Explanation:
The given data is as follows.
= 98.70 kPa = 98700 Pa,
T =
= (30 + 273) K = 303 K
height (h) = 30 mm = 0.03 m (as 1 m = 100 mm)
Density = 13.534 g/mL = 
= 13534 
The relation between pressure and atmospheric pressure is as follows.
P = 
Putting the given values into the above formula as follows.
P = 
= 
= 102683.05 Pa
= 102.68 kPa
thus, we can conclude that the pressure of the given methane gas is 102.68 kPa.
Answer: physical
Explanation: freezing and melting are physical changes.
Answer:
1. The gas law used: Dalton's law of partial pressure.
2. Pressure of nitrogen = 331 mmHg
Explanation:
From the question given above, the following data were obtained:
Total pressure (Pₜ) = 592 mmHg
Pressure of Oxygen (Pₒ) = 261 mmHg
Pressure of nitrogen (Pₙ) =?
The pressure of nitrogen in the sample can be obtained by using the Dalton's law of partial pressure. This is illustrated below:
Pₜ = Pₒ + Pₙ
592 = 261 + Pₙ
Collect like terms
592 – 261 = Pₙ
331 = Pₙ
Pₙ = 331 mmHg
Therefore, the pressure of nitrogen in the sample is 331 mmHg
Transverse wave, motion in which all points on a wave oscillate along paths at right angles to the direction of the wave's advance. Surface ripples on water, seismic S (secondary) waves, and electromagnetic (e.g., radio and light) waves are examples of transverse waves.
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M