Assume the atomic mass of element X is 22.99 amu. A 17.15−g sample of X combines with 14.17 g of another element Y to form a com
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Answer:
<em><u>= - 0.38 eV</u></em>
Explanation:
Using Bohr's equation for the energy of an electron in the nth orbital,
E = -13.6
Where E = energy level in electron volt (eV)
Z = atomic number of atom
n = principal state
Given that n = 6
⇒ E = -13.6 ×
<em><u>= - 0.38 eV</u></em>
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<em>Hope this was helpful.</em>
<em><u></u></em>
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.