Answer:
V = 22.34 L
Explanation:
Given data:
Volume of O₂ needed = ?
Temperature and pressure = standard
Number of molecules of water produced = 6.0× 10²³
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of water:
1 mole contain 6.022× 10²³ molecules
6.0× 10²³ molecules × 1 mole / 6.022× 10²³ molecules
0.99 mole
Now we will compare the moles of oxygen and water.
H₂O : O₂
2 : 1
0.996 : 0.996
Volume of oxygen needed:
PV = nRT
V = nRT/P
V = 0.996 mol × 0.0821 atm.L/mol.K × 273.15 K / 1 atm
V = 22.34 L
Answer:
to become a noble gas element P will have 2 electrons in it's outer most energy level if it has one energy level
and eight in the last energy level if more than one
Answer:
N2 + 3H2 ———> 2NH3
As we know 1000 grams ammonia is 58.82 moles so according to unitary method,
2 mole NH3 formed by 1 mole N2 hence 58.82 NH3 will be given by 29.41 moles N2.
No. Of moles = given mass/molar mass
Implies that
Mass of nitrogen required = 29.41*28 = 823.48 grams.
Explanation:
. The members of theautosome pairs are truly homologous; that is, each member of a pair contains a full complement of the same gene.The sex chromosomes, on the other hand, do not constitute a homologous pair.
