Using ideal gas equation,
P\times V=n\times R\times T
Here,
P denotes pressure
V denotes volume
n denotes number of moles of gas
R denotes gas constant
T denotes temperature
The values at STP will be:
P=100 kPa
T=293 K
R=8.314472 L kPa K⁻¹ mol⁻¹
Number of moles of gas=3.43 mole
Putting all the values in the above equation,
V=83.55 L
So the volume will be 83.55 L.
83.55 L of radon gas would be in 3.43 moles at room temperature and pressure (293 K and 100 kPa).
I think it would be C) The surrounding soil can become very fertile
2.10 x 10^-10 M. Ans
pH + pOH = 14
Where, pOH is the power of hydroxide ion concentration and pH is the power of concetration of the H+ ion.
Now, pOH = 14 - 4.32
= 9.68
Now, the concentration of [H+] is 10-7 M, then pH is 7 and for [OH-] = 10-7 M, the pOH is also 7.
Now, pOH = -log[OH-]
[OH-] = 10^- pOH
= 10^-9.68
= 2.10 x 10^-10 M
2 C2H2 + 5 02 > 4 CO2 + 2 H2O
Products - Reactants ( all units are kJ/mo1):
(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1
-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1
answer: -1255.6 kJ/mo1
