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.
Answer:
2.0000 g
Explanation:
It will show the number out to the 4th decimal place. Exactly 2 g will be displayed as 2.0000 g.
Answer:
<h3>The answer is option B</h3>
Explanation:
The pH of a solution can be found by using the formula
pH = - log [ H+ ]
To find the hydrogen ion concentration substitute the pH into the above formula and solve for the [ H+ ]
From the question
pH = 4.25
So we have
4.25 = - log [ H+ ]
<u>Find the antilog of both sides</u>
That's
<h3>
![[ H+ ] = {10}^{ - 4.25}](https://tex.z-dn.net/?f=%5B%20H%2B%20%5D%20%3D%20%20%7B10%7D%5E%7B%20-%204.25%7D%20)
</h3>
We have the final answer as
<h2>
![[ H+ ] = 5.6 \times {10}^{ - 5} \: M](https://tex.z-dn.net/?f=%5B%20H%2B%20%5D%20%3D%205.6%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%205%7D%20%20%5C%3A%20M)
</h2>
Hope this helps you
<span>Answer:
Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass.
râšM = constant
Therefore for two gases the ratio rates is given by:
r1 / r2 = âš(M2 / M1)
For Cl2 and F2:
r(Cl2) / r(F2) = âš{(37.9968)/(70.906)}
= 0.732 (to 3.s.f.)</span>