Answer:
The new volume of a gas at 750 mmhg and with a volume of 2. 00 l when allowed to change its volume at constant temperature until the pressure is 600 mmhg is 2.5 Liters.
Explanation:
Boyle's law states that the pressure of a given amount of gas is inversely proportional to it's volume at constant temperature. It is written as;
P ∝ V
P V = K
P1 V1 = P2 V2
Parameters :
P1 = Initial pressure of the gas = 750 mmHg
V1 = Initial pressure of the gas = 2. 00 Liters
P2 = Final pressure of the gas = 600 mmHg
V2 = Fimal volume of the gas = ? Liters
Calculations :
V2 = P1 V1 ÷ P2
V2= 750 × 2. 00 ÷ 600
V2 = 1500 ÷ 600
V2 = 2.5 Liters.
Therefore, the new volume of the gas is 2. 5 Liters.
<span>Yes. If you put solutions containing the same concentration table salt, glucose and starch in different bags and place the bags in water (i.e., 0M) solution, then the bag with salt will contain more water after 15 minutes than the bag with glucose, which will contain more than the bag with starch.</span>
Explanation:
Reaction equation is as follows.

Here, 1 mole of
produces 2 moles of cations.
![[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58](https://tex.z-dn.net/?f=%5BNa%5E%7B%2B%7D%5D%20%3D%202%5BNa_%7B2%7DSO_%7B3%7D%5D%20%3D%202%20%5Ctimes%200.58)
= 1.16 M
= 0.58 M
The sulphite anion will act as a base and react with
to form
and
.
As, 
= 
=
According to the ICE table for the given reaction,

Initial: 0.58 0 0
Change: -x +x +x
Equilibrium: 0.58 - x x x
So,
![K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B%5BHSO%5E%7B-%7D_%7B3%7D%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BSO%5E%7B2-%7D_%7B3%7D%5D%7D)


x = 0.0003 M
So, x =
= 0.0003 M
= 0.58 - 0.0003
= 0.579 M
Now, we will use
= 0.0003 M
The reaction will be as follows.

Initial: 0.0003
Equilibrium: 0.0003 - x x x


= 
= 
Therefore, 
As, x <<<< 0.0003. So, we can neglect x.
Therefore, 
= 
x = 
x =
= 
![[H^{+}] = \frac{10^{-14}}{[OH^{-}]}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B%5BOH%5E%7B-%7D%5D%7D)
= 
=
M
Thus, we can conclude that the concentration of spectator ion is
M.
FALSE
There are other limiting factors like lack of space, diseases, and com petition.