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3 years ago

8

Oxygen, with an atomic number of 8, is a neutral atom and would have _______ electrons in the first electron shell and _

____ electrons in the second electron shell

1 answer:

motikmotik3 years ago

8 0

2 in the first shell and 6 in the second

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In the reaction CuO(s) + CO2(g) → CuCO3(s), a. CO2 is the Lewis acid and CuCO3 is its conjugate base. b. O2– acts as a Lewis bas

adoni [48]

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

According to Lewis acid-base concept:

The substance which is donating electron pair is considered as Lewis base and the substance which is accepting electron pair is considered as Lewis acid.

For the given chemical reaction:

$CuO(s)+CO_2(g)\rightarrow CuCO_3(s)$

$CO_2$ is accepting electron pair and is getting converted to $CO_3^{2-}$ . Thus, it is considered as Lewis acid.

$O^{2-}$ present in CuO is a Lewis base because it is donating electron pair.

Thus, the correct answer is Option d.

4 0

3 years ago

22. What is a comet?

Lunna [17]

Answer:

A ball of ice blasted into outer space

Explanation:

It is a ball of ice that is going so fast it leaves a "tail"

4 0

3 years ago

At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate

Minchanka [31]

Explanation:

Since, it is shown that the reaction has been reversed. Therefore, value of $K_{c}$ will become $\frac{1}{K_{c}}$ .

Hence, new $K_{c'} = \frac{1}{K_{c}}$

= $\frac{1}{0.0500}$

= 20

Also, the number of moles of each reactant has been halved. So, $K_{c''}$ for the reaction $MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g)$ will also get halved.

Therefore, $K_{c''}$ = $K_{c'}$ = $(20)^{0.5}$

= 4.47

As the value of $\Delta H$ is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.

As a result, $K_{c}$ will also increase with increase in temperature.

6 0

3 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago

Identify the reactants in the equation: 6C0, + 6H2O - CHUO+ 60,

Effectus [21]

Answer:

The correct answer is d. 6H20 + 6CO2.

The reactant in the chemical reaction 6H2O + 6CO2 ---> C6H12O6 + 6O2 is 6H20 + 6CO2. Remember that the reactant is always at the left side of the equation. So the correct answer is 6H20 + 6CO2 since it's in the left of the equation. I hope this answer helped you.

Explanation:

5 0

3 years ago