Answer:
0.73L
Explanation:
The following data were obtained from the question :
V1 = 0.65 L
P1 = 3.4 atm
T1 = 19°C = 19 + 273 = 292K
V2 =?
P2 = 3.2 atm
T2 = 36°C = 36 + 273 = 309K
The bubble's volume near the top can be obtain as follows:
P1V1 /T1 = P2V2 /T2
3.4 x 0.65/292 = 3.2 x V2 /309
Cross multiply to express in linear form as shown below:
292 x 3.2 x V2 = 3.4 x 0.65 x 309
Divide both side by 292 x 3.2
V2 = (3.4 x 0.65 x 309) /(292 x 3.2)
V2 = 0.73L
Therefore, the bubble's volume near the top is 0.73L
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:
while cooking or washing something lol
They are more likely to be covalent as covalent compounds usually tend to be radical on the color spectrum with deep purples and blacks and even earwax colored.