Balanced chemical equation for the reaction is:
2S (g) + (g)+ 2O (l) ⇒
Moles of formed is 5.75 moles.
Moles of oxygen used is 5.75 moles in the reaction.
Explanation:
Data given:
moles of S = 11.5 moles
moles of = ?
Moles of needed =?
balanced equation with states of matter =?
Balanced chemical reaction under STP condition is given as:
2S(g) + (g) + 2O (l) ⇒
From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:
so, from 11.5 moles of S, x moles of is formed
2x = 11.5
x = 5.75 moles of sulphuric acid formed.
From the balanced reaction 1 mole of oxygen reacted to form 1 mole of sulphuric acid.
when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.
The balanced equation for the reaction is as follows
2H₂ + O₂ --> 2H₂O
stoichiometry of H₂ to O₂ is 2:1
number of H₂ moles - 30.0 g / 2 g/mol = 15 mol
number of O₂ moles - 80.0 g / 32 g/mol = 2.5 mol
limiting reactant is the reagent in which only a fraction is used up in the reaction
if H₂ is the limiting reactant
if 2 mol of H₂ requires 1 mol of O₂
then 15 mol of H₂ requires 1/2 x 15.0 = 7.5 mol of O₂
but only 2.5 mol of O₂ is required
this means that O₂ is the limiting reagentt and H₂ is in excess
The bonding of two atoms will most likely occur if A. a more stable state can result from the union.
Otherwise, there would be no point in making two atoms bond.
It should be scientific law since a hypothesis is like a prediction and a theory could be used to explain something although it might not be true. There’s many theories just to explain one thing. So it should be scientific law
Answer:
Volume of sample after droping into the ocean=0.0234L
Explanation:
As given in the question that gas is idealso we can use ideal gas equation to solve this;
Assuming that temperature is constant;
Lets and are the initial gas parameter before dropping into the ocean
and and are the final gas parameter after dropping into the ocean
according to boyle 's law pressure is inversly proportional to the volume at constant temperature.
hence,
P1=1 atm
V1=1.87L
P2=80atm
V2=?
After putting all values we get;
V2=0.0234L
Volume of sample after droping into the ocean=0.0234L