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
The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g
From Avogadro's hypothesis,
6.02×10²³ atoms = 1 mole of Cr
But:
1 mole of Cr = 52 g
Thus, we can say that:
6.02×10²³ atoms = 52 g of Cr
With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:
6.02×10²³ atoms = 52 g of Cr
Therefore,
5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³
5.15×10²¹ atoms = 0.44 g of Cr
Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g
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<h3>Solution:</h3>
f(x) = 3x - 1 = 0
f(x) = x = 1/3
By putting the value of x in g(x), we get
g(x) = 2(1/3)^2 - 3
g(x) = 2(1/9) - 3
g(x) = 2/9 - 3
g(x) = -25/9
<h2>Answer:</h2>
g(x) = -25/9
Answer:
The atomic radius of krypton is similar to that of bromine. However, the effective nuclear charge of krypton is greater than that of bromine.
Explanation:
Ionizing an atom require moving an electron from the electron cloud of the atom to a point infinitely far away from the atom. The first ionization energy of this atom is the energy change in this process.
The electron and the nucleus are oppositely-charged. There is an electrostatic force between the two. Removing the electron requires overcoming this attraction. The size of the energy input depends on the electrostatic potential energy of the electron (the gravitational potential energy is much smaller than the electrostatic potential energy.) The separation between the electron and the nucleus is much larger than their radii. Both objects can be considered as point charges. Coulomb's Law gives the electrostatic potential energy of the two point charge that are close to each other.
,
where
is Coulomb's constant,
and 
are the two charges, and
is the separation between the two charges.
Krypton and bromine are right next to each other in the same period. Their atomic radii will be similar to each other. The separation between the outermost electron and the nucleus will also be similar for the two elements.
The first charge can be the electron. However, data show that for elements after helium, the second charge is smaller than the sum of charges on all protons in the nucleus. It turns out that the inner shell electrons (all of which are also negative) repel electrons in the outermost valence shell. The effective nuclear charge 
of a neutral atom is <em>approximately</em> the same as the number of protons minus the number of non-valence electrons. That number will be slightly larger for krypton than for bromine. As a result, the electrostatic potential energy on a 4p (the outermost orbital for both Kr and Br) electron of krypton will be more negative than that on a 4p electron in bromine. Removing that electron will take more energy in Kr than in Br. The first ionization energy of Kr is hence greater than that of Br.
