If the molecule C6H12 does not contain a double bond, and there are no branches in it, what will its structure look like?

The d-metals iron, copper, and manganese form cations with different oxidation states. For this reason, they are found in many o

choli [55]

Answer:

They have electrons in their 3d- and 4s-orbital for bond formation.

Explanation:

d- metals or transition metal are metal which form ion with partially filled d-orbital. Examples are iron and manganese.

The metals have 2 electrons in their 4s orbital. If only this is used for bonding, they will form compounds where they have oxidation State of +2 as seen in MnO.

If two 4s and one of 3d electrons are used, oxidation state of +3 is formed as seen in FeCl3.

If two 2s electron I used with two 3d electrons, compound with oxidation state of +4 is formed as seen in MnO2

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How much energy must be supplied to break a single 21Ne nucleus into separated protons and neutrons if the nucleus has a mass of

love history [14]

Answer:

1)There is 2.68 * 10^-11 J of energy needed

2) The nuclear binding energy for 1 mol of Ne is 1.6 *10^13 J/mol

Explanation:

Step 1: Data given

The nucleus of a21Ne atom has a amass of 20.98846 amu.

Step 2: Calculate number of protons and neutrons

The number of electrons and protons in an 21Ne atom = 10

The number of neutrons = 21 -10 =11

Step 3: mass of the atom

Mass of a proton = 1.00727647 u

Mass of a neutron = 1.0086649 u

The mass of the atom = mass of all neutrons + mass of protons

Mass of atom = 11*1.0086649 + 10*1.00727647 = 21.1680786 amu

Step 4: Calculate change of mass

The change in mass = Mass of atom - mass of neon

Δmass = 21.1680786 - 20.98846

Δmass = 0.1796186

Step 5: Calculate mass for a single nucleus

The change of mass for a single nucleus is = Δmass / number of avogadro

Δmass of nucleus = 0.1796186 / 6.022*10^23

Δmass of nucleus =2.98 * 10^-25 grams = 2.98 * 10^-28 kg

Step 6: Calculate energy to break a Ne nucleus

Calculate the amount of energy to break a Ne nucleus

ΔEnucleus = Δmass of nucleus * c²

⇒ with c = 2.9979 *10^8 m/s

ΔEnucleus = 2.98 * 10^-28 kg * (2.9979*10^8)² = 2.68 * 10^-11 J

What is the nuclear binding energy for 1 mol of Ne?

ΔE= ΔEnucleus * number of avogadro

ΔE= 2.68 * 10^-11 J * 6.022*10^23

ΔE= 1.6 *10^13 J/mol

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3 years ago

Consider the pka (3.75) of formic acid, h-cooh as a reference. with appropriate examples, show how inductive, dipole, and resona

Luden [163]

Formic acid is the simplest carboxylic acid with a structure of HCOOH and has a pka of 3.75. The pka refers to the acidity of the molecule, which in this example refers to the molecules ability to give up the proton of the O-H. A decrease in the pka value corresponds to an increase in acidity, or an increase in the ability to give up a proton. When an acid gives up a proton, the remaining anionic species (in this case HCOO-) is called the conjugate base, and an increase in the stability of the conjugate base corresponds to an increase in acidity.

The pka of a carboxylic can be affected greatly by the presence of various functional groups within its structure. An example of an inductive effect changing the pka can be shown with trichloroacetic acid, Cl3CCOOH. This molecule has a pka of 0.7. The decrease in pka relative to formic acid is due to the presence of the Cl3C- group, and more specifically the presence of the chlorine atoms. The electronegative chlorine atoms are able to withdraw the electron density away from the oxygen atoms and towards themselves, thus helping to stabilize the negative charge and stabilize the conjugate base. This results in an increase in acidity and decrease in pka.

The same Cl3CCOOH example can be used to explain how dipoles can effect the acidity of carboxylic acids. Compared to standard acetic acid, H3CCOOH with a pka of 4.76, trichloroacetic acid is much more acidic. The difference between these structures is the presence of C-Cl bonds in place of C-H bonds. A C-Cl bond is much more polar than a C-H bond, due the large electronegativity of the chlorine atom. This results in a carbon with a partial positive charge and a chlorine with a partial negative charge. In the conjugate base of the acid, where the molecule has a negative charge localized on the oxygen atoms, the dipole moment of the C-Cl bond is oriented such that the partial positive charge is on the carbon that is adjacent to the oxygen atoms containing the negative charge. Therefore, the electrostatic attraction between the positive end of the C-Cl dipole and the negative charge of the anionic oxygen helps to stabilize the entire species. This level of stabilization is not present in acetic acid where there are C-H bonds instead of C-Cl bonds since the C-H bonds do not have a large dipole moment.

To understand how resonance can affect the pka of a species, we can simply compare the pka of a simple alcohol such as methanol, CH3OH, and formic acid, HCOOH. The pka of methanol is 16, suggesting that is is a very weak acid. Once methanol gives up that proton to become the conjugate base CH3O-, the charge cannot be stabilized in any way and is simply localized on the oxygen atom. However, with a carboxylic acid, the conjugate base, HCOO-, can stabilize the negative charge. The lone pair electrons containing the charge on the oxygen atom are able to migrate to the other oxygen atom of the carboxylic acid. The negative charge can now be shared between the two electronegative oxygen atoms, thus stabilizing the charge and decreasing the pka.

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3 years ago

2H2 + O2 --> 2H2O can be classified as what type of reaction?

Sveta_85 [38]

Combustion reaction
Key: O2
O2 is normally in a chemical formula when you are used to burn anything, so basically, anything with O2 involves burning.

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A gas fills a balloon at a temperature of 27 C and 101.325 kPa of pressure. What will the pressure of the balloon be if the gas

JulijaS [17]

135.1‬kPa

Explanation:

Given parameters:

T1 = 27°C

P1 = 101.325 kPa

T2 = 127°C

Unknown:

P2 = ?

Solution:

Using a derivative of the combined gas law where we assume that the gas has a constant volume, we can solve for the unknown.

At constant volume:

$\frac{P1}{T1} = \frac{P2}{T2}$

P1 is the initial pressure

T1 is the initial temperature

P2 is the final pressure

T2 is the final temperature

Take the given temperature to K

T1 = 27 + 273 = 300K

T2 = 127 + 273 = 400K

Input the variables:

$\frac{101.325}{300} = \frac{P2}{400}$

P2 = 135.1‬kPa

learn more:

Boyle's law brainly.com/question/8928288

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3 years ago