A sample of O2 with an initial temperature of 50.0 oC and a volume of 105 L is cooled to -25 oC. The new pressure is 105.4 kPa and the new volume is 55.0 L. What was the initial pressure of the sample? Show your work, using the G. U.E. S.S. method.

71.92 kPa

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

T1 = 50°C = 50 + 273 = 323K

V1 = 105L

T2 = -25°C = -25 + 273 = 248K

P2 = 105.4 kPa

P1 = ?

V2 = 55.0 L

Using P1V1/T1 = P2V2/T2

P1 × 105/323 = 105.4 × 55/248

105P1/323 = 5797/248

0.325P1 = 23.375

P1 = 23.375 ÷ 0.325

P1 = 71.92 kPa

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l