![\tt -\dfrac{1}{2}\dfrac{d[N_2O]}{dt}=\dfrac{1}{2}\dfrac{d[N_2]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Ctt%20-%5Cdfrac%7B1%7D%7B2%7D%5Cdfrac%7Bd%5BN_2O%5D%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B2%7D%5Cdfrac%7Bd%5BN_2%5D%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B1%7D%5Cdfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
<h3>Further explanation</h3>
Reaction
2N2O(g) — 2N2(g) + O2(g)
Required
relative rate
Solution
The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.
so the relative rates for the reaction above are :
Answer:
Change in volume on changing temperature from 33
to 5 is 5.49 mL
Explanation:
Initial volume of gas = V = 60 mL
Assuming final volume of gas to be V' mL
Initial temperature = T = 33 = 306 K
Final temperature = T' = 5 = 278 K
The relationship between volume and temperature of gas at constant pressure is shown below
Change in volume on changing temperature = 5.49 mL
Answer:
0.00335 moles
Explanation:
From the question, Using
PV = nRT................... Equation 1
Where P = pressure, V = Volume, n = number of moles of argon gas, R = Molar gas constant, T = Temperature.
make n the subject of the equation
n = PV/RT............... Equation 2
Given: P = 1 atm (standard pressure), T = 273 K (standard temperature), V = 75 mL = 0.075 dm³
Constant: R = 0.082 atm·dm³/K·mol
Substitute into equation 2
n = (1×0.075)/(273×0.082)
n = 0.075/22.386
n = 0.00335 moles
Answer: 8 moles
Explanation:
Nc2H6= 4 mol
2C2H6 + 7O2 → 4CO2+6H2O
CO2=4/2⋅4
NCO2= 8 moles
( I write this on paper so the letters and format might be confusing) sorry!!
