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ozzi
3 years ago
5

How many atoms are in 1.4 mol of phosphorus trifluoride (PF3)?

Chemistry
1 answer:
joja [24]3 years ago
5 0

Number of atoms in 1.4 mol of Phosphorus trifluoride (PF₃) : 8.428 x 10²³

<h3>Further explanation  </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance  

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

1.4 mol of Phosphorus trifluoride (PF₃), number of atoms :

\tt N=1.4\times 6.02\times 10^{23}\\\\N=8.428\times 10^{23}

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When nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic
BigorU [14]

Theoretical yield = 2.397

The product could be sodium carbonate

percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃

=  0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.0226 × 62 g = 1.401 g

mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

To learn more about percentage yield visit:

brainly.com/question/22257659

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3 years ago
The reaction A(B) = 2B(g) has an equilibrium constant of K = 0.045. What is the equilibrium constant for the reaction B(g) =1/2A
Mamont248 [21]

Answer:

The  K_c for the reaction B(g) = \frac{1}{2}A will be 4.69.

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The reverse will be 2B\leftrightharpoons A

Then,      K_c = \frac{[A]}{[B]^2}

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The equilibrium constant for B(g) = \frac{1}{2}A will be

               K_c = \sqrt{K_c}

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Therefore, K_c for the reaction B(g) = \frac{1}{2}A will be 4.69.

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