Hard water<span>... is </span>water<span> that contains an appreciable quantity of dissolved minerals (like calcium and magnesium). </span>Soft water<span>... is treated </span>water<span> in which the only ion is sodium. As rainwater falls, it is naturally </span>soft<span>. </span>
Answer:
Rate of formation of SO₃ ![[\frac{d[SO_{3}] }{dt}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%5D)
= 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒ ![-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B2%7D%20%5D%7D%7Bdt%7D%20%3D%20-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B3%7D%20%5D%7D%7Bdt%7D)
-----------------------------(1)
Given that the rate of disappearance of oxygen = ![-\frac{d[O_{2} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D)
= 3.64 x 10⁻³ M/s
So the rate of formation of SO₃ = ?
from equation (1) we can write
![\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%20%3D%202%20%5B-%5Cfrac%7Bd%5BO_%7B2%7D%5D%20%7D%7Bdt%7D%20%5D)
⇒ ![\frac{d[SO_{3}] }{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D)
= 2 x 3.64 x 10⁻³ M/s
⇒ = 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃ = 7.28 x 10⁻³ M/s
Answer:Speak English
Explanation: No speak Spanish
The equation is:
Ca(OH)₂(s) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂<span>O(l)
</span>
n=mass in g/M.M
15 g Ca(OH)₂ is n=15 g/ 74.1 g/mol=0.2024 mol of Ca(OH)₂
no. of mol of HCl:
n=0.5 mol/L*0.075L=0.0375 mol
This could react with 0.0375/2= 0.01875 mol of Ca(OH)₂ We have a lot more than that.
Therefore, HCl is the limiting reagent and determines how much CaCl₂ forms.
Based on the balanced reaction, 2 moles of HCl gives 1 mole of CaCl₂
no. of mol of CaCl₂= 0.0375/2= 0.01875 mol
mass in g=n*MM= 0.01875*111= 2.08 g
