Which of the following alkenes cannot be converted into an alkyne by reacting it with bromine followed by an excess of sodium amide and then with water? X X II III DIVE, III and IV II &IX

Answers

Answer 1

Among the following alkenes (D) IV cannot be converted into an alkyne by reacting it with bromine followed by an excess of sodium amide and then with water.

An alkene must first be brominated by an addition reaction that produces a vicinal dibromide in order to transform it into an alkyne. All alkenes that have at least one double bond double substituted render them unsuitable for this method due to the requirement of this vicinal dibromide. This is so that the alkane can undergo a double elimination and produce an alkene, which needs at least one C-H and C-X bond. This hinders the conversion of molecule IV into an alkyne.

The dibromo alkane would change into the provided alkene by losing one unit of HBr if it were to react with bromine. However, the second beta-elimination cannot take place since there isn't hydrogen that is physically linked to the alkene carbon. So (D) IV is the right option.

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Which Of The Following Alkenes Cannot Be Converted Into An Alkyne By Reacting It With Bromine Followed

Related Questions




A vitamin supplement was found to weigh 900 mg. It contained 50. 51% fluorine, and the remaining amount was iron. What mass of each element could be recovered from this vitamin?

Answers

A vitamin supplement is a dietary supplement containing one or more essential vitamins, typically in the form of a pill, capsule, or tablet.

What are Vitamins?

Vitamins are a group of organic compounds that are essential for normal cellular functioning. They are required for a wide range of bodily functions, such as metabolism, growth, development, and immunity. They can be found naturally in food sources or taken as dietary supplements. There are 13 essential vitamins, including vitamins A, C, D, E, K, and the B vitamins.

The mass of fluorine that can be recovered from the vitamin is 455.9 mg (50.51% of 900 mg). The mass of iron that can be recovered from the vitamin is 444.1 mg (the remaining amount of 900 mg).

To calculate this, we need to use the percent composition formula:

Mass of Element = (Percent Composition/100) x Total Mass

Mass of Fluorine = (50.51/100) x 900 mg = 455.9 mg

Mass of Iron = (100 - 50.51)/100 x 900 mg = 444.1 mg.

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How many moles are in 68 grams of potassium sulfide (K2S)?

Answers

Answer:No of moles of potassium sulfide (K2S)=0.61671moles

Explanation:

No of moles is given as   Mass/ molar mass

Here Mass of potassium sulfide (K2S) =68 grams

Molar mass of potassium sulfide (K2S)  = 39.0983 x 2 + 32.065 =110.2616 g/mol

No of moles =68 grams /110.2616 g/mol

=0.61671moles

Magnesium (Mg) has nine electrons. Which of the following shows the correct electron configuration for an atom of Mg?

2, 2, 5
7, 2
1, 8
2, 7

brailyist please help me

Answers

Answer:

2,7

Explanation:

Two in the first shell and 7 in the outer and last shell.

HOPE THIS HELPED

How many molecules are in 5.4 moles of HCl?

Answers

Answer:

3.25 × 10²⁴ molecules

Explanation:

The number of molecules can be found by using the formula

N = n × L

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

We have

N = 5.4 × 6.02 × 10²³

We have the final answer as

3.25 × 10²⁴ molecules

Hope this helps you

13 moles of water to molecules

Answers

I’m confused what are you asking exactly?

What are 6 substances you might find in the air?

Answers

Standard Dry Air is made up of nitrogen, oxygen, argon, carbon dioxide, neon, helium, krypton, hydrogen, and xenon.

Hope this helps!

Has 13 protons and 14 neutrons what is its mass number?

Answers

An atom that has 13 protons and 14 neutrons will have mass number 27.

Rutherford demonstrated that an atom's nucleus, which is made up of protons and neutrons, contains the vast bulk of the atom's mass. The total number of protons and neutrons in an atom is referred to as the mass number. The number of protons(atomic number) and neutrons and their sum can be used to compute it.

We know that,

Mass number = Number of protons + Number of neutrons

= 13 + 14 = 27.

The mass number  is 27.

Therefore  an atom that have 13 protons and 14 neutrons will have mass number 27.

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Which of the following ion has largest size?

A. F-
B. Cs+
C. Al+3
D. O-2

Answers

Answer:

The ion with the largest size is Cs+.

Explanation:

The size of an ion is determined by the number of electrons it contains and the arrangement of those electrons in its electron cloud. Ions with a larger number of electrons will generally have a larger size because their electron clouds will be more diffuse and occupy more space. Ions with a smaller number of electrons will have a smaller size because their electron clouds will be more compact.

In general, ions in the same group of the periodic table will have a similar number of valence electrons and will be similar in size. For example, the F- ion and the O-2 ion both belong to Group 17 and have 7 valence electrons, so they are expected to be similar in size. Similarly, the Cs+ and Al+3 ions both belong to Group 1 and have a single valence electron, so they are also expected to be similar in size.

However, the Cs+ ion is much larger than the Al+3 ion because it has a larger number of total electrons. Cs+ has 55 electrons, while Al+3 has only 13 electrons. The larger number of electrons in Cs+ results in a more diffuse electron cloud and a larger overall size. Therefore, among the ions listed, Cs+ has the largest size

Iron has a BCC crystal structure, an atomic radius of 0. 124 nm, and an atomic weight of 55. 85 g/mol. Compute its theoretical density

Answers

Iron has a BCC crystal structure, an atomic radius of 0. 124 nm, and an atomic weight of 55. 85 g/mol. its theoretical density is 7.9 g /cm³.

The molar mass of the iron = 55.85 g/mol

The atomic radius = 0.124 nm

the edge length is given as :

a = (4r) / √3

a = ( 4 × 0.124 × 10⁻⁷ cm ) / √3

a = 2.86 × 10⁻⁸ cm

The density is given below :

density = ( Z × M ) / Na × a³

density = ( 2 × 55.85 ) / (6.023 ×10²³) × ( 2.86 × 10⁻⁸ )

density = 7.9 g/cm³

Thus the theoretical density is 7.9 g/cm³.

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What volume of 0.450 M Na3PO4 is required to precipitate all the lead(II) ions from 190.0 mL of 0.600 M Pb(NO3)2

Answers

We need 0.127 L of 0.450 M Na3PO4 to precipitate all the lead(II) ions from 190.0 mL of 0.600 M Pb(NO3)2.

How to determine the amount of Na3PO4 needed?

This is a precipitation reaction. To determine the amount of Na3PO4 needed to precipitate all the lead(II) ions, we need to find the number of moles of Pb(NO3)2 present in the solution.

First, we'll use the formula to convert the volume of the solution to liters:

190.0 mL * (1 L/1000 mL) = 0.19 L

Next, we'll use the molarity formula to find the number of moles of Pb(NO3)2 in the solution:

molarity (M) = moles of solute / liters of solution

moles of Pb(NO3)2 = 0.600 M * 0.19 L = 0.114 moles

Now we can use the balanced equation for the reaction to find how many moles of Na3PO4 are needed to react with all 0.114 moles of Pb(NO3)2

Pb(NO3)2 + 3Na3PO4 -> Pb3(PO4)2 + 6NaNO3

We can see that 2 moles of Pb(NO3)2 react with 3 moles of Na3PO4

So we need 0.114 moles/ 2 moles = 0.057 moles of Na3PO4

Finally we can use the molarity formula again to find the volume of 0.450 M Na3PO4 required

Molarity (M) = moles of solute / liters of solution

0.057 moles = 0.450 M * V

V = 0.057 moles / 0.450 M = 0.127 L

So we need 0.127 L of 0.450 M Na3PO4 to precipitate all the lead(II) ions from 190.0 mL of 0.600 M Pb(NO3)2

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A compound has the following percentage composition: 40.0% carbon, 6.72% hydrogen, 53.28% oxygen. How many moles of hydrogen are present in 100 g of the compound described above

Answers

The compound is Glucose.

C: 40 g divided by 12 g per mol equals 3.333 mol

H: 6.72 g/mol (1 g) = 6.72 mol

O: 53.28 g/mol (16 g) = 3.33 mol

Subtract the lesser amount of moles.

C: 3.333 / 3.33 = 1.00

H: 6.72 / 3.33 = 2.02 ≈ 2

O: 3.33/3.33 = 1.00

Statistical method

The empirical formula's molar mass is 12 g/mol + 2*1 g/mol + 16 g/mol = 30 g/,ol.

180 g/mol / 30 g/mol = 6 times the mass of the empirical formula is contained in the molar mass.

The empirical formula of compound is six times the molecular formula.

=>[tex]C_{6} H_{12} O_{6}[/tex]

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In the molecules below, areas that have a partial negative charge are pink and areas that have a partial positive charge are blue.


The attractive force between these two molecules has most likely been produced by
covalent bonds.
dipole-dipole interactions.
dipole-induced dipole interactions.
London dispersion forces.

Answers

The attractive force (intermolecular force) between these two molecules has most likely been produced by dipole-dipole interactions.

The correct option is B.

What are intermolecular forces?

Intermolecular forces are forces of attraction or repulsion that act between atoms and other kinds of nearby particles, such as atoms or ions, to mediate interactions between molecules.

Some intermolecular forces are:

covalent bonds.dipole-dipole interactions.dipole-induced dipole interactions.London dispersion forces.

Considering the given molecules in the diagram:

Each molecule is made up of two distinct components, and as a result, each molecule has a constant bond dipole.

The attractive forces are dipole-dipole attractions because the dipoles do not cancel. The dipole-dipole attractions are substantially stronger than the dipole-induced dipole and London dispersion forces.

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Define corrosion and explain the different basis of tendency or resistance of different
metals to corrosion

Answers

Corrosion is the process of chemical or electrochemical attack of materials by their external environment.

The tendency/resistance of different metals to corrosion largely depends on the composition of the metal. Iron and steel corrode in the presence of oxygen due to oxidation, while some metals such as chromium, nickel and molybdenum provide a protective layer of oxide film on the surface of the metal which is then resistant to corrosion. Other metals such as aluminum and zinc form sacrificial oxide layers which corrode and protect the metal from becoming completely damaged.

What process do orcas use oxygen for?

Answers

The process orcas uses the oxygen is from the blowholes. the blow holes are situated on the top of of the head and it uses oxygen from their.

The whales and the humans uses the lungs for the respiration or to get oxygen required for the breathing. the orcas uses the oxygen from the blow holes situated at the top of the head. The orcas lives under the water and when the orcas dive in the water they have ability to hold the breath for at the time . but this is not good for the health . it is be very stressful for the body.

Thus, the process do the orcas uses oxygen for the intake is the from the blow holes.

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These cells produce hyrdochloric acid

Answers

The best-known component of gastric juice is hydrochloric acid, the secretory product of the parietal, or oxyntic cell. It is known that the capacity of the stomach to secrete HCl is almost linearly related to parietal cell numbers.

Write a skeleton equation using the step, that shows sulfur burns in oxygen gas to
form sulfur dioxide.

Answers

Answer:

Skeleton equation: H2S(g) + O2(g) → SO2(g) + H2O(l)

Explanation:

what are the two quantities in this module for which we will develop unit factors to do dimensional analysis with chemical substances?

Answers

The two quantities in this module for which we will develop unit factors to do dimensional analysis with chemical substances are mass (usually measured in grams) and volume (usually measured in liters).

Developing Unit Factors for Mass and Volume in Chemical Substances

Unit factors are developed to convert between two different units of measurement, in this case, mass and volume. We can do this by multiplying the given value by the appropriate unit factor to convert the given value from one unit to the other. For example, if we are given a mass of 10 grams, we can convert this to liters using the unit factor 1 gram = 1 liter. Thus, 10 grams = 10 liters.

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Cause and effect: What do you think would happen if the snow and ice didn't reflect
most of the sun?

Answers

Answer:

The earth would absorb more sun and temperatures would be a lot higher.

Explanation:

Three elements in the same period are listed in order of decreasing atomic radius. Which of the following is an appropriate explanation for the non-metal in the list having the smallest atomic radius

Answers

The appropriate explanation for the non-metal is that the higher effective nuclear charge less will be the atomic radius.

Atomic radius typically decreases during a period from left to right. There are a few little outliers, such how the oxygen radius is a tiny bit bigger than the nitrogen radius. Protons are gradually added to the nucleus at the same time that electrons are gradually added to the main energy level. The enhanced positive charge of the nucleus gradually attracts these electrons closer to it. The size of the atoms shrinks as the strength of attraction between nuclei and electrons grows. Due to electron-electron repulsions that would otherwise result in the atom's size expanding, the effect becomes less pronounced as one proceeds further to the right in a period.

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10. What is the difference between a pure and non-pure substance?

Answers

Answer:a pure substance consists only of one element or one compound. a mixture consists of two or more different substances, not chemically joined together

Explanation:

What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?
Standard Temperature and Pressure (STP) = 273 K and 1.0 atm
Convert grams to moles by dividing by molar mass of O2
a
0.449 liters
6
0.432 liters
C
0.418 liters
d.
0.406 liters

Answers

Answer: The volume of 0.640 grams of [tex]O_{2}[/tex] gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of [tex]O_{2}[/tex] gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of [tex]O_{2}[/tex] (molar mass = 32.0 g/mol) is as follows.

[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol[/tex]

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

[tex]PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L[/tex]

Thus, we can conclude that the volume of 0.640 grams of [tex]O_{2}[/tex] gas at Standard Temperature and Pressure (STP) is 0.449 L.

Identify the electron configuration of the metal ion in Fe(NO3)3.
O [Ar]3d6
O [Ar]4s23d6
O [Ar]4s23d3
O [Ar]4s23d6
O [Ar]3d5

Answers

[Ar]4s23d6 the electron configuration of the metal ion in Fe(NO3)3.

What electron arrangement does As3 have?

The noble gas krypton, which is isoelectronic with As arsenic 3 ion (As3), is isoelectronic with [Ar]3d104s2sp6 because it has gained three electrons.

Transition-metal ions are frequently referred to as having dn configurations because their valence electrons are mostly found in d orbitals. For instance, it is claimed that the Co3+ and Fe2+ ions have a d6 configuration.

The number of protons in the atomic nuclei of arsenic, abbreviated As, is 33. 33 electrons would likewise be present in a neutral As atom.

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How many grams of BeF2 are present in 655 ml of a 0.442 m solution of Bef2

Answers

Answer:

13.6 g

Explanation:

The mass of BeF₂ present in 655 ml of a solution that is 0.442 M is equal to 13.82 g.

What is the molarity?

We can calculate the concentration of a solute in a solution in terms of molarity, molality, and normality.

The molarity of a particular solution can be determined from the number of moles of a solute per unit volume of the solution.

The Molarity of the particular solution can be determined from the mathematical formula mentioned below:

Molarity = Moles/Volume of the Solution

Given, the molarity of BeF₂ solution = 0.442 M

The volume of the BeF₂ solution, V = 655 ml = 0.655 L

The molar mass of the BeF₂, M = 47.01 g/mol

Molarity of BeF₂ solution = m/(M × V)

0.442 = m/ (47.01 × 0.655)

m = 13.82 g

Therefore, 13.82 g of BeF₂ is required for the given solution.

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pls can someone mention 5 example of fine chemicals​

Answers

drugs, fragrances, additives in food, photographic chemicals, pharmaceutical and biological products and intermediates.

Due to apparent brightness, on which planets of the ones listed below would I have to be standing on to observe the smallest apparent brightness of our sun?

Answers

Answer:

Neptune to be precise

Explanation:

cause it is the last planet in our solar system

What element has the electron configuration 1 s 2 2 s 2 2 p 6 3 s 2 3 p 2 ?

Answers

Phosphorus, which has the atomic number 15 and is located in period 3 and group 15 of the periodic table.

Is white rice a heterogeneous?

Answers

Answer:No,it’s not a heterogeneous

Explanation:

I don’t believe so. It would be homogeneous

what are the two quantities in this module for which we will develop unit factors to do dimensional analysis with chemical substances?

Answers

The two portions on this module for which we can broaden unit elements to do dimensional analysis with chemical materials are volume and mass.

volume is a degree of the quantity of three-dimensional space occupied by using an object or substance. it's miles regularly measured in liters, cubic centimeters, or cubic meters. volume is used to measure the potential of a field, the dimensions of an object, or the quantity of liquid or gas in a given space. it's also used to determine the mass of an item, because the density (mass in line with unit quantity) is often acknowledged.

the 2 portions of volume and mass are critical for doing dimensional analysis with chemical substances because they help to measure the amount of a substance in distinct gadgets. extent is vital for measuring the quantity of liquid materials, and mass is critical for measuring the quantity of strong substances. each of these portions are vital for accurate and specific measurements of chemical substances, making them crucial for dimensional analysis.

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A chemical supply company sells sulfuric acid (H2SO4) in a concentration of 4.00 M. What volume of this solution would you need to make 12.0 mL of a 0.50 M H2SO4 solution

Answers

you need 3.00 mL of the 4.00 M sulfuric acid solution to make 12.0 mL of a 0.50 M solution.

What is Solution?

Solution is a way to solve a problem or address an issue. It can involve finding an answer or coming up with a strategy to tackle a difficult situation. Solutions can be found through research, brainstorming, trial and error, or by using existing resources.

To solve this problem, we must use the equation:
V1C1 = V2C2
Where V1 is the volume of the concentrated solution, C1 is the concentration of the concentrated solution, V2 is the volume of the dilute solution, and C2 is the concentration of the dilute solution.
In this case, V1 is the volume of the 4.00 M solution that we need, C1 is 4.00 M, V2 is 12.0 mL, and C2 is 0.50 M.
Plugging these values into the equation, we can solve for V1:
V1 = (V2C2) / C1
V1 = (12.0 mL)(0.50 M) / (4.00 M)
V1 = 3.00 mL

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Base your answer to the following question on the elements below. (A) Boron (B) Rubidium (C) Nitrogen (D) Mercury A highly reactive element A) A B) B C) C D) D

Answers

B) Rubidium is a highly reactive element.

Rubidium is a highly reactive element due to its low ionization energy, which makes it easily lose electrons and form compounds. It is considered to be one of the most reactive alkali metals.

When it comes into contact with water or air making rubidium is highly flammable. It can ignite spontaneously in the air and reacts violently with water to produce rubidium hydroxide and hydrogen gas.

Being noticeably more reactive than potassium but less so than cesium, it is likely to react with substances that are dangerous with sodium or potassium more forcefully. When rubidium is exposed to air or dry oxygen, it ignites, mostly generating oxide.

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