Answer:
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
Explanation:
Step 1: Data given
A sample of sulfur hexafluoride gas occupies a volume of 5.10 L
Temperature = 198 °C = 471 K
The volume will be reduced to 2.50 L
Step 2 Calculate the new temperature via Charles' law
V1/T2 = V2/T2
⇒with V1 = the initial volume of sulfur hexafluoride gas = 5.10 L
⇒with T1 = the initial temperature of sulfur hexafluoride gas = 471 K
⇒with V2 = the reduced volume of the gas = 2.50 L
⇒with T2 = the new temperature = TO BE DETERMINED
5.10 L / 471 K = 2.50 L / T2
T2 = 2.50 L / (5.10 L / 471 K)
T2 = 230.9 K = -42.1
When the volume will be reduced to 2.50 L, the temperature will be reduced to a temperature of 230.9K
Compression and rarefaction. However instead of crests and troughs, longitudinal waves have compressions and rarefactions. A compression is a region in a longitudinal wave where the particles are closest together. A rarefaction is a region in a longitudinal wave where the particles are furthest apart.
Answer:
See explanation below.
Explanation:
In the older procedure, the solution is cooled so quickly that the recrystallization can be less effective. By allowing the solution to cool slowly, the maximum amount of crystal is formed, and also the impurities are trapped, so the process is more controlled than by the older one.
Answer is: pOH = 3,29.
Kb(NH₃) = 1,5·10⁻⁶.
c(NH₃) = 0,175M.
pOH = ?
Chemical reaction: NH₃ + H₂O ⇄ NH₄⁺ + OH⁻.
Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).
c(NH₄⁺) = c(OH⁻) = x.
x² = Kb · c(NH₃)
x² = 1,5·10⁻⁶ · 0,175 = 2,625 ·10⁻⁷.
x = c(OH⁻) = √2,625 ·10⁻⁷ = 5,12 · 10⁻⁴.
pOH = -log(c(OH⁻)) = 3,29.
