Answer:
Dolphins use a method called echolocation to detect things such as obstacles and prey in the water. If a dolphin swimming in seawater at 25°C sends a 220-dB click with a frequency of 120.0 Hz, and then detects the reflection of the click exactly one-twentieth of a second later, approximately how far away is the object?
Answer:
The unknown temperature is 304.7K
Explanation:
V1 = 100mL = 100*10^-3L
P1 = 99.10kPa = 99.10*10³Pa
V2 = 74.2mL = 74.2*10^-3L
P2 = 133.7kPa = 133.7*10³Pa
T2 = 305K
T1 = ?
From combined gas equation,
(P1 * V1) / T1 = (P2 * V2) / T2
Solving for T1,
T1 = (P1 * V1 * T2) / (P2 * V2)
T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)
T1 = 3022550 / 9920.54
T1 = 304.67K
T1 = 304.7K
Answer:
31.5mL
Explanation:
The following were obtained from the question:
C1 (concentration of stock solution) = 2M
V1 (volume of stock solution) =.?
C2 (concentration of diluted solution) = 0.630M
V2 (volume of diluted solution) = 100mL
Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
2 x V1 = 0.630 x 100
Divide both side by 2
V1 = (0.630 x 100) /2
V1 = 31.5mL
Therefore, 31.5mL of 2M solution of FeCl2 required