SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.
First, we will calculate the partial pressure of SF₆ using the following expression.

where,
- pSF₆: partial pressure of SF₆
- P: total pressure of air (we will assume it is 1 atm)
- ppb: concentration of SF₆ in parts per billion

Then, we will convert 1.0 cm³ to L using the following conversion factors:
- 1 cm³ = 1 mL
- 1 L = 1000 mL

Next, we will convert 46 °C to Kelvin using the following expression.

Afterward, we calculate the moles (n) of sulfur hexafluoride using the ideal gas equation.

Finally, we will convert 3.8 × 10⁻¹⁴ mol to molecules using Avogadro's number.

SF₆ in the air at a concentration of 1.0 ppb, exerts a partial pressure of 1.0 × 10⁻⁹ atm. At this concentration, 2.3 × 10¹⁰ molecules of SF₆ are contained in 1.0 cm³ of air at 46 °C.
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Answer:
17.6% en masa será la nueva concentración de la solución
Explanation:
Una disolución al 30% en masa contiene 30g de soluto en 100g de solución (Solvente + soluto). Así, la masa de solvente (Agua) es:
100g - 30g = 70g
Dado que la densidad del agua es de 1g/mL, 70g de agua ocupan 70mL.
Si el volumen se duplica, habrán 140mL de agua = 140g de solvente.
Así, la masa total de la solución será 30g + 140g = 170g y el porcentaje en masa será:
30g soluto / 170g solución × 100 =
<h3>17.6% en masa será la nueva concentración de la solución</h3>
Answer:answer in explanation try to understand dear
Explanation:Air pollutants have a negative impact on plant growth, primarily through interfering with resource accumulation. Once leaves are in close contact with the atmosphere, many air pollutants, such as O3 and NOx, affect the metabolic function of the leaves and interfere with net carbon fixation by the plant canopy.