Answer:
D= 5 g/L
Explanation:
M= D x Vm
M is the molar mass = 112g/mol
D is the density
Vm is the molar volume whuch is 22.4L at STP
now substitute;
112 = D x 22.4
D= 112\22.4
D= 5 g/L
Answer:
2800 g de ácido nítrico
Explanation:
La ecuación por la oxidación de amoniaco es:
4NH₃ + 7O₂ → 4H₂O + 2HNO₂ + 2HNO₃
Si pensamos que el oxígeno es el reactivo limitante, trabajamos con el amoniaco. Convertimos su masa a moles:
1.36 kg = 1360 g
1360 g . 1mol /17g = 80 moles
Si 4 moles de amoniaco pueden producir 2 moles de acido nítrico
80 moles producirán, (80 . 2)/4 = 40 moles.
Convertimos los moles a gramos:
40 mol . 63g /mol = 2520 g
Si le aplicamos la pureza
2520 g . 100/90 = 2800 g
<span>The rate at this temperature for the given concentrations would be..
</span>0.000056 M × s-
Answer:
71.7 L
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/Kmol)
T = temperature (K)
According to the information provided in this question;
P = 1 atm (STP)
V = ?
n = 3.2mol
T = 273K (STP)
Using PV = nRT
V = nRT/P
V = 3.2 × 0.0821 × 273/1
V = 71.7 L
