Heat flows from a region of greater potential to lower potential?
Answer:
Rate of formation of SO₃
= 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
⇒
-----------------------------(1)
Given that the rate of disappearance of oxygen =
= 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)
⇒
= 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
Yes, it is most likely for science.
Answer:
The answers are as given in the attachment
Explanation:
The application of the de brogile equation was used and appropriate substitution were made as shown in the attachment
Lead reacts very slowly with dilute hydrochloric acid to give lead chloride<span> and </span>hydrogen<span> gas. </span>
<span>lead + hydrochloric acid —> lead chloride + hydrogen
Pb(s) + 2HCl(aq) —> PbCl2(aq) + H2(g)</span>
<span>Lead reacts very slowly with dilute sulphuric acid to give </span>lead sulphate<span> and </span>hydrogen<span> gas. </span>
<span>lead + sulphuric acid —> lead sulphate + hydrogen
Pb(s) + H2SO4 (aq) —> PbSO4(aq) + H2(g)</span>
<span>Lead reacts very slowly with dilute nitric acid to give </span>lead nitrate<span> and </span>hydrogen<span> gas. </span>
<span>lead + nitric acid —> lead nitrate + hydrogen
Fe(s) + 2HNO3(aq) —> Fe(NO3)2(aq) + H2(g)</span><span> </span>