equation.
b.) Calculate the number of molecules of water produced by 11.2g of ammonium nitrate
Answer:
a) NH₄NO₃ ⇒ N₂O + 2 H₂O
b) 1.69 × 10²³ molecules
Explanation:
Step 1: Write the balanced equation
NH₄NO₃ ⇒ N₂O + 2 H₂O
Step 2: Convert 11.2 g of NH₄NO₃ to moles
The molar mass of NH₄NO₃ is 80.04 g/mol.
11.2 g × 1 mol/80.04 g = 0.140 mol
Step 3: Calculate the moles of H₂O produced
0.140 mol NH₄NO₃ × 2 mol H₂O/1 mol NH₄NO₃ = 0.280 mol H₂O
Step 4: Calculate the number of molecules in 0.280 moles of water
We will use Avogadro's number.
0.280 mol × 6.02 × 10²³ molecules/1 mol = 1.69 × 10²³ molecules
Consider the balanced chemical equation below.
3 A ⟶ C + 4 D
How many moles of D would be produced if 9 moles of A were used?
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What is the cell potential of an electrochemical cell that has the half-reactions shown below?
Ag⁺ + e⁻ → Ag
Fe → Fe³⁺ + 3e⁻
Answer:
E°(Ag⁺/Fe°) = 0.836 volt
Explanation:
3Ag⁺ + 3e⁻ => Ag°; E° = +0.800 volt
Fe° => Fe⁺³ + 3e⁻ ; E° = -0.036 volt
_________________________________
Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...
E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt
A ball is thrown straight up into the air with a speed of 21 m/s. If the ball has a mass of 0.1 kg, how high does the ball go? Acceleration due to gravity is g = 9.8 m/s^2
Answer:
answer = 22.5m
Explanation:
using
[tex]s = \frac{ {v}^{2} - {4}^{2} }{2a} [/tex]
s= (0²-21²)/2(-9.8)
s= -441/19.6
s= 22.5m
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Suppose a 48. L reaction vessel is filled with 1.6 mol of Br2 and 1.6 mol of OCl2. What can you say about the composition of the mixture in the vessel at equilibrium
Answer:
There will be very little of BrOCl BrCl
Explanation:
Based on the equilibrium:
Br2(g) + OCl2(g) ⇄ BrOCl(g) + BrCl(g)
The equilibrium constant, Kc, is:
Kc = 1.58x10⁻⁵ = [BrOCl] [BrCl] / [Br2] [OCl2]
As Kc is <<< 1, in equilibrium, the concentration of products will remain lower regard to the concentration of the reactants. That means, right answer is;
There will be very little of BrOCl BrClExplain carefully what happen to the propanol-water system if approximately
50% of propanol by mass is fractionally distilled. What will be the distillate and
the residue?
Answer:
In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C. You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures. I've also included on the same diagram a vapor composition curve in exactly the same way as we looked at on the previous pages about phase diagrams for ideal mixtures.
All --- is a good conductor of heat
3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If 0.2815 g of barium sulfate was obtained, what was the mass percentage of barium in the sample
Answer:
[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]
Explanation:
From the question we are told that:
Mass of mixture [tex]m=3.455g[/tex]
Mass of Barium [tex]m_b=0.2815g[/tex]
Equation of Reaction is given as
[tex]Ba2+ + H2SO4 => BaSO4 + 2 H+[/tex]
Generally the equation for Moles of Barium is mathematically given by
Since
[tex]Moles of Ba^{2+} = Moles of BaSO_4[/tex]
Therefore
[tex]Moles of Ba^{2+} = \frac{mass}{molar mass of BaSO4}[/tex]
[tex]Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol[/tex]
Generally the equation for Mass of Barium is mathematically given by
[tex]Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}[/tex]
[tex]Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g[/tex]
Therefore
[tex]Ba\ percentage\ in\ Mass = mass of Ba^{2+}/mass of sample * 100%[/tex]
[tex]Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%[/tex]
[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]
Pls pls pls help me
Given 0.60 mol CO2, 0.30 mol CO, and 0.10 mol H20, what is the partial pressure of the CO if the total pressure of the mixture was 0.80 atm?
Answer:
Explanation:
/ means divided by
* means multiply
1. formula is
partial pressure = no of moles(gas 1)/ no of moles(total)
0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->
.3/(.6+.3+.1) =
.3/1 =
.3 =
partial pressure of CO
2.
.3 * .8 atm = .24
khanacademy
quizlet
The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.
Dalton's Law of Partial pressureDalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.
Dalton's Law of partial pressure using mole fraction of gas
Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure
Now, we have to find the first mole fraction of CO
Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]
= [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]
= [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]
= [tex]\frac{0.30}{1}[/tex]
= 0.3
Now, put the value in above equation, we get that
Partial pressure of carbon monoxide (CO)
= Mole fraction of carbon monoxide (CO) × Total pressure
= 0.3 × 0.8
= 0.24 atm
Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.
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Assign priorities in the following set of substituents according to Cahn-Ingold-Prelog rules.
-OCH3 -Br -Cl -CH2OH
А B C D
(Provide your ranking through a string like abcd, starting with the one with the highest priority).
Answer:
Assign priorities in the following set of substituents according to Cahn-Ingold-Prelog rules.
-OCH3 -Br -Cl -CH2OH
Explanation:
To give priorities for the substituents that are attached to chiral carbon and to assign either R or S-configuration the following rules were proposed:
1. The atom with the highest atomic number is given first priority.
2. If the Groups attached to chiral carbon are having the same first atom, then check for the atomic number of the second atom.
Among the given groups,
-Br has the highest atomic number, so it is given first priority.
Then, -Cl.
Then, -OCH3
and the last one is -CH2OH.
Hence, the order is :
BCAD.
Many reactions involve a change in hybridization of one or more atoms in the starting material. In this reaction, determine the hybridization of the indicated atoms in the organic starting material, and determine if they have changed hybridization during the reaction.
Answer:
Please find the complete question and its solution file in the attachment.
Explanation:
Using the balanced equation for the combustion of ethane: 2C2H6 + 7O2 → 4CO2 + 6H2O, how many moles of O2 needed to produce 12 moles of H2O?
Answer:
14 moles of oxygen needed to produce 12 moles of H2O.
Explanation:
We are given that balance eqaution
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
We have to find number of moles of O2 needed to produce 12 moles of H2O.
From given equation
We can see that
6 moles of H2O produced by Oxygen =7 moles
1 mole of H2O produced by Oxygen=[tex]\frac{7}{6}[/tex]moles
12 moles of H2O produced by Oxygen=[tex]\frac{7}{6}\times 12[/tex]moles
12 moles of H2O produced by Oxygen=[tex]7\times 2[/tex]moles
12 moles of H2O produced by Oxygen=14 moles
Hence, 14 moles of oxygen needed to produce 12 moles of H2O.
The amount of oxygen required for the combustion of ethane to produce 12 moles of water is 14 moles.
How are the moles produced in reaction calculated?The moles of oxygen produced in the reaction can be given from the stoichiometric law of the balanced chemical equation.
The balanced chemical equation for the combustion of ethane is:
[tex]\rm 2\;C_2H_6\;+\;7\;O_2\;\rightarrow\;4\;CO_2\;+\;6\;H_2O[/tex]
The 6 moles of water are produced from 7 moles of oxygen. The moles of oxygen required to produce 12 moles of water are:
[tex]\rm 6\;mol\;H_2O=7\;mol\;Oxygen\\12\;mol\;H_2O=\dfrac{7}{6}\;\times\;12\;mol\;O_2\\ 12\;mol\;H_2O=14\;mol\;O_2[/tex]
The moles of oxygen required to produce 12 moles of water are 14 moles.
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Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 35. g of hydrochloric acid is mixed with 73.0 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Round your answer to significant digits.
Answer:
No mass of HCl could be left over by the chemical reaction because is the limting reactant and it is all consumed.
Explanation:
Our reactants are: HCl and NaOH
Products are: NaCl and H₂O
This is a neutralization reaction that can also be called an acid base reaction, an acid and a base react to produce water and a neutral salt, in this case where we have strong acid and base.
Ratio is 1:1. We convert mass to moles:
35 g . 1 mol / 36.45 g = 0.960 moles of HCl
73 g . 1 mol / 40 g = 1.82 moles of NaOH
As ratio is 1:1, for 0.960 moles of HCl we need 0.960 moles of NaOH and for 1.82 moles of NaOH, we need 1.82 moles of acid.
As we only have 0.960 moles of HCl and we need 1.82 moles, no acid remains after the reaction goes complete. HCl is the limiting reactant, so the acid, it is all consumed.
Name the following aldehyde PLEASE PLEASE HELP
Answer:
Explanation:
Answer is D 2,5-dimethylheptanal
You should accern the lowest possible number close to the parent name
Classify each aqueous solution as acidic, basic, or neutral at 25 °C.
Acidic
Basic
Neutral
pH - 3.41
pH = 10.25
pH = 7.00
[H+] -3.5 x 10-5
[H+] - 6.7 x 10-9
[OH-]-5.8 x 10-4
[H0] -1.0 x 10-7
[OH-] - 4.5 x 10-13
Answer:
pH - 3.41 = acidic
pH = 10.25 = basic
pH = 7.00 = neutral
[H+] -3.5 x 10-5 = acidic
[H+] - 6.7 x 10-9 = basic
[OH-]-5.8 x 10-4 = basic
[H0] -1.0 x 10-7 = neutral
[OH-] - 4.5 x 10-13 = acidic
Explanation:
Let us note that from the pH scale, a pH of;
0 - 6.9 is acidic
7 is neutral
8 - 14 is basic
But pH= - log [H^+]
pOH = -log [OH^-]
Then;
pH + pOH = 14
Hence;
pH = 14 - pOH
For [H+] -3.5 x 10-5
pH = 4.46 hence it is acidic
For [H+] - 6.7 x 10-9
pH = 8.17 hence it is basic
[OH-]-5.8 x 10-4
pH= 10.76 hence it is basic
[H0] -1.0 x 10-7
pH = 7 hence it is neutral
[OH-] - 4.5 x 10-13
pH = 1.65 hence it is acidic
