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abruzzese [7]
3 years ago
15

How do we observe in experiments?

Chemistry
1 answer:
anyanavicka [17]3 years ago
6 0

Answer:

by watching and writing down the information that is provided

Explanation:

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Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -
kogti [31]
The given complex ion is as follow,

                                              [Ru (CN) (CO)₄]⁻

Where;
            [ ]  =  Coordination Sphere

            Ru  =  Central Metal Atom  =  <span>Ruthenium

            CN  =  Cyanide Ligand

            CO  =  Carbonyl Ligand

The charge on Ru is calculated as follow,

                               Ru + (CN) + (CO)</span>₄  =  -1
Where;
            -1  =  overall charge on sphere

             0  =  Charge on neutral CO

            -1  =  Charge on CN

So, Putting values,


                               Ru + (-1) + (0)₄  =  -1

                               Ru - 1 + 0  =  -1

                               Ru - 1  =  -1

                               Ru  =  -1 + 1

                               Ru  =  0
Result:
          <span>Oxidation state of the metal species in each complex [Ru(CN)(CO)</span>₄]⁻ is zero.
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What are the different rows called
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Vertical-horizon.
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A phosphorus atom and a chlorine atom are likely to form a covalent bond because both elements:
Jobisdone [24]
A phosphorus atom and a chlorine atom are likely to form a covalent bond because both elements are nonmetals. 
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How do radiography and sonography compare?
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Radiography techs are trained to operate medical equipment that uses X-rays and other types of radiation. Ultrasound techs, referred to professionally as diagnostic medical sonographers, operate equipment that uses high-frequency sound waves to create images.

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3 years ago
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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
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