Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
We’d have to be very careful because if we had our skeletons on the outside it’d be very easy to injure ourselves
Your answer is False I think
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 activation energy for the decomposition = 33813.28 J/mol
Explanation:
Using the expression,
Wherem
is the activation energy
R is Gas constant having value = 8.314 J / K mol
Thus, given that,
= ?
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (5 + 273.15) K = 278.15 K
T = (25 + 273.15) K = 298.15 K
So,




<u>The activation energy for the decomposition = 33813.28 J/mol</u>