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?
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<u>Answer:</u> The concentration of reactant after the given time is 0.0205 M
<u>Explanation:</u>
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 11.0 min = 660 s (Conversion factor: 1 min = 60 s)
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = 0.400 M
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![4.50\times 10^{-3}s^{-1}=\frac{2.303}{660s}\log\frac{0.400}{[A]}](https://tex.z-dn.net/?f=4.50%5Ctimes%2010%5E%7B-3%7Ds%5E%7B-1%7D%3D%5Cfrac%7B2.303%7D%7B660s%7D%5Clog%5Cfrac%7B0.400%7D%7B%5BA%5D%7D)
![[A]=0.0205M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0205M)
Hence, the concentration of reactant after the given time is 0.0205 M
Some fossils provide clues to the environment and climate of the time the organism lived.
T = final temperature of the block
T₀ = initial temperature of the block = 23.4 °C
Q = energy lost from the wooden block = - 759 J
c = specific heat capacity of wood = 1.716 J/(g °C)
m = mass of the wooden block = 27.2 g
Heat lost from the block is given as
Q = m c (T - T₀)
inserting the values
- 759 = (27.2) (1.716) (T - 23.4)
T = 7.1 °C
