I will have to say that this statement is true
<span>The pressure inside a coke bottle is really high. This helps keep the soda carbonated. That is, the additional pressure at the surface of the liquid inside the bottle forces the bubbles to stay dissolved within the soda. </span><span>When the coke is opened, there is suddenly a great pressure differential. The initial loud hiss that is heard is this pressure differential equalizing itself. All of the additional pressure found within the bottle pushes gas out of the bottle until the pressure inside the bottle is the same as the pressure outside the bottle. </span><span>However, once this occurs, the pressure inside the bottle is much lower and the gas bubbles that had previously been dissolved into the soda have nothing holding them in the liquid anymore so they start rising out of the liquid. As they reach the surface, they pop and force small explosions of soda. These explosions are the source of the popping and hissing that continues while the soda is opened to the outside air. Of course, after a while, the soda will become "flat" when the only gas left dissolved in the liquid will be the gas that is held back by the relatively weak atmospheric pressure.</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
The experimental mole ratio of silver chloride to barium chloride is calculated as below
fin the mole of each compound
mole= mass/molar mass
moles of AgCl = 14.5g/142.5 g/mol = 0.102 moles of AgCl
moles of BaCl2 = 10.2 g/208 g/mol = 0.049 moles of BaCl2
find the mole ratio by dividing each mole with the smallest mole(0.049)
AgCl= 0.102/0.049 =2
BaCl2 = 0.049/0.049 =1
therefore the mole ratio AgCl to BaCl2 is 2 :1
Answer:
a mixture
Explanation:
in a mixture the substances havent chemically combined so it is easy to just separate them through physical means
