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kicyunya [14]
3 years ago
11

What is extremely lightweight and exists in a cloud orbiting the nucleus

Chemistry
2 answers:
lorasvet [3.4K]3 years ago
8 0
Electrons is the answer. Hope that helped
lana [24]3 years ago
7 0

Answer:

atomic particles

Explanation:

electrons are extremely lightweight and exist in a cloud orbiting the nucleus period the electron cloud has a radius 10,000 times greater than the nucleus according to the Los Alamos national laboratory.

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What is the name given to an ion that has lost electrons? someone help me please D:
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The name given to an ion that has lost electrons is cation. Details about cations can be found below.

<h3>What is a cation?</h3>

A cation is a positively charged ion i.e. the one that would be attracted to the cathode in an electrolytic procedure.

Generally, an ion is an electrically charged atom. An ion can either be positively charged or negatively charged. The latter is called an anion while the former is called a cation.

A cation is formed when an atom loses electrons to become positively charged while an anion is formed when an atom accepts electrons to become negatively charged.

Therefore, it can be said that the name given to an ion that has lost electrons is cation.

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3 years ago
A chemist titrates 110.0 mL of a 0.7684 M methylamine (CH3NH2) sotion with 0.4469 M HNO3 solution at 25 °C. Calculate the pH at
Firdavs [7]

Answer: The pH at equivalence point for the given solution is 5.59.

Explanation:

At the equivalence point,

            n_{HNO_{3}} = n_{CH_{3}NH_{2}}

So, first we will calculate the moles of CH_{3}NH_{2} as follows.

      n_{CH_{3}NH_{2}} = 0.764 M \times \frac{110 ml}{1000 ml/L}      

                     = 0.0845 mol

Now, volume of HNO_{3} present will be calculated as follows.

          Volume = \frac{\text{no. of moles}}{\text{Molarity}}

                        = \frac{0.0845}{0.4469 M}

                        = 0.1891 L

Therefore, the total volume will be the sum of the given volumes as follows.

                    110 ml + 189.1 ml

                  = 299.13 ml

or,               = 0.2991 L

Now, [CH_{3}NH_{3}^{+}] = \frac{0.0845 mol}{0.2991 L}

                        = 0.283 M

Chemical equation for this reaction is as follows.

     CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons CH_{3}NH_{2} + H_{3}O^{+}

As,      k_{a} = \frac{k_{w}}{k_{b}}        

                     = \frac{10^{-14}}{10^{-3.36}}

                     = 2.29 \times 10^{-11}

Now,   [HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}

                      = \sqrt{2.29 \times 10^{-11} \times 0.283}

                      = 2.546 \times 10^{-6}

Now, pH will be calculated as follows.

              pH = -log [H_{3}O^{+}]

                    = -log (2.546 \times 10^{-6})

                    = 5.59

Thus, we can conclude that pH at equivalence point for the given solution is 5.59.

6 0
3 years ago
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