16(2)/74.1=.431
.431x100= 43.1%
Answer:
See explanation below
Explanation:
In this case, we have the equilibrium reaction which is:
H₂ + I₂ <------> 2HI Kp = 54
Now, we have the partial pressures of each element in equilibrium, therefore, we can use the expression of equilibrium in this case to calculate the remaining pressure:
Kp = PpHI² / PpH₂ * PpI₂
Solving for the partial pressure of iodine:
PpI₂ = PpHI² / PpH₂ * Kp
Replacing the given values, we have:
PpI₂ = (2.1)² / 0.933 * 54
PpI₂ = 4.41 / 50.382
PpI₂ = 0.088 atm
Answer:
Here you go! Hopefully this helps your question.
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Explanation:
None needed
Answer:
Empirical formula is PNCl₂
Explanation:
Percent composition means that 100 g of compound has x g of each element.
In 100 g of compound x, we have 26.73 g of P, 12.09 g of N, 61.18 g of Cl.
So, let's make some rules of three:
In 100 g of compound we have 26.73 g of P, 12.09 g of N, 61.18 g of Cl
In 579.43 g of compound we have:
(579.43 . 26.73) / 100 = 155 g of P
(579.43 . 12.09) / 100 = 70 g of N
(579.43 . 61.18) / 100 = 354 g of Cl
Let's convert the mass of the elements in moles.
155 g of P / 30.97 g/mol = 5 P
70 g / 14 g/mol = 5 N
354 g / 35.45 g/mol = 10 Cl
Answer:
pls complete your question
