Answer:
0.41kg/sec
Explanation:
PV= nRT
Given : V= 505 L
P=0.88 atm
R= 0.08206 Latm/K*mol
T= 172 .0C = 172+273 = 445 K
n = PV /RT = 0.88 * 505 / 0.08206 * 445 = 12.17 moles per sec of N2 are consumed
As per reaction : N2 + 3H2 ----> 2NH3
1 mole N2 is consumed to produce 2 moles NH3
moles of NH3 produced per sec :
(2 moles NH3/1mol N2) * 12.17 moles N2 = 24.34 moles NH3 per sec
grams of NH3 produced per sec =
24.34 moles NH3 per sec * molar mass NH3 = 24.34 moles NH3 per sec * 17.031 g/mol = 414.5 g NH3 per sec
rate in Kg/sec = 414.5 g NH3 per sec * (1kg /1000g) = 0.4145 Kg/sec
= 0.41kg/sec
A. sulfur dioxide ...................................................
Answer:
Ionic bonds form when a nonmetal and a metal exchange electrons, while covalent bonds form when electrons are shared between two nonmetals.
Explanation:
Since this is an equilibrium problem, we apply le chatelier principle. This principle states that whenever a system at equilibrium is disturbed due to change in several factors, it would move in a way to annul such change.
C2H4(g) + Cl2 ⇔ 2C2H4Cl2(g)
When the concentration of C2H4 is increased, there is more reactant sin the system. In order to annul this change, the equilibrium position will shift to the right favoring product formation.
When the concentration of C2H4Cl2 is increased, there is more product in the system. To annul this change, the equilibrium position will shift to the left, favoring reactant formation.
The answer is 69.048%.
Mass of potassium dichromate (K₂Cr₂O₇) is 294 g, which is a 100%.
Given mass of chromium is 52 g. So, the mass percent of chromium in the compound is 17.687%:
52 g : x% = 294 g : 100%
x = 52 ÷ 294 × 100 %
x = 17.687%
Given mass of potassium is 39 g. <span>So, the mass percent of potassium in the compound is 13.265%:
39 g : x% = 294 g : 100%
x = </span>39 ÷ <span>294 × 100 %
x = 13.265%
Therefore, </span>the mass percent of oxygen in the compound is <span>69.048%:
100% - </span>17.687% - <span>13.265% = 69.048%.</span>
