Answer:
The estimated feed rate of logs is 14.3 logs/min.
Explanation:
The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.
That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.
Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.
To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.
The specific gravity of the wood chips is 0.494.
The average volume of a log is
The weight of one log is
To provide 3617 ton/day of wood chips, we need
The feed rate of logs is 14.3 logs/min.
5 plates is the highest amount that can be served
There’s only 5 sandwiches so 7 is automatically ruled out, there’s 14 corn cobs and 5 sandwiches only need 10 so it works out
<span>when it comes to adding or subtracting numbers, his final answer should have the same number of decimal places as the least precise value.
For example if you add 2 numbers; 10.443 + 3.5 , 10.443 has 3 decimal places and 3.5 has only one decimal place.
Therefore 3.5 is the less precise value.
So when adding these 2 values the final answer should have only one decimal place.
after adding we get 13.943 but it can have upto one decimal place. then the second decimal place is less than 5 so the answer should be rounded off to 13.9.
the answer is the same number of decimal places as the least precise value</span>
Answer:
Explanation:
Hello!
In this case, since the dissolution of copper (I) chloride is:
And its equilibrium expression is:
![Ksp=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
We can represent the molar solubility via the reaction extent as 
, however, since there is 0.050 M KCl we immediately add such amount to the chloride ion concentration since KCl is readily ionized; therefore we write:
Thus, solving for , we obtain:
By using the quadratic equation, we obtain:
Clearly, the solution is 
because no negative results are
allowed. Therefore, the molar solubility is:
Best regards!
