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brilliants [131]
3 years ago
14

Identify the conjugate base in each pairs

Chemistry
1 answer:
DiKsa [7]3 years ago
8 0

Answer: 1) RCOO^-

2) H_2PO_4^-

3) RNH_2

4) HCO_3^-

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.

1) RCOOH\rightarrow RCOO^-+H^+

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

2) H_3PO_4\rightarrow H_2PO_4^-+H^+

Here, H_3PO_4 is loosing a proton, thus it is considered as an acid and after losing a proton, it forms H_2PO_4^- which is a conjugate base.

3) RNH_3^+\rightarrow RNH_2+H^+

Here, RNH_3^+ is loosing a proton, thus it is considered as an acid and after losing a proton, it forms RNH_2 which is a conjugate base.

4) H_2CO_3\rightarrow HCO_3^-+H^+

Here, H_2CO_3 is loosing a proton, thus it is considered as an acid and after losing a proton, it forms HCO_3^- which is a conjugate base.

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The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

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Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
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<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

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rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
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