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Misha Larkins [42]
2 years ago
10

Make a claim for why caco3 reacts more vigorously with hcl than with hc2h3o2

Chemistry
1 answer:
Maru [420]2 years ago
3 0

CaCO₃ reacts more vigorously with HCl than with CH₃COOH because HCl is a strong acid and acetic acid is a weak acid.

<h3>What is calcium carbonate?</h3>

Calcium carbonate (CaCO₃) is a salt of strong base and weak acid so that it can react with any acid.

Among hydrochloric acid (HCl) and acetic acid (CH₃COOH), HCl is a strong acid means it completely dissociates into their respective ions and acetic acid is a weak acid. So due to strong basic nature of calcium carbonate it vigorously reacts with strong HCl acid.

Hence due to strong acidity of HCl, CaCO₃ vigorously react with HCl.

To know more about acidity, visit the below link:

brainly.com/question/19584961

#SPJ1

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Which compound has both ionic and covalent bonding?.
Nikitich [7]

Answer: Calcium carbonate is another example of a compound with both ionic and covalent bonds. Here calcium acts as the cation, with the carbonate species as the anion. These species share an ionic bond, while the carbon and oxygen atoms in carbonate are covalently bonded

Explanation:

4 0
3 years ago
Easy Chem, Will Give brainliest
Crazy boy [7]

Answer:

3.94 L

Explanation:

From the question given above, the following data were obtained:

Mass of O₂ = 5.62 g

Volume of O₂ =?

Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:

Mass of O₂ = 5.62 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mole of O₂ =?

Mole = mass / molar mass

Mole of O₂ = 5.62 / 32

Mole of O₂ = 0.176 mole

Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.

Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP

5 0
3 years ago
How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg(s)? Show all calculations leading to an answer. Li
Vaselesa [24]

Balanced chemical reaction happening here is:

3Mg(s) + N₂(g) → Mg₃N₂(s)        


 <u>moles of product formed from each reactant:</u>


2.0 mol of N2 (g) x <u> 1 mol Mg₃N₂      </u>  = <u>2 mol Mg₃N₂</u>

                                    1 mol N2

and


8.0 mol of Mg(s) x <u> 1 mol Mg₃N₂      </u>   = 2.67 mol Mg₃N₂

                                 3 mol Mg


Since N2 is giving the least amount of product(Mg₃N₂) ie. 2 mol Mg₃N₂

N2 is the limiting reactant here and Mg is excess reactant.


Hence mole of product formed here is 2 mol Mg₃N₂    


molar mass of Mg₃N₂    

= 3 Mg + 2 N

= 101g/mol  


mass of product(Mg₃N₂) formed  

= moles x Molar mass

= 2 x 101

= 202g Mg₃N₂


<u>202g of product are formed from 2.0 mol of N2(g) and 8.0 mol of Mg(s).</u>


<u>   </u>   The following are indicators of chemical changes:

Change in Temperature    

Change in Color

Formation of a Precipitate



8 0
3 years ago
Helllllllpppppppppppp<br><br>l​
Ber [7]

Answer:

hi                          

Explanation:

6 0
3 years ago
Which equation is associated with the first ionization of sodium?
cupoosta [38]
I believe the correct answer is A. Na(g) — Na+(g) + e-
7 0
4 years ago
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