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.
Explanation:
Half life is simply the amount of time it takes for half of a substance to decompose.
Options;
- Carbon-14 has a half-life of 5,730 years. A 30 gram sample will be 10 grams after 5,730 years. This is incorrect. After 5730 years, 15g of the sample ought to remain.
- Nickel-59 has a half-life of 76,000 years. A sample would go through 3 half-lives in 228,000 years. This is correct. 3 * 76000 = 228,000
- Hafnium-182 has a half-life of 9 million years. A 38 gram sample would be 4.75 grams in 27 million years. This is incorrect. Mass after 3 half lives (27/9) = 9.5 (38 / 2 / 2)
- Iron-60 has a half-life of 1.5 million years. In 6 million years a 40 gram sample would be reduced to 10 grams. This is incorrect. Mass after 4 half lives (6 / 1.5) = 2.5 gram (40 / 2 / 2 /2 / 2)
- Lead-202 has a half-life of 52,500 years. The original sample must have been 120 grams if you have a 60 gram sample after 105,000 years. This is incorrect. Original sampe = 240 gram. So after 2 half lives (105,000/52500), mass left = 60 (240 / 2 /2)
When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.
0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3
Since the molar mass of NaNO3 is 85 g/mol, the mass is
0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed
Answer:
Option C = 4.25 g
Explanation:
Ounce and grams are unit of mass. Ounce is larger unit while gram is smaller unit. The one ounce is consist of 28.35 g or we can say that one ounce is equal to 28.35 g. In order to convert the given ounce value into grams the value is multiply with 28.35 g.
Given data:
Mass = 0.15 ounce
Mass in gram = ?
Solution:
One ounce is equal to 28.35 g, so
0.15 × 28.35 = 4.25 g
a. 0.51 moles of Mg
b. 0.25 moles of F
Explanation:
a. To find the number of moles knowing the mass we use the following formula:
number of moles = mass / molecular weight
number of moles of Mg = 12.15 / 24 = 0.51 moles
b. To find the number of moles knowing the number of atoms we use Avogadro's number to illustrate the following reasoning:
if in 1 mole of F there are 6.022 × 10²³ atoms of F
then in X moles of F there are 1.5 × 10²³ atoms of F
X = (1 × 1.5 × 10²³) / 6.022 × 10²³
X = 0.25 moles of F
Learn more about:
Avogadro's number
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