Sodium metal is quite reactive; sodium ions (as in NaCl) are quite unreactive. Cl^- ions are not reactive; they are stingily attracted to positive ions such as Na^+with which they form ionic bonds.
Answer:
las plantas
Explanation:
Las plantas son los seres vivos capaces de transformar energía luminosa utilizando agua y dióxido de carbono (CO₂) en energía química en forma de moléculas llamadas carbohidratos. El proceso a través del cual realizan esto se denomina fotosíntesis, y para ello requieren un pigmento de color verde denominado clorofila, el cual es necesario para captar la energía luminosa proveniente del Sol.
The freezing point depression is calculated through the equation,
ΔT = (kf) x m
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
5.88 = (1.86)(m)
m is equal to 3.16m
Recall that molality is calculated through the equation,
molality = number of mols / kg of solvent
number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.
Answer:
a) FePO4(s)⇄Fe^3+(aq) + PO4^3-(aq)
b) ZnCO3(s)⇄Zn^2+(aq) + CO3^2-(aq)
c) NH4Cl(s)⇄ NH4^+(aq) + Cl^-(aq)
Explanation:
An ionic solid simply means a solid substance that is held together by ionic bonds. When an ionic substance is added to water, the ions interact with the dipoles in water and is pulled apart to form the constituent cation and anion present in the ionic solid. This is the process that we have referred to as dissolution.
The Ksp of an ionic solid is obtained from the chemical equation that shows the dissolution of an ionic solid in water. The Ksp is actually an equilibrium constant that shows the extent of dissolution of an ionic solid in water.
a) FePO4(s)⇄Fe^3+(aq) + PO4^3-(aq)
b) ZnCO3(s)⇄Zn^2+(aq) + CO3^2-(aq)
c) NH4Cl(s)⇄ NH4^+(aq) + Cl^-(aq)
<span>Answer:
To determine hybridization you count the # of un-bonded PAIR of electrons. Theny you count bonded domains (a double bond still counts as one bonded domain, so does a triple bond).
For the first Carbon on top, you see it bonded to 2 oxygens and 1 carbon.However, when you count up the valence electrons that would be present, there is supposed to be 2 more electrons (ONE electron pair) on carbon. To do hybridization you must also be familiar with drawing lewis structures. So when you draw the pair of un-bonded electrons on carbon, you see that there is ONE un-bonded electron pair, and THREE bonded domains.
ONE plus THREE = FOUR.
so that would be sp3.</span>