Answer:
Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?
Explanation:
Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?Disociaciones ácido / base utilizando la definición de Arrhenius.
ACIDOS
HF (aq) —->?
<span>XY4Z2-->Square planar (Electron domain geometry: Octahedral) sp3d2
XY4Z-->Seesaw (Electron domain geometry: Trigonal bipyramidal) sp3d
XY5Z-->Square pyramidal (Electron domain geometry: Octahedral) sp3d2
XY2Z3-->Linear (Electron domain geometry: Trigonal bipyramidal) sp3d
XY2Z-->Bent (Electron domain geometry: Trigonal planar) sp2
XY3Z-->Trigonal pyramidal (Electron domain geometry: Tetrahedral) sp3
XY2Z2-->Linear (Electron domain geometry: Tetrahedral) sp3
XY3Z2-->T shaped (Electron domain geometry: Trigonal bipryamidal) sp3d
XY2-->Linear (Electron domain geometry: Linear) sp
XY3 Trigonal planar (Electron geometry: Trigonal planar) sp2
XY4-->Tetrahedral (Electron domain geometry: tetrahedral) sp3
XY5-->Trigonal bipyramidal (Electron domain geometry: Trigonal bipyramidal) sp3d
XY6-->Octahedral (Electron domain geometry: Octahedral) sp3d2</span>
The third substance or agent which produce the film between the interface of two immiscible liquids and thus stabilize the system are known as an emulsifying agent.
Since the solubility of the liquids depends on the polarity of the mixing liquids the thumb rule of solubility is like dissolves like that means polar liquid dissolves in polar liquid only and vice versa. For two immiscible liquids, the emulsifying agent is used which does not chemically change the polarity of liquids but acts as bridge between immiscible liquids, the polar end of the emulsifier attach to the polar liquid and the non-polar end of the emulsifier attach to the non-polar end and thus help in dissolving.
Therefore, the one end of the emulsifier is polar and the other end is non-polar
<u>Answer:</u> The amount of water produced in the given reaction is 
grams.
<u>Explanation:</u>
Let us assume that the initial amount of acetylene gas given be 'x' grams. Now, to calculate the number of moles, we will use the formula:
....(1)
Given mass of acetylene = x grams
Molar mass of acetylene = 26 g/mol
Putting values in above equation, we get:
The reaction of combustion of acetylene is given by the equation:
By Stoichiometry of the reaction:
2 moles of acetylene produces 2 moles of water.
So, 
moles of acetylene will produce = 
moles of water.
Now, to calculate the amount of water produced, we use equation 1:
Molar mass of water = 18 g/mol
Moles of water = 
Putting values in equation 1, we get:
grams
