Answer:
The law of reflection states that the angle of incidence and the angle of reflection are always equal.
Answer:
the pressure at c = 0.27 atm
Explanation:
Given that:
number of moles (n) = 1.0 moles
Value of gamma in the monoatomic gas (γ) = 5/3
During an isothermal expansion, the volume at b is = 2.5 times the volume at a ; this implies that:
∴ To calculate the pressure at c from a; the process is adiabatic compression; so we apply:
![\frac{P_c}{P_a}=[\frac{V_a}{V_c}]^{(2/3)](https://tex.z-dn.net/?f=%5Cfrac%7BP_c%7D%7BP_a%7D%3D%5B%5Cfrac%7BV_a%7D%7BV_c%7D%5D%5E%7B%282%2F3%29)
![\frac{P_c}{1.0 atm}=[\frac{1}{2.5}]^{(2/3)](https://tex.z-dn.net/?f=%5Cfrac%7BP_c%7D%7B1.0%20atm%7D%3D%5B%5Cfrac%7B1%7D%7B2.5%7D%5D%5E%7B%282%2F3%29)
Thus, the pressure at c = 0.27 atm
Answer:
40% of the ammonia will take 4.97x10^-5 s to react.
Explanation:
The rate is equal to:
R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]
R = k´ * [NH3]
k´ = 10200 s^-1
Because k´ is the psuedo first-order rate constant, we have the following:
b/(b-x) = 100/(100-40) ; 40% ammonia reacts
b/(b-x) = 1.67
log(b/(b-x)) = log(1.67)
log(b/(b-x)) = 0.22
the time will equal to:
t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s
The combustion of ammonia in presence of excess oxygen yields NO2 and H2O.
The molar mass of ammonia is 17.02 g/mol
Therefore, moles of ammonia in 43.9 g
= 43.9 /17.02
= 2.579 moles
From the equation the mole ratio of ammonia to nitrogen iv oxide is 4:4
The molar mass of NO2 is 46 g/mol
The number of moles of NO2 is the same as that of ammonia since they have equal ratio,
= 2.579 moles
Therefore, mass of NO2
= 2.579 moles ×46
= 118.634 g
≈ 119 g
