The answer is C good luck
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>
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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>
Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, or even in opposition to, external forces like gravity.
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
1)False
2)Meters allow scientists to be more accurate.
3)Air resistance.
4)C
5)Two stars very close together.
6)Rolling Friction
7)A elevator coming to a stop in order to stop at the top floor.
8)Size of the ofject
