Let us calculate the structure of the electric shells of the Al atom. It has an atomic number of 13, so it has 13 electrons. The first 2 go to the first hell. The next 8 need to go to the second shell and the last 3 ones would go to the outermost shell. The outer shell, that is the most important one for chemical reactions, has thus 3 electrons. An atom always tries to have a completed outer shell (with either 2 or 8 atoms). It is easier for a cell to have a charge of +3 than a charge of -5 (smaller absolute value) and thus the Aluminum atom will try to get rid of the 3 electrons. In this process, it loses negative charge thus it will become positively charged. Hence, the correct answer is that it will prefer to lose 3 electrons and become positively charged.
Probably true because chemicals do control weed and it hurts the environment.
The phase diagram of CO2 has a melting curve that slopes up and to the right, in contrast to the phase diagram of water, which has a more conventional shape. It is impossible for liquid CO2 to exist at pressures lower than 5.11 atm because the triple point is 5.11 atm and 56.6 °C.
Due to the fact that ice is less thick than liquid water, the phase diagram of water has an odd melting point that drops with pressure. Carbon dioxide cannot exist as a liquid at atmospheric pressure, according to the phase diagram of the gas. Thus, gaseous carbon dioxide directly sublimes from solid carbon dioxide.
Learn more about solid carbon dioxide.
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Let us assume that there is a 100g sample of Opal. The masses of each element will be:
29.2g Si
33.3g O
37.5g H2O
Now we divide each constituent's mass by its Mr to get the moles present
Si: (29.2 / 28) = 1.04
O: (33.3 / 16) = 2.08
H2O: (37.5 / 18) = 2.08
Now we divide by the smallest number and obtain:
Si: 1
O: 2
H2O: 2
Thus, the empirical formula of Opal is:
SiO2 . 2H2O
Answer: Tin (Sn)
Explanation: The electron configuration for tin (Sn) is shown in the picture. It's last electrons are:
5s^2 4d^10 5p^2
The valence electrons are in the 5th electron shell and include 2 each in the 5s and 5p orbitals.
