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LenKa [72]
3 years ago
13

Help, please? will mark brainliest​

Chemistry
1 answer:
alexira [117]3 years ago
8 0

Answer:

Sulfur would gain electrons

Explanation:

Atoms want to have a complete out valence shell and because sulfur only needs 2 more electrons to complete the outer shell it would take 2 more.

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When is a covalent bond described as polar? Choose one: when electrons are transferred from one atom to another if covalently bo
melomori [17]

Answer:

if electrons are shared unequally between bonded atoms

Explanation:

A polar covalent bond is a bond that is formed due to the unequal distribution of electrons between two partially charged atoms. This is observed when the difference in electronegativity between the bond atoms is between 0.5 and 1.7.

A polar bond is a covalent bond between two atoms where the electrons that form the bond are unevenly distributed. This causes the molecule to have a slight electric dipole moment where one end is slightly positive and the other is slightly negative.

The charge of the electric dipoles is less than a full unit charge, so they are considered partial charges and are called delta plus (δ +) and delta minus (δ-).

Because positive and negative charges are separated at the bond, molecules with polar covalent bonds interact with the dipoles of other molecules. This produces intermolecular dipole-dipole forces between the molecules.

6 0
3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
3 years ago
Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h
bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

We know that,

The relation between the C_p\text{ and }C_v for an ideal gas are :

C_p-C_v=R

As we are given :

C_p=28.253J/K.mole

28.253J/K.mole-C_v=8.314J/K.mole

C_v=19.939J/K.mole

Now we have to calculate the entropy change of the gas.

\Delta S=-n\times C_v\ln \frac{T_2}{T_1}

\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. (w=-pdV)

(C) Heat during the process will be,

q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0
3 years ago
What product is obtaned when carbon is
kirza4 [7]

Answer:

carbon dioxide

Explanation:

Carbon burns in oxygen to form carbon dioxide. Since hydrocarbon fuels only contain two elements, we always obtain the same two products when they burn. In the equation below methane (CH 4) is being burned. The oxygen will combine with the carbon and the hydrogen in the methane molecule to produce carbon dioxide (CO 2) and water (H 2O).

Carbon, as graphite, burns to form gaseous carbon (IV) oxide (carbon dioxide), CO2. ... When the air or oxygen supply is restricted, incomplete combustion to carbon monoxide, CO, occurs. 2C(s) + O2(g) → 2CO(g) This reaction is important. When one mole of carbon is exposed to some energy in the presence of one mole of oxygen gas, one mole of carbon dioxide gas is produced. This reaction is a combustion reaction.

6 0
3 years ago
Which statement is true of diffusion?
raketka [301]

Answer:

Molecules move from areas of high concentration to areas of low concentration.

4 0
3 years ago
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