Answers
Answer:
a) But-1-ene
b) E-But-2-ene
c) Z-But-2-ene
d) 2-Methylpropene
Explanation:
In this case, if we want to draw the isomers, we have to check the formula [tex]C_4_H8[/tex] in this formula we can start with a linear structure with 4 carbons. We also know that we have a double bond, so we can put this double bond between carbons 1 and 2 and we will obtain But-1-ene.
For the next isomer, we can move the double bond to carbons 2 and 3. When we do this can have two structures. When the methyl groups are placed on the same side we will obtain Z-But-2-ene. When the methyls groups are placed on opposite sides we will obtain E-But-2-ene.
Finally, we can use a linear structure of three carbons with a methyl group in the middle with a double bond, and we will obtain 2-Methylpropene.
See figure 1 to further explanations.
I hope it helps!
Related Questions
Dynamic equilibrium between liquid and gas is established when __________ . a. the vapor pressure of the liquid equals the external pressureb. the liquid and vapor are at the same temperaturec. a liquid vaporizes and condenses at the same rated. the liquid is entirely evaporated
Answers
Answer:
The correct answer is option c. "a liquid vaporizes and condenses at the same rate"
Explanation:
In chemistry, dynamic equilibrium is defined as a state where the reaction rate of a forward reaction is equal to the reaction rate of the backward reaction. In this case, the reactions that participate in the system are the vaporization of a liquid and the condensation of a gas. Therefore, dynamic equilibrium is established when a liquid vaporizes and condenses at the same rate.
Thinking about simple organic molecules made with only carbon and hydrogen and
no other elements (hydrocarbons), what are four different ways to make new
molecules? (Hint: Two versions of the same "way" only count as one thing!)
Answers
Answer:
Explanation:
Simple organic molecules having only carbon and hydrogen are alkanes, alkenes and alkynes.
Ways of making these molecules are referred to as synthesis
Synthesis of
(1) alkanes can be achieved by halogenation of an alkene or alkyne as shown in the reaction below
C₂H₄ + H₂ ⇒ C₂H₆
The reaction above shows an alkene (ethene) undergoing halogenation to form an alkane (ethane). This reaction occurs in the presence of a platinum or nickel catalyst
C₂H₂ + 2H₂ ⇒ C₂H₆
The reaction above shows an alkyne (ethyne) undergoing halogenation to form an alkane (ethane). This reaction also occurs in the presence of a nickel or platinum catalyst
(2) Alkanes can be achieved by Wurtz reaction; treating alkyl halides with sodium metal as shown below
CH₃CH₂Br + 2Na + BrCH₂CH₃ ⇒ CH₃CH₂CH₂CH₃ + 2NaBr
This reaction proceed in the presence of a dry ether.
(3) alkanes from Kolbe's electrolytic method; this involves the synthesis of alkanes through the electrolysis of sodium or potassium salt of a carboxylic acid
2CH₃COO⁻Na⁺ + 2H₂O ⇒ CH₃CH₃ + 2CO₂ + H₂ + 2NaOH
(4) alkenes from dehydration of alcohol, as shown in the reaction below
C₂H₅OH ⇒ C₂H₄ (-H₂O)
The reaction above proceeds by heating the alcohol (ethanol) at high temperature in the presence of strong acid such as sulphuric acid.
Calculate the wavelength of light produced if an electron moves from n=6 state to n=5 state of an electron in a hydrogen atom.
Answers
The questions asks us to calculate the wavelength of light produced if an electron moves from n = 6 to n = 5 state in a hydrogen atom. The answer would be:
λ = [tex]7.46 * 10^{-6}[/tex] m
m = [tex]7.46 * 10^{-4}[/tex] cm
Explanation:
Please see my graphic below:
The wavelength of light produced if an electron moves from n=6 state to n=5 state of an electron in a hydrogen atom is 7.59 × [tex]10^-6[/tex] m
Using the Rydberg formula;
1/λ = 1.09737 × [tex]10^-6[/tex] (1/[tex]nf^2[/tex] - 1/[tex]ni^2[/tex])
Where;
λ = wavelength
nf = final state of the electron
ni = initial state of the electron
Substituting values;
1/λ = 1.09737 × [tex]10^7[/tex] ([tex]1/5^2[/tex] - [tex]1/6^2[/tex])
1/λ = 1.09737 × [tex]10^7[/tex] (1/25 - 1/36)
1/λ = 1.09737 × [tex]10^7[/tex] (0.04 - 0.028)
1/λ = 1.09737 × [tex]10^7[/tex] (0.012)
λ = 7.59 × [tex]10^-6[/tex] m
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Classify each of the following as a pure substance or a mixture. If a mixture, indicate whether it is homogeneous or heterogeneous:_______. (a) rice pudding, (b) seawater, (c) magnesium, (d) crushed ice.
Answers
Answer:
a) Heterogeneous mixture (b) Homogenous mixture (c) Pure substance (d) Pure substance
Explanation:
Homogenous mixtures contains mixture of substances with similar proportions while Heterogenous mixture contains substances with a varying proportion.
Draw the structure of 2,4,4,5-tetramethyl-2-hexene.
Answers
Answer:
The answer to your question is given below
Explanation:
2,4,4,5-tetramethyl-2-hexene.
To draw the structure of the above compound, we must bear the following in mind:
1. The compound is an alkene i.e it contains carbon to carbon double bond (C=C)
2. The parent name of the compound is hexene i.e it contain 6 carbon atoms.
3. The double is located at carbon 2.
4. The substituent group attached to the compound is methly.
5. There are four methly group attached to the compound of which one is located at carbon 2, two at carbon 4 and one at 5.
With the above information, we can easily draw the structure of the compound.
Please see attached photo for the structure of the compound.
Write the electron configuration for the following elements. (1) Neon, (2) Zinc, (3) Tungsten, (4) Magnesium
Answers
Explanation:
Hey there!!
We generally use s,p,d,f formula.
So, let's do with it.
1.Neon
Atomic no. = 10
So, Its electronic configuration is;
[tex] = 1 {s}^{2} .2 {s}^{2} 2 {p}^{6} [/tex]
2.Zinc.
Atomic no. = 30
So, it's electronic configuration is;
[tex] = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{10} [/tex]
3. Tungsten.
Atomic no. = 74.
So, it's electronic configuration is;
[tex] = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{10}4 {p}^{6} 5 {s}^{2} 4 {d}^{10} 5 {p}^{6} 6 {s}^{2} 4 {f}^{14} 5 {d}^{4} [/tex]
4. Magnesium.
Atomic no. = 12
So, the electronic configuration is ;
[tex] = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6}3 {s}^{2} [/tex]
Hope it helps...
An iceberg has a volume of 783 m3. If the density of ice = 0.917 g/cm3, how much mass does the iceberg have in kg?
Answers
Answer:
7.18 × 10⁵ kg
Explanation:
Step 1: Given data
Volume of the iceberg (V): 783 m³
Density of ice (ρ): 0.917 g/cm³
Step 2: Convert "V" to cm³
We will use the relationship 1 m³ = 10⁶ cm³.
783 m³ × (10⁶ cm³/1 m³) = 7.83 × 10⁸ cm³
Step 3: Calculate the mass (m) of the iceberg
We will use the following expression.
ρ = m/V
m = ρ × V
m = 0.917 g/cm³ × 7.83 × 10⁸ cm³
m = 7.18 × 10⁸ g
Step 4: Convert "m" to kg
We will use the relationship 1 kg = 10³ g.
7.18 × 10⁸ g × (1 kg/10³ g) = 7.18 × 10⁵ kg
Benzoic acid is in the process of being recrystallized. Your lab partner is in a hurry and after the impure benzoic acid is dissolved, they quickly submerge the flask in an ice bath and see crystals. Besides possibly breaking the flask, what would be the problem with this technique
Answers
Answer:
Impurities will be trapped in the crystals of the benzoic acid crystallized in this manner.
Explanation:
After benzoic acid is dissolved in hot water, it should have been allowed to cool gradually before it is transferred into an ice bath.
This gradual cooling will aid the separation of impurities so that when the vessel is now submerged in an ice bath, only pure benzoic acid is recrystalized.
If the vessel is immediately submerged into an ice bath, impurities will be trapped in the crystals of the benzoic acid.
Determine whether each of the descriptions of matter describes:
1. a heterogeneous mixture,
2. a homogeneous mixture,
3. a pure substance.
4. no correct response
a. two substances present; two phases present
b. two substances present; one phase present
c. one substance present; one phase present
d. one substance present; two phases present
Answers
Answer:
a heterogeneous mixture,-two substances present; two phases present
a homogeneous mixture- two substances present; one phase present
a pure substance- one substance present- one phase present
no correct response-one substance present; two phases present
Explanation:
When two substances are mixed, the mixture may be homogenous or heterogeneous. If a mixture is homogenous, then the two substances form a single phase, e.g ethanol and water.
On the other hand, a heterogeneous mixture is divided into two distinct phases, e.g kerosene and water mixture.
A single substance can only give one phase, no substance can simultaneously give two phases of the same substance in the same physical state.
Similarly, a pure substance always consists of a single phase.
How do the chemical characteristics of carbon affect the characteristics of organic molecules?
Answers
Answer:
The chemical characteristics of carbon affect the characteristics of organic molecules due to its tetravalent nature. It has four valence electrons in which it shares with other elements in order to form an octet configuration.
Carbon atoms are also capable of forming double and triple bonds with other atoms. These properties help determine the functional group present and gives us a knowledge of the chemical features such as polarity, melting and boiling present in the compound.
What is something a scientist could add to or subtract from an unknown substance that will always reveal information about the physical and/or chemical properties of the substance?(1 point) A. heat B. light C. iodine D.vinegar
PLEASSSSSSSSE HELP IMMEDIAATLYYYYYY.......
Answers
Yeah! I have the same question!!! I think it is heat.
6) What is located in the nucleus?
Protons
Protons and neutrons
Protons and electrons
Neutrons
Answers
Answer:
Protons and Neutrons are found in the nucleus
In an NMR experiment, shielding refers to the reduced impact of the ______on a nucleus due to the presence of ______around the nucleus. Shielding can be_____by the presence of more electronegative groups around the nucleus. A shielded nucleus appears further ______ whereas a deshielded nucleus appears further _____in an NMR spectrum.
Answers
Answer:
In an NMR experiment, shielding refers to the reduced impact of the APPLIED MAGNETIC FIELD on a nucleus due to the presence of ELECTRON DENSITYaround the nucleus.
Shielding can be DECREASED by the presence of more electronegative groups around the nucleus.
A shielded nucleus appears further UPFIELD whereas a deshielded nucleus appears further downfield in an NMR spectrum.
Explanation:
1 There is reducece because alot of the electronegative atom binds more and thus reduce the electron density around the nucleus.
While a reduction in electron density bring about more nuclear magnetic field thus it resonates at higher frequency resulting downfield peak.
When There is reduce because a lot of the electronegative atom binds more and also that thus reduces the electron density around the nucleus.While a reduction in electron density brings about more nuclear magnetic field thus it resonates at higher frequency resulting downfield peak.
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4. Which statement gives you enough information to say that the atom is electrically neutral?
O A. The atom has 15 neutrons and 15 electrons.
OB. The atom has 4 neutrons and 4 protons.
O C. The atom has 7 protons and 7 electrons.
O D. The atom has 19 electrons and 19 neutrons.
Answers
Answer:
O C. The atom has 7 protons and 7 electrons.
If the proton amd electron of atoms are equal it is said to be electrically nuetral
437 mm converted to m
Answers
Answer:
0.437
Explanation:
Divide the length value by 1000 to get the answer.
Write the symbol for every chemical element that has atomic number less than 5 and atomic mass greater than 8.3 u.
Answers
Atomic m. Element Atomic no. Symbol
1.0079u Hydrogen 1 H
4.0026u Helium 2 He
6.941u Lithium 3 Li
Additional information:-★ Atomic number : The number of protons in one atom of an element is known as atomic number.
Atomic number of an element = No. of protons in one atom of element.The atomic no. of element is denoted by the letter z .The symbol for every chemical element that has an atomic number less than 5 and atomic mass greater than 8.3 u are hydrogen H, helium He, lithium Li and atomic masses 1,4 and 6.94 respectively in the periodic table.
What is a periodic table?A periodic table is an arrangement of the elements in which they are classified on the basis of the number of electrons, neutrons, and protons present in their nucleus and shell of them.
Hydrogen has one electron in it and atomic mass is also one, helium comes in the second number with 4 amu and lithium comes in the third number containing 6.94 mass with it.
Therefore, hydrogen H, helium He, lithium Li and atomic masses 1,4 and 6.94 respectively in the periodic table contain symbols for every chemical element that has an atomic number less than 5 and atomic mass greater than 8.3 u are hydrogen H.
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Calculate the mole fraction of nitric acid of a(n) 15.7% (by mass) aqueous solution of nitric acid. Calculate the mole fraction of nitric acid of a(n) 15.7% (by mass) aqueous solution of nitric acid. 2.56×10−2 0.102 5.33×10−2 5.11×10−2 The density of the solution is needed to solve the problem.
Answers
Answer:
0.0505
Explanation:
Step 1: Calculate the masses of solute and solvent
We have a 15.7% by mass nitric acid solution, that is, there are 15.7 g of nitric acid (solute) per 100 g of solution. The mass of water (solvent) is:
m(solution) = m(solute) + m(solvent)
m(solvent) = m(solution) - m(solute)
m(solvent) = 100 g - 15.7 g = 84.3 g
Step 2: Calculate the moles of nitric acid
The molar mass of nitric acid is 63.01 g/mol.
15.7 g × (1 mol/63.01 g) = 0.249 mol
Step 3: Calculate the moles of water
The molar mass of water is 18.02 g/mol.
84.3 g × (1 mol/18.02 g) = 4.68 mol
Step 4: Calculate the mole fraction of nitric acid
[tex]X(HNO_3) = \frac{nHNO_3}{nHNO_3+nH_2O} = \frac{0.249mol}{0.249mol+4.68mol} = 0.0505[/tex]
If u burn paper in the jar, is it the same mass as before u burn it or less or more.
Answers
Answer: the mass will be same
Answer:
the mass wouldnt be the same because of the ashes in the jar
Explanation:
Which of the following describes freezing? A)A substance changing from the liquid state to the solid state. B) A substance changing from the solid state to the liquid state. C) A substance changing from the liquid to the gaseous state. D) A substance changing from the gaseous state to the liquid state.
Answers
Answer:
A
Explanation:
A because water turns into ice
1. For one atom of lithium -6 I will need_____ atom of lithium -7 for an average atomic mass of 6.944? 2. For one atom of lithium -6 needed _____ Adams of lithium -7 for an average atomic mass of 6.944.
Answers
Answer:
For one atom of lithium -6 I will need_16.85____ atom of lithium -7 for an average atomic mass of 6.944 .
Explanation:
1 ) Let required atom be m .
[tex]\frac{6\times 1 + 7\times m}{1 + m} = 6.944[/tex]
6 + 7m = 6.944 + 6.944m
7 m - 6.944 m = .944
.056 m = .944
m = 16.85
Here, we are required to find how many atoms of lithium -7 is to combine with 1 atom of lithium -6 for an average atomic mass of 6.944.
For one atom of lithium -6 I will need 16.86 atoms of lithium -7 for an average atomic mass of 6.944.
Let the total number of lithium -6 and lithium -7 required be; x.
Therefore, the number of lithium -7 required is;
(x-1)
Therefore;
{(1/x) × 6 + ((x-1)/x)× 7} = 6.944
Therefore, (6 + 7x - 7)/x = 6.944
And, 7x -1 = 6.944x
0.056x = 1
And, x = 1÷0.056 = 17.86 atoms.
Therefore, the number of lithium -7 atoms needed is;
x - 1 = 17.86 - 1 = 16.85 atoms of lithium -7.
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Give the names and charges of the cation and anion in each of the following compounds:________.
(a) CuS, (b) Ag2SO4, (c) Al(ClO3)3, (d) Co(OH)2 and (e) PbCO3
Answers
Answer:
(a) [tex]Cu^{2+}S^{2-}[/tex]: copper (II) sulfide or cupric sulfide.
(b) [tex]Ag^+_2(SO_4)^{2-}[/tex]: silver sulfate.
(c) [tex]Al^{3+}(ClO_3)_3^-[/tex]: aluminum chlorate.
(d) [tex]Co^{2+}(OH)^-_2[/tex]: cobalt (II) hydroxide.
(e) [tex]Pb^{2+}(CO_3)^{2-}[/tex]: lead (II) carbonate.
Explanation:
Hello,
In this case, we proceed by exchanging the subscripts between the cation and anion for each case as follows:
(a) CuS: cation is cupper (II) and anion is sulfide, thus the name is copper (II) sulfide or cupric sulfide.
[tex]Cu^{2+}S^{2-}[/tex]
(b) Ag₂SO₄: cation is silver ion and anion is sulfate, thus the name is silver sulfate.
[tex]Ag^+_2(SO_4)^{2-}[/tex]
(c) Al(ClO₃)₃: cation is aluminum ion and anion is chlorate, thus the name is aluminum chlorate.
[tex]Al^{3+}(ClO_3)_3^-[/tex]
(d) Co(OH)₂: cation is cobalt (II) and anion is hydroxide, thus the name is cobalt (II) hydroxide.
[tex]Co^{2+}(OH)^-_2[/tex]
(e) PbCO₃: cation is lead (II) and anion is carbonate, thus the name is lead (II) carbonate.
[tex]Pb^{2+}(CO_3)^{2-}[/tex]
Regards.
What is the relationship between the bases displayed when the arrow is pointed to the left versus when it is pointed to the right?
Answers
Answer:
Options
Explanation:
Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.
Answers
Answer:
-) 3-bromoprop-1-ene
-) 2-bromoprop-1-ene
-) 1-bromoprop-1-ene
-) bromocyclopropane
Explanation:
In this question, we can start with the I.D.H (hydrogen deficiency index):
[tex]I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}[/tex]
In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:
[tex]I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1[/tex]
We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).
See figure 1
I hope it helps!
Which molecule does not exhibit hydrogen bonding?a. HFb. CH3NH2c. CH2F2d. HOCH2CH2OH
Answers
Answer:
(c) CH₂F₂
Explanation:
Hydrogen bonds are weak intermolecular forces. They are the strongest kind of intermolecular forces, although they are weaker than the covalent bonds.
Hydrogen bonds arise from molecules which contain a hydrogen atom which is bonded to one of the most electronegative elements such as N, O or F.
(a) HF, → has H-F bond
(b) CH₃NH₂, → has N-H bond
(c) CH₂F₂, → has no H-F bond ( F- C- F)
(d) HOCH₂CH₂OH, → has O-H bond
Therefore, only CH₂F₂ does not exhibit hydrogen bonding.
Gold (Au) has a density of 19.3 g/ml. A gold filling falls out of your teeth and you bring it to chemistry class. When you drop it into a graduated cylinder the water rises from 9.27 ml to 12.03 ml.
What is the mass of your gold filling?
Answers
Answer:
53.27 g.
Explanation:
From the question given above, the following data were obtained:
Density of gold = 19.3 g/mL
Volume of water = 9.27 mL
Volume of water + gold = 12.03 mL
Mass of gold =.?
Next, we shall determine the volume of the gold filling. This can be obtained as follow:
Volume of water = 9.27 mL
Volume of water + gold = 12.03 mL
Volume of gold =?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 12.03 – 9.27
Volume of gold = 2.76 mL
Finally, we shall determine the mass of the gold filling as follow:
Density of gold = 19.3 g/mL
Volume of gold = 2.76 mL
Mass of gold =.?
Density = mass/volume
19.3 = mass of gold /2.76
Cross multiply
Mass of gold = 19.3 × 2.76
Mass of gold = 53.27 g
The, the mass of the gold filling is 53.27 g.
For which reaction, carried out at standard conditions, wouldboth the enthalpy and entropy changes drive the reaction in thesame direction? Please EXPLAIN.
A. 2H2(g) + O2(g) --->2H2O(l) ΔH = -571.1 kJ
B. 2Na(s) + Cl2(g)---->2NaCl(s) ΔH = -822.0kJ
C. N2(g) + 2O2(g) ---->2NO2(g) ΔH= +67.7 kJ
D. 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ
Answers
Answer:
2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ
Explanation:
Entropy increases with increase in the number of particles from left to right in a reaction. Hence, the reaction; 2NH3(g) ----> N2(g) +3H2(g) ΔH= +92.4 kJ is favoured by increase in entropy.
Similarly, the enthalpy change for the reaction is only +92.4 KJ. Hence for this reaction, both enthalpy and entropy changes drive the reaction in the same direction.
How is Oxygen -18 different from oxygen -16?
Answers
Answer:
because they are both 2 deferent type of oxygen levels
HELP ME PLEASE, what kind of cell is?
Answers
Answer:
Explanation:
Plant cell because its shape is rectangular
Which solution below would have the greatest buffering capacity? The solution was prepared from a weak acid and the salt of its conjugate base. 0.160 M C6H5OHCOOH and 0.160 M C6H5OHCOONa, Ka = 1.05e-3 0.0892 M H2NCH2COOH and 0.0892 M H2NCH2COOK, Ka = 4.50e-3 0.0725 M H2NCH2COOH and 0.0725 M H2NCH2COONa, Ka = 4.50e-3 0.1360 M C6H5OHCOOH and 0.1360 M C6H5OHCOOK, Ka = 1.05e-3
Answers
Answer:
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
Explanation:
The first mixture
0.160 M C₆H₅OHCOOH and 0.160 M C₆H₅OHCOONa,
Ka = 1.05x 10⁻³ .
The second mixture
0.0892 M H₂NCH₂COOH and 0.0892 M H₂NCH₂COOK,
Ka = 4.5 x 10⁻³ .
The third mixture
.0725 M H₂NCH₂COOH and 0.0725 M H₂NCH₂COONa,
Ka = 4.5 x 10⁻³
fourth mixture
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
In all the mixtures the ratio of acid and its salt are same and equal to one so this ratio will not determine their relative buffering capacity .
Now we know that weak acid has more buffering capacity so mixture having acid of less Ka will have more buffering capacity .
Ka is less if
Ka = 1.05 x 10⁻³ .
Dilute acids have greater buffering capacity
So ultimate answer is
0.1360 M C₆H₅OHCOOH and 0.1360 M C₆H₅OHCOOK,
Ka = 1.05x 10⁻³ .
If u burn paper in the jar, is it the same mass as before u burn it or less or more. And why.
Answers
Answer:
the mass would not be the same because of the ashes in the jar
Explanation:
Answer: The total mass before and after burning is the same
Explanation: When the carbon and hydrogen of the hydrocarbon-containing substance (the paper) chemically combine with the oxygen, the remaining materials may appear as ash, the solid remains of a fire.
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