Answer:
Ammonium nitrate, (NH4NO3), a salt of ammonia and nitric acid, used widely in fertilizers and explosives. The commercial grade contains about 33.5 percent nitrogen, all of which is in forms utilizable by plants; it is the most common nitrogenous component of artificial fertilizers.
Answer : The correct option is, (C) 0.675 M
Explanation :
Using neutralization law,

where,
= concentration of
= 13.5 M
= concentration of diluted solution = ?
= volume of
= 25.0 ml = 0.0250 L
conversion used : (1 L = 1000 mL)
= volume of diluted solution = 0.500 L
Now put all the given values in the above law, we get the concentration of the diluted solution.


Therefore, the concentration of the diluted solution is 0.675 M
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.

- 2mol of N_2 forms 2mol of N_2O
- 1mol of N_2 forms 1mol of N_2O
Moles of N_2

Molar mass of N_2O=44g/mol
Mass of N_2O formed