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At room temperature, PF3 (phosphorus trifluoride) is a colorless gas that has a pungent smell. It is highly toxic and can cause severe irritation to the eyes, skin, and respiratory system if inhaled. PF3 is commonly used in the semiconductor industry and as a reagent in organic chemistry reactions.

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1y ago

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How many molecules of PF3 are in a gas cylinder containing 150.0 grams of PF3?

To find the number of molecules of PF3, we need to convert 150.0 grams of PF3 to moles, then use Avogadro's number (6.022 x 10^23 molecules/mol) to find the number of molecules. The molar mass of PF3 is approximately 87.97 g/mol. By dividing 150.0 g by the molar mass, we get the number of moles, which can be converted to molecules.


Which molecule has bonds that are more polar between PF3 and PCl3?

PCl3 has more polar bonds than PF3 because chlorine is more electronegative than fluorine, resulting in greater electron density towards chlorine in PCl3.


What mass of F2 is needed to produce 120g of PF3 if the reaction is 78.1?

To determine the mass of F2 needed to produce 120 g of PF3, we first need to calculate the molar mass of PF3, which is approximately 88 g/mol (phosphorus = 31 g/mol, fluorine = 19 g/mol × 3). The balanced chemical reaction for producing PF3 from P and F2 is: P + 3F2 → PF3. Since the molar ratio of PF3 to F2 is 1:3, we need 3 moles of F2 for every mole of PF3 produced. Therefore, for 120 g of PF3, which is about 1.36 moles (120 g / 88 g/mol), we would require approximately 4.08 moles of F2, or about 164 g (4.08 moles × 38 g/mol for F2).


What the equation for phosphorus triflourine?

The chemical formula is PF3. The reaction with phsphorus is 2P + 3F2 -> 2PF3


What mass of F2 is needed to produce 120g of PF3 if the reaction is 78.1 yield?

To find the mass of F2 needed to produce 120 g of PF3 at a yield of 78.1%, we first calculate the theoretical mass of PF3 that would require 100% yield. The molar mass of PF3 is approximately 88 g/mol. Therefore, the theoretical mass of PF3 produced from 120 g at 78.1% yield is ( 120 , \text{g} / 0.781 = 153.5 , \text{g} ). Assuming the reaction is balanced as ( P + \frac{3}{2} F_2 \rightarrow PF_3 ), we can then determine that the stoichiometry requires 1 mole of F2 for every mole of PF3. Hence, the mass of F2 needed is calculated using the molar mass of F2 (approximately 38 g/mol), resulting in approximately 60 g of F2 needed to produce the required PF3.