Why is the theoritical strngth of metals much higher than that observed experimentally

The summary or ending of your experiment

Katyanochek1 [597]

Answer:

Conclusion

Explanation:

I believe you were asking for the term that best matches with the description given. Typically the conclusion summarizes your experiment in a 1 to 2 paragraph format.

7 0

3 years ago

Compare which element would have larger first ionization energy: an alkali metal in Period 2 or an alkali metal in Period 4?

maria [59]

Answer:

An alkali metal present in period 2 have larger first ionization energy.

Explanation:

Ionization energy:

The amount of energy required to remove the electron from the atom is called ionization energy.

Trend along period:

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

Trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.

6 0

2 years ago

Would you weigh more or less on the moon than you do on Earth, and why?

VladimirAG [237]

Answer:You would weigh less on the moon because there is less gravity on the moon.

Explanation:

7 0

3 years ago

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What do Potassium and Krypton have in common and what are their differences based on their location on the periodic table?

Kamila [148]

Potassium (K) and Krypton (Kr) are in the same period, so they have the same number of electron shells. Potassium is more reactive than Krypton because it only has one valence electron.

K and Kr are both in Period 4, so they each have the n = 1, 2, 3, and 4 shells. Kr is less reactive than K because it already has a complete octet.

“Potassium (K) and Krypton (Kr) are in the same period, so they have the same number of electron shells. Krypton is more reactive than Potassium because it has a full outer shell” is incorrect because having a full outer shell means that the element is unreactive.

“Potassium (K) and Krypton (Kr) are in the same group, so they have the same number of valence electrons. They have different numbers of electron shells” is incorrect for three reasons:

1. K is in Group 1 and Kr is in Group 18.

2. K has one valence electron, and Kr has eight valence electrons.

3. They have the same number of electron shells.

“Potassium (K) and Krypton (Kr) are in the same group, so they have the same number of valence electrons. They have different numbers of electron shells” is incorrect for two reasons:

1. K and Kr have different numbers of valence electrons.

2. They have the same number of electron shells.

8 0

2 years ago

A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. what is the percent yield for this rea

Masja [62]

CaCO3(s) ⟶ CaO(s)+CO2(s)

moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131

From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3, therefore 0.0131 moles CO2 should also be formed.
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2

Therefore:
% Yield: 0.53/.576 x100= 92 percent yield

4 0

2 years ago

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