Suppose that the root‑mean‑square velocity vrms of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol ) in a flame is found to be 1310 m/s. What temperature T does this represent? The Boltzmann constant is k=1.38×10−23 J/K and Avogadro's number is A=6.022×1023 mol−1.

Answer : The value of temperature is, 3028.7 K

Explanation :

The formula used for root mean square speed is:

where,

= root mean square speed  = 1310 m/s

k = Boltzmann’s constant =

T = temperature = ?

M = molar mass = 44.0 g/mole  = 0.044 kg/mol

= Avogadro’s number =

Now put all the given values in the above root mean square speed formula, we get:

Therefore, the value of temperature is, 3028.7 K

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

b. examine and describe cells