The complete orbital notation diagram of an atom is shown. Five squares are shown aligned horizontally. Inside the first square
from the left is shown one upwards pointing arrow and one downwards pointing arrow. In the second square is another pair of upwards and downwards pointing arrow. Inside the third square is another pair of upwards and downwards pointing arrow. Inside the fourth square is another upwards and downwards pointing arrow. Inside the fifth square is an upwards pointing arrow. Based on the diagram, what values can be assigned to the angular momentum quantum number for the electrons in the atom? What information does this quantum number provide about the location of the electron?
1 answer:
Answer:
l = 0, 1, and 2; it is in a d orbital.
Explanation:
1. Angular momentum quantum number
One square means l = 0.
Three squares mean l = 1.
Five squares mean l = 2.
If an atom contains electrons with l = 2, it must also have electrons with l = 0 and 1.
Thus, the electrons in the atom can be assigned the angular momentum quantum numbers l = 0, 1, and 2.
Those in a set of five squares all have l = 2.
2. Location of electron
The last electron is the lone electron in the fifth square. Since l = 2, the electron is in a d orbital.
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Answer:
(a) 0.22 mol Cl₂ and 15.4g Cl₂
(b) 2.89.10⁻³ mol O₂ and 0.092g O₂
(c) 8 mol NaNO₃ and 680g NaNO₃
(d) 1,666 mol CO₂ and 73,333 g CO₂
(e) 18.87 CuCO₃ and 2,330g CuCO₃
Explanation:
In most stoichiometry problems there are a few steps that we always need to follow.
- Step 1: Write the balanced equation
- Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
- Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.
(a)
Step 1:
2 Na + Cl₂ ⇄ 2 NaCl
Step 2:
In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. <u>46g of Na react with 1 mol of Cl₂</u>. Since the molar mass of Cl₂ is 71g/mol, then <u>46g of Na react with 71g of Cl₂</u>.
Step 3:
(b)
Step 1:
HgO ⇄ Hg + 0.5 O₂
Step 2:
<u>216.5g of HgO</u> form <u>0.5 moles of O₂</u>. <u>216.5g of HgO</u> form <u>16g of O₂</u>.
Step 3:
(c)
Step 1:
NaNO₃ ⇄ NaNO₂ + 0.5 O₂
Step 2:
<u>16g of O₂</u> come from <u>1 mol of NaNO₃</u>. <u>16g of O₂</u> come from <u>85g of NaNO₃</u>.
Step 3:
(d)
Step 1:
C + O₂ ⇄ CO₂
Step 2:
<u>12 g of C</u> form <u>1 mol of CO₂</u>. <u>12 g of C</u> form <u>44g of CO₂</u>.
Step 3:
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(e)
Step 1:
CuCO₃ ⇄ CuO + CO₂
Step 2:
<u>79.5g of CuO</u> come from <u>1 mol of CuCO₃</u>. <u>79.5g of CuO</u> come from <u>123.5g of CuCO₃</u>.
Step 3: