Answer:
27 liters of hydrogen gas will be formed
Explanation:
Step 1: Data given
Number of moles C = 1.03 moles
Pressure H2 = 1.0 atm
Temperature = 319 K
Step 2: The balanced equation
C +H20 → CO + H2
Step 3: Calculate moles H2
For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2
For 1.03 moles C we'll have 1.03 moles H2
Step 4: Calculate volume H2
p*V = n*R*T
⇒with p = the pressure of the H2 gas = 1.0 atm
⇒with V = the volume of H2 gas = TO BE DETERMINED
⇒with n = the number of moles H2 gas = 1.03 moles
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 319 K
V = (n*R*T)/p
V = (1.03 * 0.08206 *319) / 1
V = 27 L
27 liters of hydrogen gas will be formed
<h2>Answer:</h2>
The correct answer is
A) Regular operation
<h2>
Explanation:</h2>
Even those workplaces that have established LO/TO processes face challenges, including: Lack of specific procedures written for each piece of equipment identifying all energy sources and energy isolation devices. Lack of comprehensive safety training for everyone in the workplace. Incorrect tag use.
So, regular operation is the primary cause of LO/TO accidents.
The average Kenectic energy
Answer:
48%
Explanation:
Based on Gay-Lussac's law, the pressure is directly proportional to the temperature. To solve this question we must assume the temperature increases and all CO2 remains without reaction. The equation is:
P1T2 = P2T1
<em>Where Pis pressure and T absolute temperature of 1, initial state and 2, final state of the gas:</em>
P1 = 10.0atm
T2 = 1420K
P2 = ?
T1 = 730K
P2 = 10.0atm*1420K / 730K
P2 = 19.45 atm
The CO2 reacts as follows:
2CO2 → 2CO+ O2
Where 2 moles of gas react producing 3 moles of gas
Assuming the 100% of CO2 react, the pressure will be:
19.45atm * (3mol / 2mol) = 29.175atm
As the pressure rises just to 24.1atm the moles that react are:
24.1atm * (2mol / 19.45atm) = 2.48 moles of gas are present
The increase in moles is of 0.48 moles, a 100% express an increase of 1mol. The mole percent that descomposes is:
0.48mol / 1mol * 100 = 48%
It would be the same thing if the Co does not have a number with it because it can’t reduce