Answer;
The total pressure is 1.107 atm.
Explanation;
The total pressure is the sum of the pressures of the three gases in the flask
Pressure (total) = 0.215 atm + 0.066 atm + 0.826 atm = 1.107 atm
= 1.107 atm.
Explanation:
It is known that efficiency is denoted by
.
The given data is as follows.
= 0.82,
= (21 + 273) K = 294 K
= 200 kPa,
= 1000 kPa
Therefore, calculate the final temperature as follows.
0.82 =
= 1633 K
Final temperature in degree celsius =
= 
Now, we will calculate the entropy as follows.

For 1 mole, 
It is known that for
the value of
= 0.028 kJ/mol.
Therefore, putting the given values into the above formula as follows.

= 
= 0.0346 kJ/mol
or, = 34.6 J/mol (as 1 kJ = 1000 J)
Therefore, entropy change of ammonia is 34.6 J/mol.
Answer:
20 kJ/mol
Explanation:
From ∆G°= -RTlnK
But
Ag2SO4(s)<----------->2Ag+(aq) + SO4^2-(aq)
Ksp= [2Ag+]^2 [SO4^2-]
But Ag+ = 0.032M
Ksp= (2×0.032)^2 (0.032)
Ksp= 1.31072×10^-4
∆G°= -RTlnK
∆G°= -(8.314× 298×(-8.93976))= 20KJmol-1( to the nearest KJ)
The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.
The mass of NaOH is 45 g
The molar mass of NaOH = 40 g/mol
The moles of NaOH = mass / molar mass
= 45 / 40
= 1.125
Thus, 1.125 mol NaOH is available
3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl
3 mol of NaOH react with 1 mol of FeCl₃
1.125 moles of NaOH will react with x moles of FeCl₃
x = 1.125 / 3
x = 0.375 mol
0.375 mol FeCl₃ can take part in reaction
The molar mass of FeCl₃ is 162 g/mol
The mass of FeCl₃ = moles × mass
= 0.375 × 162
= 60.75 g
Thus, the amount of FeCl₃, which can be consumed is 60.75 g
First, let's write down the balanced chemical reaction between the given reactants:
NO₂ + NO → N₂O + O₂
The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:
Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)
For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1