The question is incomplete, the complete question is;
The Lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group______ (red).
In this representation, each Y atom needs ______ electron(s) to complete its octet, and gains these electrons by forming______ bond(s) with atoms of H .
There are ______ unshared electron pair(s) and _______bonding electron pair(s) in the product molecule.
The bonds in the product are _________ (Ionic or Covalent)
Answer:
1) 16
2) 2 electrons
3) 2 bonds
4) 2 unshared pairs of electrons
5) 2 bonding pairs of electrons
6) The bonds in the product are covalent
Explanation:
Group sixteen elements have six electrons on their outermost shell. These include two unshared pairs of electrons and two unpaired electrons. These two unpaired electrons can now be covalently bonded to two hydrogen atoms to give H2Y. The compound H2Y has two lone pairs and two bond pairs of electrons.
H2Y can be a general formula for all hydrides of group 16. They are all very similar in structure but gradually differ in physical and chemical properties according to the graduated variation observed down the group.
True becuase dew is coming out of the air which if you look at a glass of water it has condensation on it becuase it is hot
Answer:
When we say "chlorine wants to gain one electron", we speak of the radical atom. Chlorine as a free radical, Cl⋅ , is the chlorine atom that we say has 7 valence electrons and wants its 8th to form an octet. So, Cl⋅ , chlorine radical, is less stable, and Cl− , chlorine ion, is more stable
Answer:
B.false because if the reactant concentration is disturbed the whole reaction will be affected.
Explanation:
Answer:
3.64g
Explanation:
Given parameters:
Mass of NH₃ = 18.1g
Mass of Cu₂O = 90.4g
Unknown:
Limiting reactant = ?
Mass of N₂ formed = ?
Solution:
The reaction equation is given as:
Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O
The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;
Number of moles = 
Molar mass of Cu₂O = 2(63.6) + 16 = 143.2g/mol
Molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles of Cu₂O = 
= 0.13moles
Number of moles of NH₃ = 
= 5.32moles
From this reaction;
1 mole of Cu₂O combines with 2 mole of NH₃
So 0.13moles of Cu₂O will combine with 0.13 x 2 mole of NH₃
= 0.26moles of NH₃
Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;
Mass of N₂;
Mass = number of moles x molar mass
1 mole of Cu₂O will produce 1 mole of N₂
0.13 mole of Cu₂O will produce 0.13 mole of N₂
Mass = 0.13 x (2 x 14) = 3.64g
