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3 years ago

How many orbitals are completely filled in an atom whose electron configuration is 1s^2 2s^2 2p^6 3s^1?

Chemistry

2 answers:

Nimfa-mama [501]3 years ago

6 0

Hey there!:

The 1s, 2s and 2p subshells are completely filled (a maximum of two electrons go into the 1s subshell and a maximum of two electrons go into the 2s subshell. The 2p subshell includes 3 orbitals, with 2 electrons maximum per orbital). The 3s subshell has only one of a maximum of two electrons.

Hope that helps!

nataly862011 [7]3 years ago

3 0

The number of orbitals that are completely full is 5.

The reason you may have thought it was 3 is because you're confusing orbitals with subshells.

In this electron configuration, the sub shells are: 1s^2, 2s^2, 2p^6, 3s^1

So, we have four sub shells.

Each sub shell consists of a specific number of orbitals. s subshells have one orbital, p subshells have three orbitals, d subshells have five orbitals, and f subshells have seven orbitals. Knowing this, we can look back at the electron configuration we were given and determine how many orbitals are filled.

1s^2 has one orbital, 2s^2 has one orbital, 2p^6 has three orbitals, and 3s^1 has one orbital. The first three subshells are completely full, so a total of five orbitals are full.

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3 years ago

Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 mL sample titrated wi

julia-pushkina [17]

Answer: The $K_{sp}$ for calcium hydroxide is $5.324\times 10^{-6}$

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.0983M\\V_1=11.22mL\\n_2=2\\M_2=?M\\V_2=50mL$

Putting values in above equation, we get:

$1\times 0.0983\times 11.22=2\times M_2\times 50\\\\M_2=0.011M$

The concentration of $Ca(OH)_2$ comes out to be 0.011 M.

The balanced equilibrium reaction for the ionization of calcium hydroxide follows:

$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

The expression for solubility constant for this reaction follows:

$K_{sp}=[Ca^{2+}][OH^-]^2$

Putting the values in above equation, we get:

$K_{sp}=(0.011)\times (2\times 0.11)^2$

$K_{sp}=5.324\times 10^{-6}$

Hence, the $K_{sp}$ for calcium hydroxide is $5.324\times 10^{-6}$

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Read 2 more answers

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