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ankoles [38]
3 years ago
13

Identify the brønsted-lowry acid, the brønsted-lowry base, the conjugate acid, and the conjugate base in each reaction: (a) c5h5

n(aq)+h2o(l)⇌c5h5nh+(aq)+oh−(aq) (b) hno3(aq)+h2o(l)⇌h3o+(aq)+no3−(aq) drag the appropriate items to their respective bins.
Chemistry
2 answers:
Pie3 years ago
7 0

acids give away protons (H+), bases accept protons, conjugate bases are what u get when when you take the protons from the acid, and conjugate acids are what u get when you add the protons to the base.

so for (a) the C5H5N is the base, water is the acid, C5H5NH+ is the conj acid, OH- is the conj base

(b) HNO3 is the acid, H2O is the base, hydronium ion is the conj. acid, NO3- is the conj base.

raketka [301]3 years ago
4 0

Answer: a) C_5H_5N(aq.)+H_2O(l)\rightarrow C_5H_5NH^+(aq.)+OH(aq.)

bronsted- lowry acid : H_2O

conjugate base : OH^-

bronsted- lowry base : C_5H_5N

conjugate acid : C_5H_5NH^+

b) HNO_3(aq)+H_2O(l)\rightarrow H_3O^+(aq.)+NO_3^-(aq.)

bronsted-lowry acid : HNO_3

conjugate base : NO_3^-

bronsted- lowry 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.

For the given chemical equation:

a) C_5H_5N(aq.)+H_2O(l)\rightarrow C_5H_5NH^+(aq.)+OH(aq.)

Here, H_2O is loosing a proton, thus it is considered as a brønsted-lowry acid and after losing a proton, it forms OH^- which is a conjugate base.

And, C_5H_5N is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it forms C_5H_5NH^+ which is a conjugate acid.

b) HNO_3(aq)+H_2O(l)\rightarrow H_3O^+(aq.)+NO_3^-(aq.)

Here, HNO_3 is loosing a proton, thus it is considered as a brønsted-lowry acid and after losing a proton, it forms NO_3^- which is a conjugate base.

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

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A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
Is the center of an atom is called the nucleus?​
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Explanation:

3 0
3 years ago
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

6 0
3 years ago
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