Answer:
6H20 represents six molecules of water
Producers such as plants.
pH of solution = 13.033
<h3>Further explanation</h3>
Given
2.31 g Ba(OH)₂
250 ml water
Required
pH of solution
Solution
Barium hydroxide is fully ionized, means that Ba(OH)₂ is a strong base
So we use a strong base formula to find the pH
[OH ⁻] = b. Mb where
b = number of OH⁻
/base valence
Mb = strong base concentration
Molarity of Ba(OH)₂(MW=171.34 g/mol) :

Ba(OH)₂ ⇒ Ba²⁺ + 2OH⁻(b=valence=2)
[OH⁻]= 2 . 0.054
[OH⁻] = 0.108
pOH= - log 0.108
pOH=0.967
pOH+pH=14
pH=14-0.967
pH=13.033
Answer: 0.422 M⁻¹s⁻¹
Explanation: <u>Reaction</u> <u>Rate</u> is the speed of decomposition of the reactant(s) per unit of time.
A <u>Rate</u> <u>Law</u> relates concentration of reactants, rate reaction and rate constant:
![r=k[A]^{x}[B]^{y}](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D)
where
[A] and [B] are reactants concentration
x and y are reaction order, not related to the stoichiometric coefficients
k is rate constant
r is rate
Before calculating rate constant, first we have to determine reaction order.
In this question, the reactio order is 2. So, the rate law for it is
![-\frac{d[A]}{dt} =k[A]^{2}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dk%5BA%5D%5E%7B2%7D)
and the integrated formula is
![\frac{1}{[A]} =\frac{1}{[A]_{0}} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_%7B0%7D%7D%20%2Bkt)
in which
[A]₀ is initial concentration of reactant
Then, using initial concentration at initial time and final concentration at final time:



k = 0.422
The rate constant for the reaction is 0.422 M⁻¹.s⁻¹
Answer:
400 ml at 20⁰C
Explanation:
VP ∝ Temp => Increasing temperature increases VP or decreasing temperature decreases VP. VP is independent of the quantity of liquid in consideration. Therefore, the sample at the lower temperature would have the lowest VP; i.e., 400-ml at 20⁰C.