For each of the bonds listed, tell which atom is more negatively charged a. C--H b. C--N c. C--S d. C--O

Chemistry

1 answer:

blondinia [14]3 years ago

5 0

Explanation:

The atom is considered to be more electronegative , when it has the tendency to get a slight negative i.e. , delta negative charge on it , and the other atom gets a delta positive charge .

In the periodic table ,

As we go left to right in the periodic table , the electronegativity increases ,

and

As we go top to bottom , the electronegativity decreases .

Hence , from the option given in the question ,

a. C--H , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , Carbon is more electronegative than hydrogen .

b. C--N , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , nitrogen is more electronegative than Carbon .

c. C--S , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , sulfur is more electronegative than Carbon .

d. C--O , the electronegativity of both the atom is almost the same , but looking at the treads of the periodic table , oxygen is more electronegative than Carbon .

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nadya68 [22]

Answer:Strontium (Sr) and Gallium (Ga)

Explanation:

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Consider the following reaction, which is spontaneous at room temperature. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Is ΔS positive o

Ugo [173]

Answer:

∆H = negative and ∆S = positive.

Explanation:

The reaction given in the question is spontaneous at room temperature ,

hence ,

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∆G = ∆H -T∆S

The value of ∆G can be negative , if ∆H has a negative value and ∆S has a positive value , because , T∆S , has a negative sign .

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3 years ago

Which of these describes an endothermic reaction?

neonofarm [45]

B) energy is absorbed by the reaction
is right answer.

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3 years ago

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kkurt [141]

<u>Answer:</u> The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$