Answer:
2AlCl3 + Ca3N2 - 2AlN+ 3CaCl2
You are looking for ADHESION
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
Answer:
36.4 atm
Explanation:
To find the pressure, you need to use the Ideal Gas Law. The equation looks like this:
PV = nRT
In this equation,
-----> P = pressure (atm)
-----> V = volume (L)
-----> n = moles
-----> R = constant (0.0821 L*atm/mol*K)
-----> T = temperature (K)
Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.
P = ? atm R = 0.0821 L*atm/mol*K
V = 5.00 L T = 393 °C + 273.15 = 312.45 K
n = 7.10 moles
PV = nRT
P(5.00 L) = (7.10 moles)(0.0821 L*atm/mol*K)(312.45 K)
P(5.00 L) = 182.130
P = 36.4 atm
Seven elements make up the compound
