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Blizzard [7]
3 years ago
9

On July 8th, I will be 8.51472x10^8 seconds old. How old will I be in years? (Assume 365 days in a year.) Group of answer choice

s
Chemistry
1 answer:
KatRina [158]3 years ago
5 0

Answer:

26.98

almost 27 years

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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
Read 2 more answers
A mushroom would most likely eat
dmitriy555 [2]
Dead plants hope this helps 
5 0
3 years ago
t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol
spayn [35]

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

                        H_2(g)+I_2(g)\rightarrow 2HI(g)

Initial conc.    0.10        0.10      0.50

At eqm.        (0.10-x)  (0.10-x)   (0.50+2x)

As we are given:

K_c=5.00\times 10^2

The expression for equilibrium constant is:

K_c=\frac{[HI]^2}{[H_2][I_2]}

Now put all the given values in this expression, we get:

5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}

x = 0.0713  and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M

Thus, the concentration of HI (g) at equilibrium is, 0.643 M

8 0
3 years ago
BRANILIESTTT ASAP!! PLEASE HELP ME :)
Svet_ta [14]

Answer:

As temperature increases, the molecules and atoms move faster.

Explanation:

Molecules and atoms move according to how hot or cold the temperature is. As the temperature rises, the molecules and atoms will begin to move faster. As the temperature cools, the molecules and atoms will begin to move slower.

Hope this helps! :)

3 0
3 years ago
What kingdom would an organism be in if it was unicellular and contained no nucleus
Ksivusya [100]
Kingdom Monera, organisms in this Kingdom are prokaryotes, they don't have a nuclear envelope, thus they have a nucleoid. 

7 0
3 years ago
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