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2 years ago

Which molecule could form a coordinate covalent bond with a proton (h )?

Chemistry

2 answers:

Luda [366]2 years ago

8 0

Answer:

A chlorine (Cl-) molecule could make a coordinated covalent bond with a proton (H +, in this case) forming hydrogen chloride (HCl).

Explanation:

Covalent bonds occur when there is electron sharing between atoms, making all atoms that participate in this type of bond stable, but in the case of coordinated or dative covalent bond, the atom that has available pairs of electrons and is not performing Sharing them with another atom can “donate” these free electrons to any other atom, thereby performing a simple covalent bond and a coordinated or dative covalent bond, where electron (s) transfer from one atom to the other. This type of chemical bond is quite common and often occurs with chlorine, sulfur, phosphorus, among other various chemical elements. As an example we can cite the bond between chlorine and hydrogen to form hydrogen chloride.

For a coordinated covalent bond to occur it is necessary for one of the atoms to share a pair of electrons with another atom, and at the same time assign one or more electrons to another atom, which only participates in the bond receiving the electrons.

Reil [10]2 years ago

5 0

Molecule is the smallest particle of a compound or an element with similar properties as the element or the compound. Covalent bond (dative bond) is a bond that is formed due to the sharing of electrons between atoms. Coordinate covalent bond is a type of covalent where electrons shared during bond formation comes from one atom. In this case a molecule of H2O would form a coordinate bond with H+ because the oxygen contains a lone pair of electrons forming H3O+.

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Which missing item would complete this beta decay reaction?<br> PLATO

goldfiish [28.3K]

Answer: $_{-1}^0\beta$

Explanation:

General representation of an element is given as: $_Z^A\textrm{X}$

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

Beta decay : In this process, a neutron gets converted into a proton and an electron releasing a beta-particle. The beta particle released carries a charge of -1 units.

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta$

The given reaction would be represented as:

$_{43}^{98}\textrm{Tc}\rightarrow_{44}^{98}\textrm{Pb}+_{-1}^0\beta$

7 0

2 years ago

Read 2 more answers

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −

Grace [21]

Answer : The value of $K_{eq}$ is, 11.2

The value of $\Delta G_{rxn}$ is -9.04 kJ/mol

Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$\Delta G^o=-RT\times \ln K_{eq}$

$-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=11.2$

Thus, the value of $K_{eq}$ is, 11.2

Now we have to calculate the $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

The given reaction is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

$\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})$

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -9.04 kJ/mol

3 0

2 years ago

For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati

AveGali [126]

Answer:

a. withdraws electrons inductively

b. donates electrons by hyperconjugation

c. donates electrons by resonance

d. withdraws electrons inductively

Explanation:

a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:

F > Cl > Br>I

b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.

c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.

d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.

3 0

2 years ago

The standard free energy ( Δ G ∘ ′ ) (ΔG∘′) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . −12.6 kJ⋅mol−1. The Δ G ΔG

Dmitrij [34]

Answer:

The concentration of [ADP] = 21.896*10^-6 μM

Explanation:

Given Data:

creatine + ATP -----------> ADP + creatine phosphate

ΔG∘ = -12.6 KJ/mole = -12600 J/mole

ΔG = -0.1 KJ/mole = -100 J/mole

[Creatine phosphate] = 25 mM = 25*10^-3 M

[Creatine] = 17 mM = 17*10^-3 M

[ATP] =5mM = 5*10^-3M

Calculating the concentration of [ADP] using the formula;

ΔG = ΔG∘ + RTlnQc

Substituting, we have

-12600 = -100 + 8.314*298lnQc

-12600+100 = 8.314*298lnQc

-12500 = 2477.57lnQc

lnQc = -12500/2477.57

lnQc = -5.045

Qc = e^ -5.045

Qc = 6.44*10^-3

But,

Qc = [Creatine phosphate]*[ADP]/[creatine]*[ATP]

6.44*10^-3 = 25*10^-3*[ADP]/ (17*10^-3* 5*10^-3)

6.44*10^-3 = 25*10^-3[ADP]/8.5*10^-5

6.44*10^-3 * 8.5*10^-5 = 25*10^-3[ADP]

5.474*10^-7 = 25*10^-3[ADP]

[ADP] = 5.474*10^-7 /25*10^-3

= 2.1896 *10^-5 M

= 21.896*10^-6 μM

Therefore, the concentration of [ADP] = 21.896*10^-6 μM

3 0

2 years ago

Consider the reaction 2Al(OH)3(s)→Al2O3(s)+3H2O(l) with enthalpy of reaction ΔHrxn∘=21.0kJ/mol What is the enthalpy of formation

bekas [8.4K]

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8 0

2 years ago

Read 2 more answers