(1) FeO (3) Fe3O
(2) Fe2O3 (4) Fe3O=
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Explanation:
This Metal belongs to Group I (Alkali Metals).
Alkali Metals present in Group I are considered as the most reactive metals in periodic table. There reactivity increases tremendously down the group. The reactivity is mainly due to less ionization energies. Therefore, going from top to bottom along the group the ionization energies decreases hence, increasing there reactivity respectively.
Alkali metals when reacted with water undergoes an exothermic reaction resulting the formation of corresponding metal hydroxide and hydrogen gas i.e.
2 M + 2 H₂O → 2 MOH + H₂
Also, these metals are in solid state at room temperature (i.e. 25 °C) and their boiling points are as follow,
Lithium = 180.5 °C
Sodium = 97.79 °C
Potassium = 63.5 °C
Rubidium = 39.48 °C
Cesium = 28.44 °C
What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3?
The balanced chemical reaction is:
N2 + 3H2 = 2NH3
We are given the amount of H2 being reacted. This will be our starting point.
26.3 g H2 (1 mol H2 / 2.02 g H2) 2 mol O2/3 mol H2) ( 17.04 g NH3 / 1mol NH3) = 147.90 g O2
Percent yield = actual yield / theoretical yield x 100
Percent yield = 79.0 g / 147.90 g x 100
Percent yield = 53.4%
The atomic number of sodium is 11. Sodium is a metal and it has the electronic configuration: 2, 8,1. This configuration implies that, the electrons in an atom of sodium are distributed into three different shells and the outermost shell has one electron in it. If sodium decide to give away the electron on its outermost shell, it will have 10 electrons left and those electrons will be distributed in only the first two shells and then it will be described as an ion. The third shell will not exist again because the electron there has been given away. Thus, the sodium ion is going to have a smaller atomic radius because its size has been reduced. This implies that the sodium ion will have a smaller radius than the sodium atom.