I think the answer is C but don’t quote me on that
H20* SOO N34 Thats my answer
Answer: 0.07868 mol H₂O
Explanation:
1) Chemical equation:
Cu₂O +H₂ → 2Cu + H₂O
2) mole ratios:
1 mol Cu₂O : 1 mol H₂ : 2 mol Cu : 1 mol H₂O
3) Convert 10.00 g of Cu to grams, using the atomic mass:
Atomic mass of Cu: 63.546 g/mol
number of moles = mass in grams / atomic mass = 10.00g / 63.546 g/mol
number of moles = 0.1574 mol
4) Use proportions
2mol Cu 0.1574 mol Cu
--------------- = ---------------------
1 mol H₂O x
⇒ x = 0.1574 mol Cu × 1 mol H₂O / 2mol Cu = 0.07868 mol H₂O
That is the answer
The reason that some of the elements of period three and beyond are steady in spite of not sticking to the octet rule is due to the fact of possessing the tendency of forming large size, and a tendency of making more than four bonds. For example, sulfur, it belongs to period 3 and is big enough to hold six fluorine atoms as can be seen in the molecule SF₆, while the second period of an element like nitrogen may not be big to comprise 6 fluorine atoms.
The existence of unoccupied d orbitals are accessible for bonding for period 3 elements and beyond, the size plays a prime function than the tendency to produce more bonds. Hence, the suggestion of the second friend is correct.