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Eddi Din [679]
3 years ago
7

Sodium (na) , atomic number 11, has a tendency to lose an electron in the presence of chlorine. after losing the electron, na wi

ll have ___ protons in its nucleus
Chemistry
1 answer:
Vinvika [58]3 years ago
4 0

The atomic number is equal to number of protons present in the nucleus of an atom or vice-versa. For any atom, there is no change in number of protons either by adding or removal of electron.

Since, sodium is a metal, it has tendency to lose electron in the presence of non-metal i.e. chlorine. Thus, after lose of an electron, there is no change in the number of protons or number of protons will remain same.

Therefore, atomic number of metal is 11 which is also equal to the number of protons implies sodium will have 11 protons in its nucleus.

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A student weighs out a 2.23 g sample of KF, transfers it to a 300 ml volumetric flask, adds enough water to dissolve it and then
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How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?
Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

Cu(OH)_2 -> Cu^{2+} + 2OH^-

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles

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