Answer:
when the ethoxide ion is used, 2,3-dimethylbut-2-ene is obtained as the major product while using tert-butoxide yields 2,3-dimethylbut-1-ene
Explanation:
The predominant reaction product may be a Saytzeff or non-Saytzeff product. The product that actually predominates highly depends on the structure of the alkyl halide and the nature of base used in the reaction. A tertiary alkyl halide such as 2-bromo-2,3-dimethylbutane can undergo elimination by E1 or E2 mechanism depending on the structure of the base used in the reaction.
Dehydrohalogenation reactions yield alkenes, they are important in creating carbon-carbon double bonds in chemistry. For a tertiary alkyl halide, the use of a bulky base yields the non-Saytzeff product due to steric hinderance. Hence, when the ethoxide ion is used, 2,3-dimethylbut-2-ene is obtained as the major product while using tert-butoxide yields 2,3-dimethylbut-1-ene (non-Saytzeff product). The reactions are shown in the image attached to this answer.
If solid ammonium fluoride (NH4F) is dissolved in pure water, will the solution be acidic, neutral, or basic?
Answer:
Dissolving NH4F in water will form a weak acidic solution.
Explanation:
That it is a weak acid solution means that it has a pH below 7 but close to the value, that is, it does not contain as many acids as those substances that are around a pH of 1 to 4, generally weak acids have a pH approximately 5 to 6
The solution of solid ammonium fluoride in pure water has been slightly acidic in nature.
Ammonium fluoride has been an ionic compound formed by the interaction of cationic ammonia and anionic fluoride ions. The dissolution of ionic compounds will result in the compound in its dissociated ionic state.
The dissociation results in the formation of ammonium cation. The ammonium has been a strong acid.
The resulted anion has been fluoride. It has been a strong base, but slightly weaker than ammonia.
Thus the resultant solution will result in slightly acidic nature.
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4. If 13 percent of the carbon-14 in a sample of cotton cloth remains, what's the approximate age of the cloth? Show your work
The approximate age of the cloth is 17190 years.
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Original amount (N₀) = 100%Amount remaining (N) = 13%Number of half-lives (n) =?2ⁿ = 100 / 13
2ⁿ = 8
2ⁿ = 2³
n = 3
Finally, we shall determine the age of the cloth.
Half-life (t½) = 5730 yearsNumber of half-lives (n) = 3Time (t) =?t = n × t½
t = 3 × 5730
t = 17190 years
Thus, the approximate age of the cloth is 17190 years
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how many calories are in a 50g package of peanuts
Answer:
284 calories
Explanation:
There are 284 calories in 50 grams of peanuts.
Calorie breakdown: 73% fat, 11% carbs, 17% protein.
Write a balanced equation for the single-displacement reaction of Al(s) with CuSO4(aq). Include states of matter in your answer.
Answer:
2 Al(s) + 3 CuSO₄(aq) ⇒ 3 Cu(s) + Al₂(SO₄)₃(aq)
Explanation:
Let's consider the single displacement reaction of Al(s) with CuSO₄(aq). Copper has a higher reduction potential than aluminum, so aluminum will take the place of copper to form aluminum sulfate and metallic copper. The corresponding balanced chemical equation is:
2 Al(s) + 3 CuSO₄(aq) ⇒ 3 Cu(s) + Al₂(SO₄)₃(aq)
The chemical equation is 2 Al(s) + 3 CuSO₄(aq) ⇒ 3 Cu(s) + Al₂(SO₄)₃(aq)
Chemical equation:
here we considered the single displacement reaction of Al(s) with CuSO₄(aq). Also, Copper contained a higher reduction potential as compared to aluminum, due to this aluminum will take the place of copper to create aluminum sulfate and metallic copper. So the above should be the balance chemical equation.
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1. What form of matter is made from only one type of atom?
A molecule
B compound
C element
6.66%
D material
Answer:
A molecule is the answer.
Which type of rock is formed from existing rock or organisms?
Answer:
Sedimentary rocks
Explanation:
My explanation is that when an animal decomposes it body returns to the ground eventually being used in the rock cycle and rocks form this through the rock cycle when broken down by weathering and erosion.
Hope this helps you
Answer:
sedimentary rocks
They form from deposits that accumulate on the Earth's surface.
Silver crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?
Answer:
Edge length of the unit cell is 4.07x10⁻¹⁰m
Explanation:
In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:
a² + a² = b² = (4r)²
2a² = 16r²
a = √8 r
That means edge lenght is = √8 r
adius
As radius of Silver is 144pm = 144x10⁻¹²m:
a = √8 r
a = √8 ₓ 144x10⁻¹²m
a = 4.07x10⁻¹⁰m
Edge length of the unit cell is 4.07x10⁻¹⁰mClick the "draw structure" button to launch the drawing utility. Draw the product of the reaction of propanal with lithium aluminum hydride, followed by water.
Answer:
Product: propan-1-ol
Explanation:
IIn this case, we have to remember that [tex]LiAlH_4[/tex] is a reduction agent. So, this is a reduction reaction. The [tex]LiAlH_4[/tex] has the ability to produce hydride ions [tex]H^-[/tex]. This ion can attack the carbonyl group generating a negative charge in the oxygen. In the next step, the negative charge in the oxygen can attack a water molecule to protonate the molecule and produce propan-1-ol.
See figure 1
I hope it helps!
Which of the following is NOT one of the types of bonds? A. Ionic B. Metallic C. Covalent D. Valence
Considering the definition of bond and the different type of bonds, valence is not one of the types of bonds.
What is a chemical bondA chemical bond is defined as the force by which the atoms of a compound are held together. These are electromagnetic forces that give rise to different types of chemical bonds.
In other words, a chemical bond is the force that joins atoms to form chemical compounds and confers stability to the resulting compound.
Covalent bondThe covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas. The shared electron pair is common to the two atoms and holds them together.
Ionic bondAn ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions.
Metallic bondMetallic bonds are a type of chemical bond that occurs only between atoms of the same metallic element. In this way, metals achieve extremely compact, solid and resistant molecular structures, since the atoms that share their valence electrons.
SummaryIn summary, valence is not one of the types of bonds. The types of bonds are covalent, ionic and metallic.
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The combustion of propane may be described by the chemical equation C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g) How many grams of O2(g) are needed to completely burn 56.3 g C3H8(g)?
Answer:
204.7 g
Explanation:
(taking the atomic mass of C, H, O as 12, 1 and 16 respectively).
no. of moles of C3H8 burnt = 56.3 / (12x3 + 1x8)
= 1.27955 mol
From the equation, the mole ratio of C3H8 : O2 = 1:5
Hence,
the no. of moles of O2 required will be
=1.27955 x 5
= 6.397727 mol
Mass of O2 required = 6.397727 x (16x2)
= 204.7 g
What is the Percent composition of a pure substance that contains 7.22g of nickel, 2.53g of phosphorus and 5.25 g oxygen
Answer:
Explanation:
Total mass of substance = 7.22 + 2.53 + 5.25 g
= 15 g
percentage of nickel = 7.22 x 100 / 15
= 48.13
= 48.1 %
percentage of phosphorus = 2.53 x 100 / 15
= 16.87%
= 16.9%
percentage of oxygen = 5.25 x 100 / 15
= 35 %
The percent composition of the pure substance should be 48.1%, 16.9%, and 35%.
Calculation of the percent composition:
Total mass of substance = 7.22 + 2.53 + 5.25 g
= 15 g
Now
percentage of nickel = 7.22 x 100 / 15
= 48.13
= 48.1 %
And,
percentage of phosphorus = 2.53 x 100 / 15
= 16.87%
= 16.9%
And, finally
percentage of oxygen = 5.25 x 100 / 15
= 35 %
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Why is the separation of mixtures into pure or relatively pure substances so important when performing a chemical analysis?
Answer:
It is important to separate mixture into pure or relatively pure substances when performing a chemical analysis SO AS TO KNOW THE PROPERTIES COMING FROM EACH PART MIXTURE WHICH MAY INTERFERE WITH THE SEPARATION.
Explanation:
In chemistry, Mixture is the combination of two or more substances which are not combine chemically.
Mixture contain different substances with different physical and chemical properties.
It is important to purify the substances in a mixture so as to identify what properties are coming from each mixture and also some part of the mixture can interfere with the properties of other mixture present for skewing analysis.
4Ga + 3S2 → 2Ga2S3
1. How many grams of Gallium burned if 200.0 grams of Gallium(III)Sulfide formed?
Answer:
118.4 g
Explanation:
4 Ga + 3 S₂ → 2 Ga₂S₃
According to the equation, for every 4 moles of gallium burned, 2 moles of gallium(III) sulfide.
First, convert grams of Ga₂S₃ to moles. The molar mass is 235.641 g/mol.
(200.0 g)/(235.641 g/mol) = 0.8487 mol
Use the relationship above to convert moles of Ga₂S₃ to moles of Ga.
(0.8487 mol Ga₂S₃) × (4 mol Ga)/(2 mol Ga₂S₃) = 1.697 mol Ga
Convert moles of Ga to grams. The molar mass is 69.723 g/mol.
(1.697 mol Ga) × (69.723 g/mol) = 118.4 g
If 5.00 mL of a 0.5 M solution is diluted to a final volume of 100.0 mL, what is the concentration of the final dilute solution?
Answer:
0.025 M
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 5 mL
Initial concentration (C1) = 0.5 M
Final volume (V2) = 100 mL
Final concentration (C2) =..?
Using the dilution formula, we can obtain the final concentration of the diluted solution as follow:
C1V1 = C2V2
0.5 x 5 = C2 x 100
Divide both side by 100
C2 = (0.5 x 5)/100
C2 = 0.025 M
Therefore, the final concentration of the diluted solution is 0.025 M
The concentration of the final diluted solution is 0.025M
The dilution formula is expressed according to the formula:
[tex]C_1V_1=C_2V_2[/tex]
Given the following parameters
[tex]C_1=0.5M\\V_1=5.00mL\\V_2=100.0mL\\C_2=?[/tex]
Substitute the given parameters into the formula:
[tex]C_1V_1=C_2V_2\\0.5(5)=100C_2\\2.5=100C_2\\C_2=\frac{2.5}{100}\\C_2= 0.025M[/tex]
Hence the concentration of the final diluted solution is 0.025M
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help please i have 5 minutes to do this !!!
Answer:
A) One that occurs on its own
you mix 45 ml of .20M KOH in calorimeter. The temperature of both reactions before mixing is 21.5 C. The Cp of the calorimeter was 36 J/K. If the final temperature of the mixture is 23.6 C, what is the enthalpy change per mole of water produced?
Answer:
THE ENTHALPY CHANGE PER MOLE OF KOH IS 8400 Joules/ mole OF HEAT.
Explanation:
Heat = heat capacity * change in temperature
Heat capacity = 36 J/K
Temperature of the mixture before mixing = 21.5 C
Temperature of mixtire after mixing = 23.6 C
Calculate the change in temperature:
Change in temperature = 23.6 C - 21.5 C = 2.1 C
Heat = 36 * 2.1
Heat = 75.6 J of heat
In essence, 45 ml of 0.20 M of KOH produces 75.8 J of heat
The enthalpy change per mole of water:
It is important t obtain the number of moles involved in the reaction of 45 mL of 0.20 M of KOH
n = C V
n = 0.20 M * 45 *10^-3
n = 0.009 moles
Since number of moles = mass / molar mass
The mass of 45 ml of 0.20 M of KOH is then:
Molar mass = ( 39 + 16 + 1) g/mol = 56 g/mol
Mass = number of moles * molar mass
Mass = 0.009 * 56
Mass = 0.504 g
So therefore 0.504 g of KOH produces 75.6 J of heat
1 mole of KOH will produce x J of heat
1 mole of KOH = 56 g of KOH
0.504 g = 75.6 J
56 g = x J
x J = 56 * 75.6 / 0.504
x J = 8400 J / mole of KOH
The element potassium forms a _______ with the charge . The symbol for this ion is , and the name is . The number of electrons in this ion is
Answer:
The element potassium forms a cation with the charge +1 . The symbol for this ion is K⁺, and the name is potassium ion. The number of electrons in this ion is 18.
Explanation:
Potassium is a metal. It belongs to the group 1 elements. Metals form cations by losing electrons. Since potassium is a group element, it forms a cation by losing one electron. The charge it has is +1 due to the excess of the protons compared t the electrons by 1.
Potassium has 19 electrons. Potassium io on the other hand has 19-1 = 18 electrons.
Compound X has the formula C7H14. X reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form 3-methylhexane. Treatment of X with ozone follwed by zinc in aqueous acid gives a ketone plus formaldehyde (CH2=O). What is the structure of X?
Answer:
3-methylenehexane
Explanation:
In this case, we have two clues.
1) The hydrogenation reaction
2) The ozonolysis reaction
See figure 1.
With this in mind, lets analyze each clue. In the first reaction, we know that only 1 molecule of [tex]H_2[/tex] is added to the unknown molecule. This indicates that we only have 1 double bond in the molecule. Now, the next question is where is placed the double bond?
To answer this question, we have to use the second clue. In the ozonolysis reaction, a double bond is broken and is replaced with a carbonyl group. If, formaldehyde is formed the double bond is formed with a primary carbon. The primary carbons in the structure (given in the first reaction: 3-methylhexane) are carbons 1, 6, and 7. So, the double bond can be placed between carbons:
a) 6 and 5
b) 7 and 3
c) 1 and 2
To decide which one is the position of the double bond we have to keep in mind the second product of the ozonolysis reaction a ketone. With this in mind, the carbon bonded to the primary one (deduced by the formaldehyde) it has to be a tertiary carbon. The only option that has a primary carbon bonded to tertiary carbon is b). (See figure 2)
Finally, with this in mind the structure is 3-methylenehexane. To be sure, we can check the formula for the compound, [tex]C_7H_1_4[/tex] and the reactions. (See figure 3)
I hope it helps!
How did Ernest Rutherford change the atomic model?
A. He showed that the atom could be divided into smaller particles.
B. He showed that electrons were located within an atom's nucleus.
C. He showed that the atom contained both positive and negative
charges
D. He showed that most of an atom's mass was located in the atom's
nucleus.
Answer:
D. He showed that most of an atom's mass was located in the atom's
nucleus.
Explanation:
Ernest Rutherford changed the atomic model because of his experiment which was the gold foil experiment. A beam of alpha particles was aimed at a piece of gold foil, most particles passed through but some were scattered backward which showed that the middle of an atom (nucleus) is the where most of the mass is located.
Rutherford's model of atoms is the improved version of Thomson's model. In the model, it is stated that most of an atom's mass is located in the nucleus. Thus, option D is correct.
What is Rutherford's model?Ernest Rutherford gave the improved atomic model that postulated the failure of Thomson's model. Rutherford's model described the atom to consist of a sub-atomic particle with a positively charged nucleus.
The nucleus is in the center of the atom and had nearly all mass concentrated in it due to the presence of the protons and neutrons. The electrons were called negatively charged species that were present in the shells around the nucleus like the planets around the Sun.
Therefore, Rutherford's model showed mass concentrated in the center of the nucleus.
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When silver nitrate is added to the Fe/SCN equilibrium, why is the colorless intense and a precipitate forms?
Answer:
Here's what I get
Explanation:
You have an equilibrium reaction between Fe³⁺/ SCN⁻ and FeSCN²⁺.
[tex]\underbrace{\hbox{Fe$^{3+}$}}_{\text{pale yellow-green}} +\underbrace{\hbox{SCN$^{-}$}}_{\text{colourless}} \, \rightleftharpoons \, \underbrace{\hbox{Fe(SCN)$^{2+}$}}_{\text{deep blood red}} \\[/tex]
When you add AgNO₃, the Ag⁺ reacts with the SCN⁻. It forms a colourless precipitate of Ag(SCN).
Ag⁺(aq) + SCN⁻(aq) ⟶ AcSCN(s)
According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
If you add Ag⁺ to the equilibrium solution, it removes the SCN⁻ [as an Ag(SCN) precipitate].
The system responds by trying to replace the missing SCN⁻:
The Fe(SCN)²⁺ dissociates to form SCN⁻, so the position of equilibrium shifts to the left,
You now have more Fe³⁺ and SCN⁻ and less of the highly coloured Fe(SCN)²⁺ at the new equilibrium.
The deep red colour becomes less intense.
When silver nitrate is added to the Fe/SCN equilibrium, the colourless intense and precipitate forms because it settles at the bottom.
What is chemical equilibrium?Chemical equilibrium is the condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.
The added silver nitrate, [tex]AgNO_3[/tex] , effectively removes thiocyanate ions, [tex]SCN^{-1}[/tex], from the equilibrium system via a precipitation reaction when the [tex]Ag^{+1}[/tex] combines with [tex]SCN^{-1}[/tex] to produce insoluble silver thiocyanate, AgSCN, which settles to the bottom of the test tube.
Ag⁺(aq) + SCN⁻(aq) ⟶ AcSCN(s)
According to Principle, when we apply stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
Adding Ag⁺ to the equilibrium solution, it removes the SCN⁻ [as an Ag(SCN) precipitate].
The system responds by trying to replace the missing SCN⁻:
The Fe(SCN)²⁺ dissociates to form SCN⁻, so the position of equilibrium shifts to the left,
You now have more Fe³⁺ and SCN⁻ and less of the highly coloured Fe(SCN)²⁺ at the new equilibrium.
The deep red colour becomes less intense.
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What is the skeleton equation of the following?
Potassium oxide
Magnesium oxide
Sulfur trioxide
Sodium chloride
Answer:
Explanation:
Skeleton equation is opposite of word equation because here you use chemical formulas to write down the components.
Potassium Oxide = K2O
Magnesium Oxide = MgO
Sulfur Trioxide = SO3
Sodium Chloride = NaCl
Q3. Which of the following objects are chiral and which are achiral? a
club b. teacup c. football d. corkscrew e. tennis racket f. shoe g. portrai
pencil (8 marks)
Q4. Write a structure for each of the compounds listed. Explain why th
name given is incorrect, and give a correct name in each case (20 mai
a. l-methylbutane
b. 1.1,3-trimethylhexane
c. 5-octyne
d. 2-ethyl-1-propanol
e. 2.2-dimethyl-3-butanol
Submit via aduwiejuah a uds.edu.gh and deadline for submission is
July, 2020 at 11:00 AM.
Answer:
See explanation
Explanation:
In this question, we have to follow the IUPAC rules. Lets analyze each compound:
a. 1-methylbutane
In this compound we have a chain of 5 carbons, so the correct name is Pentane.
b. 1,1,3-trimethylhexane
In this compound, we longest chain is made of 7 carbons, so, we have to use the name "heptane". Carbon one would be the closet one to the methyl group, so the correct name is 2,4-dimethylheptane.
c. 5-octyne
In this case, carbon 1 would be the closet one to the triplet bond. With this in mind, the correct name is oct-3-yne.
d. 2-ethyl-1-propanol
In this compound, we longest chain is made of 4 carbons, so, we have to use the name "butane". Carbon one would be the carbon with the "OH" group, so the correct name is 2-methylbutan-1-ol.
e. 2.2-dimethyl-3-butanol
In this case, carbon 1 would be the closet one to the "OH". With this in mind, the correct name is 3,3-dimethylbutan-2-ol.
See figure 1
I hope it helps!
Determine the volume in mL of 0.37 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 20.8 mL of 0.51 M CH3CH2NH2(aq). Enter your answer with one decimal place. The Kb of ethylamine is 6.5 x 10-4.
Answer:
14.3mL you require to reach the half-equivalence point
Explanation:
A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:
CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.
Initial moles of CH₃CH₂NH₂ are:
20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
0.0106moles CH₃CH₂NH₂
To reach the half-equivalence point you require:
0.0106moles ÷ 2 = 0.005304 moles HClO₄
As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:
0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
14.3mL you require to reach the half-equivalence pointWhen The Kb of ethylamine is 6.5 x 10-4 is = 14.3mL you require to reach the half-equivalence point.
What is Ethylamine?
When A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, Therefore:
Then CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
So when the reaction is 1:1, to reach the equivalence point then you instruct to that add the moles of HClO₄ equal to the moles CH₃CH₂NH₂ you add originally. Also, When the half-equivalence point requires you to add half-moles of CH₃CH₂NH₂ you add originally.
Then Initial moles of CH₃CH₂NH₂ are:
After that 20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
Then 0.0106moles CH₃CH₂NH₂To get the half-equivalence point you require is:
Then 0.0106moles ÷ 2 = 0.005304 moles HClO₄
After that As the concentration of HClO₄ is 0.37M, the volume you require to add 0.005304moles is:
Then 0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
Therefore, 14.3mL you require to reach the half-equivalence point.
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Which example involves a phase change in which heat energy is released by the substance?
Ofreezing ice cream
O cooking a pot of soup
O melting ice under sunlight
O watching frost disappear into air
Answer:
Cooking a pot of soup
Explanation:
id say that because when you freeze ice cream, its already frozen, so no heat is being released. melting ice wouldn't be the answer because, once again, it is already frozen, and no heat is being released.
Answer:
the correct answer is freezing ice cream
Explanation:
i took the test & got this question correct. also, heat energy is released when freezing because there is no heat energy involved.
A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg of CO2 and 6.048 mg of H2O. What is the empirical formula
Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
The half-life of element X is 500 years. If there are initially 8 g of X, how much will remain after 1500 years
Answer:
1 g
Explanation:
From the formula;
N/No = (1/2)^t/t1/2
Where;
N= mass of radioactive element left after a time t = the unknown
No= mass of radioactive element originally present in the sample = 8g
t= time taken for N mass of the sample to remain = 1500
t1/2= half-life of the radioactive element = 500 years
Substituting values, we have;
N/8 = (1/2)^1500/500
N/8 = (1/2)^3
N/8 = 1/8
N= 1/8 ×8
N= 1 g
Therefore; mass of radioactive element left after 1500 years is 1 g
What is the boiling point of an aqueous solution that has a vapor pressure of 23.0 torr at 25 ∘C? (P∘H2O=23.78 torr; Kb= 0.512 ∘C/m).
Answer:
Boiling point of the solution is 100.964°C
Explanation:
In this problem, first, you must use Raoult's law to calculate molality of the solution. When you find the molality you can obtain the boiling point elevation because of the effect of the solute in the solution (Colligative properties).
Using Raoult's law:
Psol = Xwater × P°water.
As vapour pressure of the solution is 23.0torr and for the pure water is 23.78torr:
23.0torr= Xwater × 23.78torr.
0.9672 = Xwater.
The mole fraction of water is:
[tex]0.9672 = \frac{X_{H_2O}}{X_{H_2O}+X_{solute}}[/tex]
Also,
[tex]1 = X_{H_2O}+X_{solute}[/tex]
You can assume moles of water are 0.9672 and moles of solute are 1- 0.9672 = 0.0328 moles
Molality is defined as the ratio between moles of solute (0.0328moles) and kg of solvent. kg of solvent are:
[tex]09672mol *\frac{18.01g}{1mol}* \frac{1kg}{1000g} = 0.01742kg[/tex]
Molality of the solution is:
0.0328mol Solute / 0.01742kg = 1.883m
Boiling point elevation formula is:
ΔT = Kb×m×i
Where ΔT is how many °C increase the boiling point regard to pure solvent, Kb is a constant (0.512°C/m for water), m molality (1.883m) and i is Van't Hoff factor (Assuming a i=1).
Replacing:
ΔT = 0.512°C/m×1.882m×1
ΔT = 0.964°C
As the boiling point of water is 100°C,
Boiling point of the solution is 100.964°C
Boiling point of the solution is 100.964°C.
Raoult's Law:It says that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present.
Using Raoult's law:[tex]P_{sol} = X_{water} * P^o_{water}[/tex]
Given:
Vapor pressure of the solution = 23.0torr Vapor pressure of pure water = 23.78torrSubstituting the values:
[tex]23.0torr = X_{water} * 23.78torr\\\\0.9672 = X_{water}[/tex]
The mole fraction of water is:[tex]0.9762=\frac{X_{water}}{X_{water}+X_{solute}}[/tex]
The sum of the mol fractions of water and solute is 1.
We can consider,
Moles of water = 0.9672
Moles of solute = 1- 0.9672 = 0.0328 moles
Calculation for Molality:It is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.
[tex]\text{Mass of solvent}=0.9672*\frac{18g/mol}{1mol} *\frac{1kg}{1000g}\\\\\text{Mass of solvent} =0.01745kg[/tex]
Molality of the solution is:[tex]\text{Molality}= \frac{0.0328mol}{0.01742kg} \\\\\\text{Molality}= 1.883m[/tex]
Calculation of Boiling point:[tex]\triangle T = K_b*m*i[/tex]
Substituting the values in the above formula:
[tex]\triangle T = 0.512^oC/m*1.882m*1\\\\\triangle T = 0.964^oC[/tex]
Thus, Boiling point of the solution is 100.964°C, since boiling point of water is 100°C.
Find more information about Boiling point here:
brainly.com/question/40140
The percent errors of your experimental values of the specific heats may be quite large. Identify several sources of experimental error.
Answer:
The various sources of such errors are given below.
Explanation:
Sources of uncertainty or error could include necessary splattering of water leading to reduced cold water density as well as elevated temperatures of equilibration.The temperature might not have been reasonably stable when developers evaluated at every phase of the investigation or research.So that the percentage of someone specific produces heat exploratory value systems inaccuracies can be somewhat massive.
When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
g compute the specific heat capacity at constant volume of nitrogen gas. the molar mass of N2 is 29.0 You warm 1.8 kg ov water at a constant volume from 21 C to 30.5 C in a kettle
Complete question:
(a) compute the specific heat capacity at constant volume of nitrogen gas. the molar mass of N₂ is 29.0 You warm 1.8 kg of water at a constant volume 1.00 L from 21 C to 30.5 C in a kettle. For the same amount of heat, how many kilograms of 21∘C air would you be able to warm to 30.5∘C ?
(b) What volume (in liters) would this air occupy at 21∘C and a pressure of 1.00 atm? Make the simplifying assumption that air is 100% N₂
Answer:
(a) The specific heat capacity of N₂ is 715.86 J/kg.K
(b) The volume the air occupy at 21∘C is 8784.29 Liters
Explanation:
Given;
M is the molar mass of N₂ = 29 x 10⁻³ kg/mol
specific heat of N₂ at constant volume, Cv = 20.76 J/mol.K
(a)
The specific heat capacity of N₂ is calculated as;
[tex]C = \frac{C_v}{M} \\\\C = \frac{20.76}{29 *10^{-3}} \\\\C = 715.86 \ J/kg.K[/tex]
(b) heat capacity of water;
Q = mcΔθ
where;
c is the specific heat capacity of water = 4200 J/kg.K
m is mass of water, = 1.8 kg
Δθ is change in temperature, = 30.5 - 21 = 9.5 °C
Q = 1.8 x 4200 x 9.5
Q = 71820 J
Mass of nitrogen gas N₂, at this quantity of heat;
[tex]m_{N_2} = \frac{Q}{C*\delta \theta} \\\\m_{N_2} = \frac{71820}{715.86*9.5}\\\\m_{N_2} = 10.56 \ kg[/tex]
The volume this air occupy at 21∘C
Apply ideal gas law;
[tex]PV = nRT = \frac{m}{M} RT[/tex]
[tex]PV = \frac{mRT}{M} \\\\V = \frac{mRT}{MP}\\\\V = \frac{10.56(kg)*8.314*10^3(L.Pa/mol.K)*294(K)}{29*10^{-3}(kg)1.01325*10^5 (Pa)}\\\\V = 8784.29 \ Liters[/tex]