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Stels [109]
3 years ago
9

A chemist is using a solution of HNO3 that has a pH of 3.75. What is [OH− ] for the solution?

Chemistry
2 answers:
snow_lady [41]3 years ago
8 0

Answer:

[OH-] = 5.62

n = -11

pOH = 10.25

Explanation: Just did it on edge

strojnjashka [21]3 years ago
7 0

Explanation:

It is known that pH is negative log of H^{+}. Also the relationship between pH and pOH is as follows.

                      pH + pOH = 14

                    3.75 + pOH = 14

                               pOH = 14 - 3.75

                                        = 10.25

Therefore,          pOH = -log [OH^{-}]

                          10.25 = -log [OH^{-}]

                     antilog 10.25 = - [OH^{-}]

                         [OH^{-}] = 5.6 \times 10^{-11}  

Thus, we can conclude that [OH^{-}] for the solution is 5.6 \times 10^{-11}.

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Answer:

Following are the responses to the given question:

Explanation:

\to C_6H_9NO

calculating the HOI:

= \frac{2C+2+N-H-X}{2} \\\\ =\frac{2\times 6+2+1-9-0}{2}\\\\ =\frac{12-6}{2}\\\\=\frac{6}{2}\\\\=3

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Please find the attachment file of the Function group:

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3 years ago
Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
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:

rate = k*[A]^m * [B]^n

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>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<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 =>

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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) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

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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Answer:

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