Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of ·0.0020Ms−1 : 2NH3 (g)→N2 (g
2 answers:
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Answer:
The percentage (by mass) of KBr in the original mixture was 33.1%.
Explanation:
The mixture of KCl and KBr has a mass of 3.595g, thus the sum of the moles of KCl (<em>x</em>) multiplied by it molar mass (74.5g/mol) and the moles of KBr (<em>y</em>) multiplied by it molar mass (119g/mol) is the total mass of the mixture:
Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence
But the 0.042 moles came from the originals KCl and KBr moles, thus
Now it is possible to propose a system of equations:
Solving the system of equations,
0.010 moles of KBr multiplied it molar mass is
Therefore, the percentage (by mass) of KBr in the original mixture was:
%
Answer:
A compressed gas cylinder is filled with 5270 g of argon gas.
The pressure inside the cylinder is 2050 psi at a temperature of 18C.
The valve to the cylinder is opened and gas escapes until the pressure inside the cylinder is 650. psi and the temperature are 26 C.
How many grams of argon remains in the cylinder?
Explanation:
First, calculate the volume of argon gas that is present in the gas cylinder by using the ideal gas equation:
Mass of Ar gas is --- 5270g.
The number of moles of Ar gas:
Temperature T=(18+273)K=291K
Pressure P=2050psi
Volume V=?
Using this volume V=22.6L
Pressure=650psi=44.2atm
Temperature T= (26+273)K=299K
calculate number of moles "n" value:
Mass of 40.7mol of Ar gas:
Answer:
The mass of Ar gas becomes 1625.8g.