Answer: 65.7 grams
Explanation:
1) ratio
Since 1 mole of CaF2 contains 1 mol of F2, the ratio is:
1 mol CaF2 : 1 mol F2
2) So, to produce 8.41 * 10^ -1` mol of F2 you need the same number of moles of CaF2.
3) use the formula:
mass in grams = molar mass * number of moles
molar mass of CaF2 = 40.1 g/mol + 2 * 19.0 g/mol = 78.1 g/mol
mass in grams = 78.1 g/mol * 8.41 * 10^ -1 mol = 65.7 grams
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.
F = ma = (kg)(m/s2) = kg ´ m/s2 N
hope this helps :D
Answer:
<h3>
moles of carbon dioxide=13.95mol</h3>
Explanation:
First wrie down the balance chemical reaction:

Combustion reaction: The reacion in which hydrocarbon is burnt in the presence of oxygen gas and it releases heat and this reaction exothermic because heat of cumbustion is negative.
eg. burning of methane
By using unitry method,
From the above balanced reaction it is clearly that,
from 1 mole of propane 3 moles of carbon dioxide is formed
there fore,
from 4.65 mole of propane
moles of carbon dioxide will form
moles of carbon dioxide=13.95mol
Molarity is expressed as
the number of moles of solute per volume of the solution. For example, we are
given a solution of 2M NaOH this describes a solution that has 2 moles of NaOH
per 1 L volume of the solution. We calculate as follows:
0.115 M = n mol KBr / .55 L solution
n = 0.06325 mol KBr
mass = 0.06325 mol KBr (119 g / mol) = 7.53 g KBr