Answer:
The simplified expression for the fraction is 
Explanation:
From the given information:
O3* → O3 (1) fluorescence
O + O2 (2) decomposition
O3* + M → O3 + M (3) deactivation
The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)
The rate of decomposition is = k₂ × cO
The rate of deactivation = k₃ × cO × cM
where cM is the concentration of the inert molecule
The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:
since cM is the concentration of the inert molecule
Answer:
31.60% phosphorus
Explanation:
To find the percent by mass, you first calculate the total mass and then the mass of phosphorus, and finally divide phosphorus mass by total.
Phosphorus mass: the molar mass of P is 30.97 g/mol, and since there's 1 mole of P, we just have 30.97 g.
Total mass: we add all the molar masses of the components together.
We have 3 moles of H, so we multiply 3 by 1.008 g/mol = 3.024 g H.
We already calculated the mass of phosphorus: 30.97 g P.
We have 4 moles of O, so we multiply 4 by 16.00 g/mol = 64.00 g O.
The total is then the sum: 3.024 + 30.97 + 64.00 = 97.994 g ≈ 97.99 g
Now, to find the percentage, we take 30.97 g P and divide by 97.99:
30.97/97.99 ≈ 0.3160 ⇒ 31.60% P
Thus, the answer is 31.60% phosphorus.
Hope this helps!
236 I DONT KNOW I AM TOTALLY THAT IS NOT THE ANSWER
<span>equilibrium constant times the concentration of water. - 100% guaranteed correct</span>
