Kc' =Kc^1/3
=3√0.0061
=0.182716013
Answer:
Q = 8.8 kJ
Explanation:
Step 1: Data given
The specific heat of a solution = 4.18 J/g°C
Volume = 296 mL
Density = 1.03 g/mL
The temperature increases with 6.9 °C
Step 2: Calculate the mass of the solution
mass = density * volume
mass = 1.03 g/mL * 296 mL
mass = 304.88 grams
Step 3: Calculate the heat
Q = m*c*ΔT
⇒ with Q = the heat in Joules = TO BE DETERMINED
⇒ with m = the mass of the solution = 304.88 grams
⇒ with c = the specific heat of the solution = 4.18 J/g°C
⇒ with ΔT = the change in temperature = 6.9 °C
Q = 304.88 g * 4.18 J/g°c * 6.9 °C
Q = 8793.3 J = 8.8 kJ
Q = 8.8 kJ
Answer:
13 mol NO
Explanation:
Step 1: Write the balanced equation
4 NH₃(g) + 5 O₂(g) ⇒ 4 NO(g) + 6 H₂O(g)
Step 2: Establish the appropriate molar ratio
According to the balanced equation, the molar ratio of O₂ to NO is 5:4.
Step 3: Calculate the number of moles of O₂ needed to produce 16 moles of NO
We will use the previously established molar ratio.
16 mol O₂ × 4 mol NO/5 mol O₂ = 13 mol NO
The answer is 60 kpa, also did you try look it up because that is what i did but i did not copy and paste it.
hope this helped have a good day