Hey there!
5 moles will be produced.
N₂ has a molar mass of 28.014 g/mol.
Convert 70g to mol:
70 ÷ 28.014 = 2.5
In N₂ there are 2 nitrogen atoms. In NH₃ there is 1 nitrogen atom.
So, there will be twice as many moles of NH₃ because every one molecule of N₂ will produce two molecules of NH₃.
2.5 x 2 = 5 moles
Hope this helps!
Answer:
4.81 moles
Explanation:
The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.
Pressure at which gauge reads zero = 14.7 psi
Pressure read by the gauge = 988 psi
Total pressure = 14.7 + 988 psi = 1002.7 psi
Also, P (psi) = P (atm) / 14.696
Pressure = 1002.7 / 14.696 = 68.2297 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 1.50 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 4.81 moles
The atomic number of an atom is determined by the number of protons it has..
It is also the whole number shown on the periodic table
Answer:
a) 90 kg
b) 68.4 kg
c) 0 kg/L
Explanation:
Mass balance:

w is the mass flow
m is the mass of salt

v is the volume flow
C is the concentration





![-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)](https://tex.z-dn.net/?f=-%5Bln%282000L%2B3%2AL%2Fmin%2At%29-ln%282000L%29%5D%3Dln%28m%29-ln%2890kg%29)
![-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)](https://tex.z-dn.net/?f=-ln%5B%282000L%2B3%2AL%2Fmin%2At%29%2F2000L%5D%3Dln%28m%2F90kg%29)
![m=90kg*[2000L/(2000L+3*L/min*t)]](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2At%29%5D)
a) Initially: t=0
![m=90kg*[2000L/(2000L+3*L/min*0)]=90kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A0%29%5D%3D90kg)
b) t=210 min (3.5 hr)
![m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A210min%29%5D%3D68.4kg)
c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.