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tia_tia [17]
3 years ago
15

What’s the relationship between atomic radius and ease of oxidation on the activity series.

Chemistry
1 answer:
user100 [1]3 years ago
3 0

The bigger the atomic radius the easier it is to oxidise the atom. Remember that an atom is oxidized by the loss of an electron.

Explanation:

The bigger the atomic radius the further away the valence electron are from the attractive force of the atomic nucleus. This means that the energy required to remove an electron from the valence shell is easier compared to an atom with a smaller atomic radius. This is because you need to overcome the attractive force of the nucleus on the electron for you to oxidize the atom.

Learn More:

For more on oxidation energy check out;

brainly.com/question/8835627

brainly.com/question/13507502

#LearnWithBrainly

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There are 1000 mililiters in a liter, so 1000 ml for every liter, you have 5 liters, so:

5L*1000 = 5000 mL

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WILL GIVE 50 POINTS AND BRAINLIEST Describe the properties of alkaline earth metals. Based on their electronic arrangement, expl
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They are highly reactive metals. They react with water but do so less readily than alkali earth metals.

Owing to their high reactivity, they are seldom found free in nature. They always occur in combined state with other highly reactive nonmetals.
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2 years ago
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3 years ago
The standard emf for the cell using the overall cell reaction below is +2.20 V:
ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: 2Al-6e^{-}\rightarrow 2Al^{3+}

Reduction: 3I_{2}+6e^{-}\rightarrow 6I^{-}

---------------------------------------------------------------------------------

Overall: 2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}

Nernst equation for this cell reaction at 25^{0}\textrm{C}-

E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}

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Plug in all the given values in the above equation -

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3 0
3 years ago
A gamma ray photon has an energy of 4.75 x 10-14 joules. What is the frequency of this radiation?
Rama09 [41]

The frequency of the radiation is equal to 7.17  \times 10^{19} Hertz.

<u>Given the following data:</u>

  • Photon energy = 4.75 \times 10^{-14} Joules

To find the frequency of this radiation, we would use the Planck-Einstein equation.

Mathematically, the Planck-Einstein relation is given by the formula:

E = hf

<u>Where:</u>

  • h is Planck constant.
  • f is photon frequency.

Substituting the given parameters into the formula, we have;

4.75 \times 10^{-14} = 6.626 \times 10^{-34} \times F\\\\F = \frac{4.75 \times 10^{-14}}{6.626 \times 10^{-34}}

Frequency, F = 7.17  \times 10^{19} Hertz

Read more: brainly.com/question/16901506

4 0
3 years ago
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