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Brut [27]
3 years ago
8

Pls help, science 8th grade question, #11 thx :)

Chemistry
1 answer:
Anarel [89]3 years ago
4 0
D, KOH is a concentrated Base and HCL is an acid. D is the only one that has these two.
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What is released in the process of an atom moving from an excited level to a lower energy level?
luda_lava [24]
The correct answer is a Photon.

One photon is released for each event. Photons are elementary particles of all electromagnetic radiation, including light.
7 0
3 years ago
Which of the following is not the same as 2.97 milligrams? 0.0000297 kg, 0.00297 g, 0.297 ch
Travka [436]

the correct answer its  0.002975

4 0
3 years ago
Read 2 more answers
When atoms gain electrons, they can become larger, because the addition of an electron increases electrostatic repulsion.
astra-53 [7]

Answer:True

Explanation: An anion has a larger radius than a neutral atom because it gains valence electrons. There are added electron/electron repulsions in the valence shell that expand the size of the electron cloud, which results in a larger radius for the anion.

hit the crown for me pls :)

Have a great day

6 0
2 years ago
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
How do the two desalination processes differ from the atmospheric water generator in terms of where the processes can be used?
-Dominant- [34]

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Hope I helped :)

4 0
3 years ago
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