Answer:
1.53 × 10²² atoms Ag
Explanation:
Step 1: Define conversions
3.271 × 10⁻²² g = 1 atom
Step 2: Use Dimensional Analysis
= 1.52858 × 10²² atoms Ag
Step 3: Simplify
We have 3 sig figs.
1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag
<span>The solid lines between N and Mg are actually ionic bonds. N has 5 valence electrons (2 of which are paired). Of the 3 that are unpaired, 2 are part of covalent bonds with adjacent carbon atoms. N accepts an extra electron to complete its octet, but gets a formal charge of -1. This allows for formation of an ionic bond with Mg, which is +2. Two of these charged N atoms therefore neutralize the charge of the central Mg. As for the coordinate (dative) covalent bonds, Mg has empty orbitals - the ionic bonds with the charged N atoms give it only 4/8 possible valence electrons.
The other two N atoms (dotted lines) have a formal charge of 0 since they form three covalent bonds with adjacent carbon atoms, but they still have a lone pair. Therefore, just to improve stability, each of these N atoms can "donate" its lone pair to Mg in order to complete its octet.
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Answer:
Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.
Explanation:
If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.
Homogeneous or heterogeneous, Hope this helps!!
Answer:
Given: 42 g of N2
Solve for O2 mass that contains the same number of molecules to 42 g of N2.
Solve for the number of moles in 42 g of N2
1 mole of N2 = (14 * 2) g = 28 g so the number of moles in 42 g of N2 is equal to 42 g / 28 g per mole = 1.5 moles
Solve for mass of 1 mole of oxygen
1 mole of O2 = 16 g * 2 = 32 g per mole
Solve for the mass of 1.5 moles of oxygen
mass of 1.5 moles of O2 = 32 g per mole * 1.5 moles
mass of 1.5 moles of O2 = 48 g
So 48 g of O2 contains the same number of molecules as 42 g of N2
