All of th above support the heliocentric theory.
The answer is D only metals are shiny and highly malleable
Answer:
Because warm air is less dense
Explanation:
F. <em>None of the above
</em>
<em>No O atoms are present</em> as reacting substances, only O_2 and H_2O molecules.
O_2 + 2H_2O + 2e^(-) → 4OH^(-)
We must use <em>oxidation numbers</em> to decide whether oxygen or water is the substance reduced.
The oxidation number of O changes from 0 in O_2 to -2 in OH^(-).
A decrease in oxidation number is <em>reduction</em>, so O_2 is the substance reduced.
The oxidation number of O is -2 in both H_2O and OH^(-), so water is <em>neither oxidized nor reduced</em>.
Answer:
The correct answers are: <u>Each oxygen of carbonate ion has -2/3 or -0.67 charge.</u>
<u>Bond order of each carbon‑oxygen bond in the carbonate ion</u> = <u>1.33</u>
Explanation:
The carbonate ion (CO₃²⁻) is an organic compound, in which a carbon atom is covalently bonded to three oxygen atoms. The net formal charge on a carbonate ion is −2.
The carbonate ion is <u>resonance stabilized</u> and has three equivalent resonating structures, which exhibits that all the three carbon-oxygen bonds in a carbonate ion are equivalent.
In the resonance hybrid of carbonate ion,<u> the negative charge is equally delocalized on all the three oxygen atoms. </u>
<u>Thus, each bonded oxygen has -2/3 or -0.67 charge.</u>
<u />
In a carbonate ion there is one double bond oxygen (C=O) and two single bonded oxygen (C-O). Bond order of 1 C=O is 2 and bond order of C-O is 1.
∴ <u>Bond order</u> = sum of all bond orders ÷ number of bonding groups = (2+1+1) ÷ 3 = <u>1.33</u>
