Asample of diborane gas a substance that bursts into flame when exposed to air, has a pressure of 345 tore at a temperature of -15c and a volume of 3.48 l. if conditions are changed so that the temperature is 36c and the pressure is 468 tore, what will be the volume of the sample?

3.072 L

We are given;

But, K = °C + 273

          = -15°C + 273

          = 258 K

                                   = 36 °C + 273

                                   = 309 K

We are required to calculate the new volume, V₂

P₁V₁/T₁ = P₂V₂/T₂

V₂ =(P₁V₁T₂) ÷ (T₁P₂)

   = (345 torr × 3.48 L × 309 K) ÷ (258 K × 468 torr)

   = 3.072 L

Therefore, the new volume of the sample is 3.072 L

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.