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evablogger [386]
3 years ago
14

Are there any molecules that form a square?

Chemistry
1 answer:
djyliett [7]3 years ago
7 0
Yes there are.  Perhaps the most famous and obvious is squaric acid.
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What is the ph of a 0.45 m solution of aniline (c6h5nh2)? (pkb  9.40)?
mote1985 [20]

Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+ H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).

pKb(C₆H₅NH₂) = 9.40.

Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.

c₀(C₆H₅NH₂) = 0.45 M.

c(C₆H₅NH₃⁺) = c(OH⁻) = x.

c(C₆H₅NH₂) = 0.45 M - x.

Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).

4·10⁻¹⁰ = x² /  (0.45 M - x). 

Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.

pOH = -log(0.0000134 M.) = 4.87.

pH = 14 - 4.87 = 9.13.

7 0
3 years ago
Define the following: Bronsted-Lowry acid - Lewis acid- Strong acid - (5 points) Problem 6: Consider the following acid base rea
Allushta [10]

Answer: Yes, HCl is a strong acid.

acid = HCl , conjugate base = Cl^- , base = H_2O, conjugate acid = H_3O^+

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

Yes HCl is a strong acid as it completely dissociates in water to give H^+ ions.

HCl\rightarrow H^++Cl^-

For the given chemical equation:

HCl+H_2O\rightarrow H_3O^-+Cl^-

Here, HCl is loosing a proton, thus it is considered as an acid and after losing a proton, it forms Cl^- which is a conjugate base.

And, H_2O is gaining a proton, thus it is considered as a base and after gaining a proton, it forms H_3O^+ which is a conjugate acid.

Thus acid =  HCl

conjugate base = Cl^-

base = H_2O

conjugate acid = H_3O^+.

8 0
3 years ago
Which of the following is a limitation of Bohr's model?
Murljashka [212]

Answer:

 The answer is:

The first option

Explanation:

6 0
3 years ago
A(n) ________ acid is an acid that can leave solution and enter the atmosphere. a(n) ________ acid is an acid that can leave sol
Aneli [31]

A volatile acid is an acid that can leave solution and enter the atmosphere.


7 0
3 years ago
Ammonia, NH 3 , may react with oxygen to form nitrogen gas and water. 4 NH 3 ( aq ) + 3 O 2 ( g ) ⟶ 2 N 2 ( g ) + 6 H 2 O ( l )
Alex Ar [27]

Answer:

The limiting reactant is NH₃

0.0186moles of N₂ are the one produced by the limiting reactant

0.020 moles of N₂ are the one produced by the reactant in excess

Explanation:

This is the reaction

4NH₃ + 3O₂  → 2N₂ + 6H₂O

We should calculate the moles of each reactant

Mass / Molar mass = Moles

3.55 g / 17g/m = 0.208 moles NH₃

5.33 g / 32g/m = 0.166 moles O₂

4 moles of ammonia react with 3 moles of oxygen

0.208 moles of ammonia react with (0.208  .3)/4 = 0.156 moles O₂

We have 0.166 moles of O₂ and we need 0.156 moles, so O₂ is the reactant in excess.

3 moles of O₂ react with 4 moles of NH₃

0.166 moles of O₂ react with (0.166 . 4)/ 3 = 0.221 moles

We have 0.208 moles NH₃ and we need 0.221, so NH₃ is the limiting reactant.

To know the moles of N₂, let's apply the Ideal Gas Law

P.V =n.R.T

1atm . 0.450L = n . 0.082 . 295K

0.450 / (0.082 .295) = 0.0186 moles

If we have 100 % yield reaction:

4 moles NH₃ make 2 moles N₂

0.208 moles NH₃ make (0.208  .2)/4 = 0.104 moles

So the % yield reaction is.

0.104 moles ___ 100%

0.0186 moles ___ 17.9%

0.0186moles of N₂ are the one produced by the limiting reactant.

3 moles of O₂ produce 2 moles N₂

0.166 moles O₂ produce  (0.166  .2)/3 = 0.111 moles

Now, we apply the yield.

100% ____ 0.111 moles

17.9% = 0.020 moles

8 0
3 years ago
Read 2 more answers
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