Answer:
D
Explanation:
I think it is d sorry if wrong hoped I helped
The grams of solid copper oxide must be used to prepare a solution of 0.125m concentration is 5.26 g.
According to the definition of molar concentration of a substance dissolved in a solution is defined as the ratio of the number of moles to the volume of the solution.
C = n/V
The number of moles is equal to the given mass divided by the molar mass.
n = m/Mm = n ×m
Given,
The volume of the solution of copper oxide = 0.53
Molar mass of copper oxide = 79.5
Concentration of copper oxide = 0.125
CuO = cVM
= 0.125 × 0.53 × 79.5
= 5.26g
Thus, we concluded that the grams of solid copper oxide must be used to prepare a solution of 0.125m concentration is 5.26 g.
DISCLAIMER: The above question is wrong. The correct question is
Question: In lab you have to prepare 530. 00 ml solution of 0. 125 m copper (ii) oxide. How many grams of solid copper oxide must be used to prepare a solution of this concentration?
learn more about molar concentration :
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I believe it is less dense.. that is why ice floats on water
Answer:
Therefore it takes 8.0 mins for it to decrease to 0.085 M
Explanation:
First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.
A→ product
Let the concentration of A = [A]
![\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]](https://tex.z-dn.net/?f=%5Ctextrm%7Brate%20of%20reaction%7D%3D-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3Dk%5BA%5D)
![k=\frac{2.303}{t} log\frac{[A_0]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%20log%5Cfrac%7B%5BA_0%5D%7D%7B%5BA%5D%7D)
[A₀] = initial concentration
[A]= final concentration
t= time
k= rate constant
Half life: Half life is time to reduce the concentration of reactant of its half.
Here 
To find the time takes for it to decrease to 0.085 we use the below equation
![\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}](https://tex.z-dn.net/?f=%5CRightarrow%20t%3D%5Cfrac%7B2.303%7D%7Bk%7D%20log%5Cfrac%7B%5BA_0%5D%7D%7B%5BA%5D%7D)
Here , 
, [A₀] = 0.13 m and [ A] = 0.085 M
Therefore it takes 8.0 mins for it to decrease to 0.085 M
