Answer:
When copper(II) chloride and sodium carbonate solutions are combined, solid copper(II) carbonate precipitates, leaving a solution of sodium chloride. Write the conventional equation, total ionic equation, and net ionic equation for this reaction.
Explanation:
The word equation for the reaction is:
Copper (II) chloride(aq) + sodium carbonate (aq) ->sodium chloride (aq) + copper carbonate(s)
The balanced chemical equation of the reaction is:

The complete ionic equation is:

The net ionic equation is obtained from the complete ionic equation after removing the spectator ions:

<h3>
Answer:</h3>
82.11%
<h3>
Explanation:</h3>
We are given;
- Theoretical mass of the product is 137.5 g
- Actual mass of the product is 112.9 g
We are supposed to calculate the percentage yield
- We need to know how percentage yield is calculated;
- To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.
Thus;
% yield = (Actual mass ÷ Experimental mass) × 100%
= (112.9 g ÷ 137.5 g) × 100%
= 82.11%
Therefore, the percentage yield of the product is 82.11 %
The electron-group arrangement of CO₃²⁻ is trigonal planar. The molecular shape is trigonal planar, and the ideal bond angle(s) is CO₃²⁻ is 120°
<h3>What is the molecular geometry of a compound?</h3>
The position of the compound's electrons and nuclei can be seen in the molecular geometry. It demonstrates how the form of the complex is created by the interaction of electrons and nuclei.
Here, according to the VSEPR theory, the shape of the carbonate ion is trigonal planar. The carbon will be in the center.
Thus, the electron-group arrangement and the shape of the carbonate ion are trigonal planar. The bond angle will be 120°.
To learn more about molecular geometry, refer to the link:
brainly.com/question/16178099
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Answer:
(a) sp³ sp³
H₃<u>C</u> - <u>C</u>H₃
(b) sp³ sp²
H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂
sp²
(c) sp³ sp
H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH
sp sp³
(d) sp³ sp²
H₃<u>C</u> - <u>C</u>H=O
Explanation:
Alkanes or the carbons with all the single bonds are sp³ hybridized.
Alkenes or the carbons with double bond(s) are sp² hybridized.
Alkynes or the carbons with triple bond are sp hybridized.
Considering:
(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.
Hence,
sp³ sp³
H₃<u>C</u> - <u>C</u>H₃
(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.
Hence,
sp³ sp²
H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂
sp²
(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.
Hence,
sp³ sp
H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH
sp sp³
(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is bonded by double bond to oxygen.
Hence,
sp³ sp²
H₃<u>C</u> - <u>C</u>H=O