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scZoUnD [109]
2 years ago
10

What happens to the air temperature of a decending mass of air

Chemistry
1 answer:
Aleksandr [31]2 years ago
4 0
When air descends it compresses. When the air compresses the molecules in the air knock into each other more often, increasing the tempertature.

Hope this helps
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Jamie had $37 in his bank account on Sunday.
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In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o
Kitty [74]

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.

V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1}  )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w

The specific gravity of the wood chips is 0.494.

The average volume of a log is

V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2}  \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}

The weight of one log is

M=\rho*V=0.494*\rho_w*12.57  ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm

To provide 3617 ton/day of wood chips, we need

n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min

The feed rate of logs is 14.3 logs/min.

7 0
3 years ago
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7. How many moles of argon are there in 20.0 L, at 25 degrees Celsius and 96.8 kPa?
suter [353]
<h3>Answer:</h3>

              0.8133 mol

<h3>Solution:</h3>

Data Given:

                 Moles  =  n  =  ??

                 Temperature  =  T  =  25 °C + 273.15  =  298.15 K

                  Pressure  =  P  =  96.8 kPa  =  0.955 atm

                  Volume  =  V  =  20.0 L

Formula Used:

Let's assume that the Argon gas is acting as an Ideal gas, then according to Ideal Gas Equation,

                  P V  =  n R T

where;  R  =  Universal Gas Constant  =  0.082057 atm.L.mol⁻¹.K⁻¹

Solving Equation for n,

                  n  =  P V / R T

Putting Values,

                  n  =  (0.955 atm × 20.0 L) ÷ (0.082057 atm.L.mol⁻¹.K⁻¹ × 298.15 K)

                 n  =  0.8133 mol

4 0
3 years ago
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