Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
Water, H2O, HOH these are all the same compound just worded differently
Answer:
A net ionic equation shows only the chemical species that are involved in a reaction, while a complete ionic equation also includes the spectator ions.
Brainlist pls!
102 grams of ammonia is formed when 3 moles of nitrogen and 6.7 moles of hydrogen reacts.
Explanation:
The equation given is of Haeber's process in which the nitrogen is limiting factor in the ammonia formation and hydrogen if in excess gets delimited.
We know that 1 mole of Nitrogen gives 2 moles of ammonia.
We have 3 moles of nitrogen here,
So, 6 moles of ammonia will be form
so from the formula
no of moles=mass/atomic mass
mass= no. of moles*atomic mass
= 6*17
= 102 grams of ammonia will be formed.
So, 6 moles or 102 grams of ammonia is formed when 3 mole of nitrogen and 6.7 mole of hydrogen reacts.
