Answer:
P_2 =0.51 atm
Explanation:
Given that:
Volume (V1) = 2.50 L
Temperature (T1) = 298 K
Volume (V2) = 4.50 L
at standard temperature and pressure;
Pressure (P1) = 1 atm
Temperature (T2) = 273 K
Pressure P2 = ??
Using combined gas law:
According to the reversible reaction equation:
2Hi(g) ↔ H2(g) + i2(g)
and when Keq is the concentration of the products / the concentration of the reactants.
Keq = [H2][i2]/[Hi]^2
when we have Keq = 1.67 x 10^-2
[H2] = 2.44 x 10^-3
[i2] = 7.18 x 10^-5
so, by substitution:
1.67 x 10^-2 = (2.44 x 10^-3)*(7.18x10^-5)/[Hi]^2
∴[Hi] = 0.0033 M
₁₇Cl 1s²2s²2p⁶3s²3p⁵, 1 unpaired electron in 3pz orbital.
3-pentanone to form 3-pentanol
Here 3-pentanone reacts with H2/Pt to given 3-pentanol
So here with one step we can convert given ketone to alcohol
The reaction will be