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Andrei [34K]
3 years ago
9

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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7 0
3 years ago
2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'
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Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

\displaystyle M_1V_1= M_2V_2

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}

Convert this value to mL:
\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

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2 years ago
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