Answer:
6 stages
Explanation:
Given:
Copper ore contains:
4.3 wt% water
10.3 wt% CuSO₄
85.3 wt% rock
Feed rate = 11.7 tons per hour
solution produced is to be 10 wt% copper sulfate
90 wt% water
Now,
The CuSO₄ supplied through the feed = 10.3 wt% CuSO₄ × Feed rate
= 0.103 × 11.7
= 1.2051 tons per hour
also,
At each stage 2 tons of solution is collected
therefore,
The CuSO₄ collected in each stage
= weight of solution × 10 wt% copper sulfate
= 2 tons per hour × 0.1
= 0.2 tons per hour
also,
98% of the CuSO₄ is to be recovered
thus,
the amount of CuSO₄ to be recovered
= 0.98 × CuSO₄ supplied through the feed
= 0.98 × 1.2051 tons per hour
= 1.180998
Therefore,
The number of stages required = 
= 
= 5.90499 ≈ 6 stages
False, carbon is an allotrope that can be in the forms of graphite, diamond etc
This is an acid-base reaction where HF is the acid and H2O is the base (it's amphoteric and can be an acid or a base). The products would then H3O+ (the conjugate acid) and F- (the conjugate base). Now, we can simply construct a reaction using the found products and reactants. This acid-base reaction would be HF + H2O <--> H3O+ + F-.
Hope this helps!
<u>Answer:</u> The amount remained after 151 seconds are 0.041 moles
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
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 = 151 sec
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = 0.085 moles
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}](https://tex.z-dn.net/?f=4.82%5Ctimes%2010%5E%7B-3%7D%3D%5Cfrac%7B2.303%7D%7B151%7D%5Clog%5Cfrac%7B0.085%7D%7B%5BA%5D%7D)
![[A]=0.041moles](https://tex.z-dn.net/?f=%5BA%5D%3D0.041moles)
Hence, the amount remained after 151 seconds are 0.041 moles
