S and S²⁻ do not have the outer subshell fully filled with electrons.
Explanation:
We look at electronic configurations:
Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² - the outer subshell 4s² is fully-filled with electrons
S 1s² 2s² 2p⁶ 3s² 3p⁴ - the outer subshell 3p⁴ is not fully-filled with electrons
Zn²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s⁰ - here the 4s subshell is higher in energy than 3d subshell so will consider 3d¹⁰ the out subshell which is fully-filled with electrons
S²⁻ 1s² 2s² 2p⁶ 3s² 3p² - the outer subshell 3p² is not fully-filled with electrons
Ca²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ - the outer subshell 3p⁶ is fully-filled with electrons
Learn more about:
electron configurations
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Answer:
0.17 moles
Explanation:
In the elements of the periodic table, the atomic mass = molar mass. <u>Ex:</u> Atomic mass of Carbon is 12.01 amu which means molar mass of Carbon is also 12.01g/mol.
In order to find the # of moles in a 12 g sample of NiC-12, we will need to multiply the number of each atom by its molar mass and then add the masses of both Nickel and C-12 found in the periodic table:
- Molar Mass of Ni (Nickel): 58.69 g/mol
- Molar Mass of C (Carbon): 12.01 g/mol
Since there's just one atom of both Carbon and Nickel, we just add up the masses to find the molar mass of the whole compound of NiC-12.
- 58.69 g/mol of Nickel + 12.01 g/mol of Carbon = 70.7 g/mol of NiC-12
There's 12g of NiC-12, which is less than the molar mass of NiC-12, so the number of moles should be less than 1. In order to find the # of moles in NiC-12, we need to do some dimensional analysis:
- 12g NiC-12 (1 mol of NiC-12/70.7g NiC-12) = 0.17 mol of NiC-12
- The grams cancel, leaving us with moles of NiC-12, so the answer is 0.17 moles of NiC-12 in a 12 g sample.
<em>P.S. C-12 or C12 just means that the Carbon atom has an atomic mass of 12amu and a molar mass of 12g/mol, or just regular carbon.</em>
Answer:
6.78 × 10⁻³ L
Explanation:
Step 1: Write the balanced equation
Mg₃N₂(s) + 3 H₂O(g) ⇒ 3 MgO(s) + 2 NH₃(g)
Step 2: Calculate the moles corresponding to 10.2 mL (0.0102 L) of H₂O(g)
At STP, 1 mole of H₂O(g) has a volume of 22.4 L.
0.0102 L × 1 mol/22.4 L = 4.55 × 10⁻⁴ mol
Step 3: Calculate the moles of NH₃(g) formed from 4.55 × 10⁻⁴ moles of H₂O(g)
The molar ratio of H₂O to NH₃ is 3:2. The moles of NH₃ produced are 2/3 × 4.55 × 10⁻⁴ mol = 3.03 × 10⁻⁴ mol.
Step 4: Calculate the volume corresponding to 3.03 × 10⁻⁴ moles of NH₃
At STP, 1 mole of NH₃(g) has a volume of 22.4 L.
3.03 × 10⁻⁴ mol × 22.4 L/mol = 6.78 × 10⁻³ L
Answer:
i think so 1 and 3 is the correct answer...................
Hello!
Answer:
<span>Calculate Density Using Volume of a Cylinder.
That is one great way to measure the volume!
Hoping this helps you! ;D</span>
