Answer:
From the image the answer is 24.
The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

where,
- i: van 't Hoff factor (1 for non-electrolytes)
- Kf: cryoscopic constant
- m: molality
The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.
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Answer:
They all have: nitrogen :780,900 755,100
oxygen: 209,500 231,500
argon 9,300 12,800
carbon dioxide
386 591
Answer:
El termopar B presenta un mayor grado de dispersión y también es más preciso. ... (c) La estimación para T = 175 ° C es probablemente la más cercana al valor real, porque el ... (cm3). Flujo de masa. Velocidad. (kg / min). Diferencia. Duplicar. (Di). Yo y yo. 2. 1 ... atm de gas. 2. 2. 2 f. 3. 2 f f. 30 14,7 lb 20 pulg. 4 14,7 lb 24 pulg 392 lb 7,00 10 lb pulg.
Answer:
126 moles
Explanation:
2S +3 o2=2so3
So if 2 moles of so3 required 3 moles of oxygen
. So 84 moles of so3 will require 84*3/2=126 moles of oxygen