You have 4 moles of a gas in a 50 l container held at 2 atm pressure. currently the temperature is 27 ºc. r = 0.0821 l*atm/(mol*k) • what is the temperature in kelvins? k = 27 +273.15 = 300.15 • assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºc? the pressure should decrease as well, i believe • if you increase the number of moles to 6 moles, increase temperature to 400k and reduce the volume to 25 l, what will the new pressure be? i have no idea what formula to use

The formula used for determining gas pressure, volume and temperature interaction would be PV=nRT. 

•What is the temperature in Kelvins?
You already right at this part. Kelvin temperature formula from celsius should be:
K= C+273.15= 
K= 27 +273.15 = 300.15 
It is important to remember that the formula in this question is using Kelvin unit at temperature, not Celcius or Fahrenheit.

•Assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºC?
In this case, the temperature is decreased from 27C into -15C and you asked the change in the pressure.
Using PV=nRT formula, you can derive that the temperature will be directly related to pressure. If the temperature decreased, the pressure will be decreased too. 

 If you increase the number of moles to 6 moles, increase temperature to 400K and reduce the volume to 25 L, what will the new pressure be?
PV=nRT
P= nRT/V
P= 6 moles* 0.0821 L*atm/(mol*K) * 400K/25L= 7.8816 atm

Ithink it's carbonic acid

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.