Answer:
Biphenyl
Explanation:
The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.
The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.
Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.
How many oxygen atoms are there in 0.25 mole of CO2?
Answer:
It is known that 1 mol of a molecule contains 6.023×1023 6.023 × 10 23 number of molecules. So, 0.25 moles of CO2 C O 2.
The correct option for the given question about Mole Concept is Na / 2 atom of oxygen.
What is mole?In chemistry, a mole is a unit. A mole of a substance is the mass of the substance containing precisely 12 gram of the same number of atoms as 12C.The mole is the SI unit used to measure a substance's quantity. It has the symbol mol.
How to solve this question?
In 1 mole number of molecules of CO₂ = Na (Avagadro Number)
In 0.25 mole number of molecules of CO₂ = 0.25 × Na molecules
In 1 molecule number of oxygen atom = 2 atom
In 0.25 × Na molecules number of oxygen atom = 2 × 0.25 × Na atom
In 0.25 × Na molecules number of oxygen atom = 0.5 × Na atom
So in 0.25 mole of CO₂ number of oxygen atom = Na / 2 atom
Thus we can conclude that in 0.25 mole of CO₂ number of oxygen atom will be Na / 2, where Na is Avagadro Number.
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Predict whether reactants or products will be favored at equilibrium for the below reaction.
Kp= 2.2 x 10^6 at 298K
2COF2 (g) + ⇌ CO2(g) + CF4(g)
Answer:
The products will be favored at equilibrium.
Explanation:
The balanced chemical equation for the reaction is the following:
2 COF₂ (g) + ⇌ CO₂(g) + CF₄(g)
The reactant is COF₂ (left side) and the products are CO₂ and CF₄ (right side).
The equilibrium constant is given by the ratio between the partial pressures (P) of products and reactants, because they are in the gas phase. Thus, the expression of the equilibrium constant is the following:
[tex]Kp = \frac{P(CO_{2}) P(CF_{4}) }{P(COF_{2} )^{2} } = 2.2 x 10^{6}[/tex]
Since Kp>>>>1 ⇒ (P(CO₂) x P(CF₄)) > (P(COF₂))²
So, the partial pressures of the products (CO₂ and CF₄) are higher than the partial pressure of the reactant (COF₂).
Therefore, products will be favored at equilibrium at 298 K.
Tech A says that hydrocarbons are a result of complete combustion. Tech B says that a catalytic converter creates a chemical reaction, changing carbon monoxide and hydrocarbons to water and carbon dioxide. Who is correct
Answer:
Neither Tech A nor B is correct
Explanation:
Combustion is a chemical reaction that occurs when a chemical molecule(s) interacts quickly with oxygen and produces heat.
When hydrocarbon undergoes a complete combustion reaction, they produce water and CO2.
Tech B is also incorrect because the main purpose of a catalytic converter is to accelerate and speed up the chemical reaction rates, Hence, they are not involved in chemical reaction formation. Catalytic converters are utilized as a control device in exhaust emission to lessen the effect of toxic gas fumes.
Which of the following statements is correct concerning the class of reactions to be expected for benzene and cyclooctatetraene?
A) Both substances undergo addition reactions.
B) Both substances undergo substitution reactions.
C) Benzene undergoes addition; cyclooctatetraene undergoes substitution.
D) Benzene undergoes substitution; cyclooctatetraene undergoes addition.
Answer:
Both substances undergo substitution reactions.
Explanation:
Let us go back to the idea of aromaticity. Aromatic substances are said to possess (4n + 2) π electrons according to Huckel rule.
Aromatic substances are unusually stable and the aromatic ring can not be destroyed by addition reactions.
Since both benzene and cyclooctatetraene are both aromatic, they do not undergo addition reactions whereby the aromatic ring is destroyed. They both undergo substitution reaction in which the aromatic ring is maintained.
According to the EPA Lead and Copper Rule (LCR), the action level for Pb in drinking water (the level at which threat to human health requires public notification and action towards mitigation) is 15 ppb. If you were to add enough phosphate to the system
saturated with respect to Pb3(PO4)2(s), would the [Pb2+] be below the action limit?
Answer:
The right answer is "105.17 ppb".
Explanation:
According to the question,
The amount of [tex]Pb^{2+}[/tex] in ppb will be:
= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]
= [tex]105.17 \ ppb[/tex]
Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.
A solution of the primary standard potassium hydrogen phthalate (KHP), KHC8H4O4 , was prepared by dissolving 0.4877 g of KHP in about 50 mL of water. Titration of the KHP solution with a KOH solution of unknown concentration required 38.91 mL to reach a phenolphthalein end point. What is the concentration of the KOH solution
Answer:
The concentration of KOH is 0.06137 M
Explanation:
Step 1: Data given
Molar mass of KHP = 204.22 g/mol
Mass of KHP = 0.4877 grams
Volume of water = 50 mL
Volume of KOH solution = 38.91 mL
Step 2: The balanced equation
C8H5KO4 + KOH ⇒ C8H4K2O4+ H2O
Step 3: Calculate number of moles of KHP
Moles = Mass / molar mass
Moles KHP = 0.4877 grams / 204.22 g/mol
Moles KHP = 0.002388 moles
Step 4: Calculate moles of KOH
For 1 mol KHP we need 1 mol KOH to produce 1 mol C8H4K2O4 and 1 mol H2O
For 0.002388 moles KHP we need 0.002388 moles KOH
Step 5: Calculate the concentration of KOH
Concentration = moles / volume
Concentration of KOH = 0.002388 moles / 0.03891 L
Concentration of KOH = 0.06137 M
The concentration of KOH is 0.06137 M
Low-density polyethylene is formed because _______ polymerization is very unpredictable and difficult to control.
dehydration-condensation
anionic-initiated
radical-initiated
esterification
Answer:
radical-initiated
Explanation:
Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.
Which of these is an organic compound?
CaO
H2SO4
C3H8
H2O2
Answer:
which class are you please mention
At 25 oC the solubility of chromium(III) iodate is 2.07 x 10-2 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect). [a]
Answer:
5.0 × 10⁻⁶
Explanation:
Step 1: Write the balanced equation for the solution of chromium(III) iodate
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
Step 2: Calculate the solubility product constant (Ksp)
To relate Ksp and the solubility (S), we will make an ICE chart.
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product constant is:
Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶
When naming organic compounds, there are strict rules regarding punctuation.
1. A comma is used to separate two numbers.
2. A hyphen is used to separate a number from a letter.
Rewrite the name of this compound using hyphens and commas as appropriate.
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
2,3,3-trimethylhexane
Explanation:
IUPAC nomenclature provides a universally acceptable method of naming organic compounds from its structure.
According to this system of nomenclature;
A comma is used to separate two numbers.
A hyphen is used to separate a number from a letter.
Applying these rules, the name of the compound shown in the question should be written as 2,3,3-trimethylhexane.
convert 36.52 mg to ?___ g
Answer:
0.0365 is your answer
Explanation:
hope it will help uAnswer:
365.2gㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ
For a different reaction, the plot of the reciprocal of concentration versus time in seconds was linear with a slope of 0.056 M-1 s -1 . If the initial concentration was 2.2 M, calculate the concentration after 100 seconds. Show your work.
Answer:
[tex]C_t=0.165M[/tex]
Explanation:
From the question we are told that:
Slope [tex]K=0.056 M-1 s -1[/tex]
initial Concentration [tex]C_1=2.2M[/tex]
Time [tex]t=100[/tex]
Generally the equation for Raw law is mathematically given by
[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]
[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]
[tex]C_t=0.165M[/tex]
The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.
What is the specific rate constant?The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.
Given,
Slope (k)= 0.056 per M per s
Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M
Time (t) = 100 seconds
The concentration of the reaction after 100 seconds can be given by,
[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]
Substitute values in the above equation:
[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]
Therefore, after 100 seconds the concentration is 0.165 M.
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Evaluate the validity of the following statement: Spontaneous processes are ones that occur quickly and have a low activation energy. A) False. Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly. B) False. Spontaneous processes can occur slowly, but always have a low activation energy. C) False. Spontaneous reactions can react slowly and can have a high activation energy. D) False. Spontaneous processes always react slowly and always have a high activation energy. E) True.
Answer:
Spontaneous processes are ones that occur quickly and have a low activation energy. - False -
Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly - False
Spontaneous processes can occur slowly, but always have a low activation- false
Spontaneous reactions can react slowly and can have a high activation energy - True
Spontaneous processes always react slowly and always have a high activation energy- False
Explanation:
A spontaneous reaction is reaction that proceeds on its own without us having to do a thing at all!
A spontaneous reaction may be fast or slow depending on the activation energy of the reaction. A spontaneous reaction having a high activation energy will be slow. However, if the spontaneous reaction has a low activation energy then it will be fast.
We have to note here that a spontaneous reaction proceeds without a prolonged input energy. Sometimes energy may be supplied to the reaction at the beginning for instance in the case of the combustion of hydrocarbons.
So, spontaneous processes are not necessarily fast. Some of them may have a very high activation energy such as in the rusting of iron hence they are slow.
A nuclease enzyme breaks the covalent bond originally connecting the phosphate to the 5' carbon in a nucleic acid. After allowing this enzyme to completely digest the nucleic acid down to monomers, you perform tests to determine where the phosphate is attached to each monomer. Where do you expect to find this phosphate
Answer:
The phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.
Explanation:
The structure of nucleic acid polymers is built up from monomers of nucleotides.
A nucleotide consists of a sugar backbone which is either a ribose or deoxyribose sugar, a nitogenous base which is either a purine or pyrimidine, and a phosphate group. The nitrogenous base is attached to the carbon number 1 or C-1 of the sugar backbone by a covalent bond. The phosphate group on the other hand is covalently attached to the carbon number 5 or 5' carbon of the sugar backbone.
When polymers of nucleic acids are formed, the phosphate at the 5' carbon of the sugar backbone is covalently linked in a phosphodiester bond to the 3' carbon of the sugar backbone in another nucleotide molecule, thus extending the strands of the nucleic acid molecule.
Nucleases are enzymes that break down the phosphodiseter bonds in nucleic acids resulting in nucleotide monomers. After complete digestion ofmthe nucleic acid polymer by nucleases, the phosphate will remain attached to the 5' carbon of the deoxy or the ribose sugar in the nucleic acid monomers.
GM 2 all ,What is an atom define it .Good Day
Answer:
An atom is the smallest particle of an element that can take part in chemical reaction.
Explanation:
hope it will help u Amri
WHAT WOULD THE RIGHT OPTION??
how many resonance structures of benzene are known?
A) 3
B) 4
C) 5
D) 6
Answer:
C) 5
Step-by-step explaination:
Benzene has 5 resonance structures.
Many important analgesic compounds are derived from simple aromatic starting materials.
a. True
b. False
Which type of organic compound is shown below?
A. Carboxylic acid
B. Ester
C. Amine
D. Alcohol
Answer:
I think its A maybe am not sure
How many chromosomes do we not understand?
Answer:
we don't understand why humans have only 46 chromosomes
Answer:
46 chromosomes is what we don't understand
Which is the electronic configuration for oxygen?
A metal (C = 0.2158 cal/g· °C) is removed from a hot (350. °F) oven in which it had achieved thermal equilibrium. The metal is placed into 200. mL acetic acid. The temperature of the acid increases to 90.3 °C from 24.3 °C. What is the mass of the metal? (dacetic acid = 1.04 g/cm3; Cs, acetic acid = 2.055 J/g·°C) Group of answer choices 120. g 362 g 1452 g 347 g 281 g
Answer:
362g
Explanation:
heat lost by metal= heat gained by acetic acid
tfs are the same so you cando delta T
convert Cal/gc to J/gc
thectgod ig follow
Classify these bonds as ionic, polar covalent, or nonpolar covalent. You are currently in a sorting module.
Ionic Polar Covalent Nonpolar covalent
C-O , Mg-F , Cl-Cl
Answer:
C-O: polar covalent
Mg-F: ionic
Cl-Cl: nonpolar covalent
Explanation:
Ionic bonds are formed between an atom of a metallic element and another atom of a non-metallic element. Thus, Mg-F is an ionic bond, in which Mg is the metal and F is the nonmetal.
Covalent bonds are formed between two non-metallic elements. So, C-O and Cl-Cl are covalent bonds, because C, O, and Cl are nonmetals.
In C-O, the atom of oxygen (O) has more electronegativity than the atom of carbon (C). Thus, O will attract the electrons with more strength and a difference in charge will be established between the two bonded atoms. So, this covalent bond is polar.
In Cl-Cl, both atoms have the same electronegativity because they are from the same chemical element (Cl). Thus, this bond is nonpolar.
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2). This mixture is stored at 25.0 ∘C. The vapor pressure and the densities for the two pure components at 25.0 ∘C are given in the following table. What is the vapor pressure of the stored mixture?
Answer:
The answer is "170.9 mm Hg".
Explanation:
[tex]\text{Mass of acetone = volume} \times density[/tex]
[tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]
[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]
[tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]
[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]
[tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]
[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]
[tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]
[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]
Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).
The vapor pressure of the stored mixture is: 170.03 mmHg
In the given information, there is some information that is still missing.
The parameters that we are being given include:
The volume of acetone = 70.0 mLThe volume of ethyl acetate = 75.0 mLThe standard temperature for the mixture = 25° CThe first step we need to take is to determine the mass and number of moles of each compound (i.e. for acetone and ethyl acetate)
For us to do that:
We need the density of acetone and ethyl acetate, which is not given:
Assuming that at a standard condition of vapour pressure:
230 mmHg of acetone has a density of 0.791 g/mL95.38 mmHg of ethyl acetate has a density of 0.900 g/mLThen;
Using the relation:
[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]
Mass of acetone = Density of acetone × volume of acetone
Mass of acetone = 0.791 g/mL × 70.0 mL
Mass of acetone = 55.37 g
Mass of ethyl acetate = Density of ethyl acetate × volume of ethyl acetate
Mass of ethyl acetate = 0.900 g/mL × 75.0 mL
Mass of ethyl acetate = 67.5 g
At standard conditions;
For acetone, molar mass = 58.08 g/molFor ethyl acetate, molar mass = 88.11 g/molNow, using the formula for calculating the numbers of moles which can be expressed as:
[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]
For acetone:
[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]
[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]
For ethyl acetate:
[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]
[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]
Now, we will determine the mole fraction of each compound.
The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.
Using the formula:
[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]
For Acetone:
[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]
[tex]\mathbf{mole \ fraction =0.5545 }[/tex]
For ethyl acetate:
[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]
[tex]\mathbf{mole \ fraction =0.4455}[/tex]
Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.
Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.
∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.
where:
Vapour pressure of the solution = (mole fraction × vapor pressure) of solventFor acetone;
Vapor pressure = 0.5545 × 230 mmHg
Vapour pressure = 127.54 mmHg
For ethyl acetate:
Vapour pressure = 0.4455 × 95.38 mmHg
Vapour pressure ==42.49 mmHg
Thus, the vapor pressure of the stored mixture is
= (127.54 + 42.49 ) mmHg
= 170.03 mmHg
Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg
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Wet helium gas is placed into a balloon at 24.4 degrees Celsius and a pressure of 765.3 mm Hg. What volume (in L) does the dry gas occupy if the water vapor pressure is 24.3 torr and the mass of dried helium gas in the balloon is 0.498 g
Answer:
Hence the Volume of Gas = 3.04 L.
Explanation:
pressure of dry gas = 765.3 - 24.3 = 741 mmhg
Temperature of gas = 24.4+273.15 = 297.55 k
No of mol of gas = 0.498/4 = 0.1245 mol
R = gas constant = 0.0821 l.atm.k-1.mol-1
From ideal-gas equation
PV = nRT
(741/760) x v = 0.1245 x 0.0821 x 297.55
V = Volume of Gas = 3.04 L
A system fitted with a piston expands when it absorbs 53.1 ) of heat from the surroundings. The piston is working against a pressure of 0.677 atm. The final volume is 63.2 L. What was the initial volume of the system if the internal energy of the system decreased by 108.3 J?
a. 65.6 L
b. 64.0 L
c. 70.8 L
d. 60.8 L
e. 54.4L
Answer:
d. 60.8 L
Explanation:
Step 1: Given data
Heat absorbed (Q): 53.1 JExternal pressure (P): 0.677 atmFinal volume (V2): 63.2 LChange in the internal energy (ΔU): -108.3 JStep 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
34. 3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 11.21°C. Calculate ∆H, in kJ/mol, for the dissolution of the unknown solid. (The specific heat of the solution is 4.184 J/g･°C and the density of the solution is 1.20 g/mL).
Answer:
ΔH = 2.68kJ/mol
Explanation:
The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:
q = m*S*ΔT
Where q is heat of reaction in J,
m is the mass of the solution in g,
S is specific heat of the solution = 4.184J/g°C
ΔT is change in temperature = 11.21°C
The mass of the solution is obtained from the volume and the density as follows:
150.0mL * (1.20g/mL) = 180.0g
Replacing:
q = 180.0g*4.184J/g°C*11.21°C
q = 8442J
q = 8.44kJ when 3.15 moles of the solid react.
The ΔH of the reaction is:
8.44kJ/3.15 mol
= 2.68kJ/mol
Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.
Answer:
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
Explanation:
The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.
The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.
The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.
From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.
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You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (density -0.779 g/mL, MW - 84.16 g/mol) and 1.6575 grams of propylbenzene (density = 0.862 g/mL, MW = 120.2 g/mol). What is the volume percent composition of cyclohexane in the mixture?
Answer:
66.67%
Explanation:
From the given information:
mass of cyclohexane = 2.9949 grams
density of cyclohexane = 0.779 g/mL
Recall that:
Density = mass/volume
∴
Volume = mass/density
So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL
= 3.8445 mL
Also,
mass of propylbenzene = 1.6575 grams
density of propylbenzene = 0.862 g/mL
Volume of propylbenzene = 1.6575 g/ 0.862 g/mL
= 1.9229 mL
The volume % composition of cyclohexane from the mixture is:
[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]
[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]
[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]
= 66.67%
What is the volume of 6 moles of hydrogen gas, H2 (g)
Answer:
Molar volume, or volume of one mole of gas , depends on pressure and temperature, and is 22.4 liters - at 0 °C (273.15 K) and 1 atm (101325 Pa), or STP (Standard Temperature and Pressure), for every gas which behaves similarly to an ideal gas. The ideal gas molar volume increases to 24.0 liters as the temperature increases to 20 °C (at 1 atm).
11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
