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NeX [460]
2 years ago
6

Methane (CH4) reacts with excess oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O). What is the percent yield of c

arbon dioxide if the actual yield is 29.8 grams and the theoretical yield is 0.814 moles?
Chemistry
1 answer:
kotegsom [21]2 years ago
6 0
I'm pretty sure that it's either 83.2% or 91.4%
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Which of the following electron configurations gives the correct arrangement of the four valence electrons of the carbon atom in
Juliette [100K]

Answer:

D) 2s^12p^3

Explanation:

Carbon.

The electronic configuration is -  

1s^22s^22p^2

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.

But in methane, CH_4 it forms 4 bonds. So, 1 electron each from 2s orbital jumps to the next orbital in the p subshell.

Thus, the configuration is:-

1s^22s^22p^2

Thus, the valence electron configuration is:-

2s^12p^3

7 0
2 years ago
For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate
makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=2.87-(-1.66)=4.53V

Hence, the standard electrode potential of the cell is 4.53 V.

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