Answer:
2.55 moles
Explanation:
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
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.
Every organic molecules/compound contains carbon (c).
Some other very abundant are hydrogen, nitrogen, oxygen, phosphorus, and sulfur.
I learned this with the acronym CHNOPS.
C - Carbon
H - Hydrogen
N - Nitrogen
O - Oxygen
P - Phosphorus
S - Sulfur
Hope this helps!
0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
To learn more about stoichiometric quantities visit:
<h3>
brainly.com/question/28174111</h3>
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Explanation:
hexaaqua iron (III) trinitrate
