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3 years ago

What is the structure of carbon

Chemistry

1 answer:

expeople1 [14]3 years ago

3 0

Answer:

With an atomic number of 6 (six electrons and six protons), the first two electrons fill the inner shell, leaving four in the second shell. Therefore, carbon atoms can form four covalent bonds with other atoms to satisfy the octet rule.

Explanation:

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Negatively charged particles in the outermost energy level of the electron cloud.

balandron [24]

Answer:

Electrons are negatively charged particles that surround the atom's nucleus. Electrons were discovered by J. J. Thomson in 1897. *Protons are positively charged particles found within atomic nuclei.

Explanation:

6 0

3 years ago

Compare the molecular shape and hybrid orbitals of PF3 and PF5 molecules

MaRussiya [10]

PF3 is phosphorustrifluoride with p in the center and three f bonding with it.PF5 is phosphoruspentafluoride with p in the middle and 5 f linking to bond

5 0

3 years ago

Read 2 more answers

Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict wheth

Elena-2011 [213]

Answer:

$\Delta S_{surr} = + 0.32113\: kJ/K$

Explanation:

Given: Entropy of surrounding: ΔSsurr = ?

Temperature: T= 355 K

The change in enthalpy of reaction: ΔH = -114 kJ

Pressure: P = constant

As we know, ΔH = -114 kJ ⇒ negative

Therefore, the given reaction is an exothermic reaction

Therefore, Entropy of surrounding at constant pressure is given by,

$\Delta S_{surr} = \frac{-\Delta H}{T}$

$\therefore \Delta S_{surr} = -\left (\frac{-114 kJ}{355 K} \right ) = + 0.32113\: kJ/K > 0$

In the given reaction:

2NO(g) + O₂(g) → 2NO₂(g)

As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.

$\therefore \Delta S_{system} < 0$

As we know, for a spontaneous process, that the total entropy should be positive.

$\Delta S_{total} = \Delta S_{surr} + \Delta S_{system} > 0$

Therefore, at the given temperature,

if $\Delta S_{surr} > \Delta S_{system} \Rightarrow \Delta S_{total} > 0$ then the given reaction is spontaneous

if $\Delta S_{surr} < \Delta S_{system} \Rightarrow \Delta S_{total} < 0$ then the given reaction is non-spontaneous

6 0

3 years ago

Calculate the standard enthalpy of formation of liquid methanol, CH3OH(l), using the following information: C(graphite) + O2 --&

Darya [45]

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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

The formation reaction of $CH_3OH$ will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C(graphite)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ/mole$

(2) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ/mole$

(3) $CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-726.4kJ/mole$

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1) $C(graphite)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_1=-393.5kJ/mole$

(2) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$

(3) $CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g)$ $\Delta H_3=726.4kJ/mole$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

6 0

3 years ago

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is

lions [1.4K]

Carbon disulfide is prepared by heating sulfur and charcoal. the chemical equation is S2(g) + C(s) <---> CS2 (g) Kc = 9.40 at 900 K.
Hope this helps:)

4 0

3 years ago