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2 years ago

1. A gas has a pressure of 799.0 mm Hg at 50.0 degrees C. What is the temperature at standard Pressure?

Chemistry

2 answers:

larisa [96]2 years ago

7 0

Answer:

A = 674.33mmHg

B = 0.385atm

Explanation:

Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn

Data:

P1 = 799mmHg

T1 = 50°C = (50 + 273.15) = 323.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2

P2 = (P1 × T2) / T1

P2 = (799 × 273.15) / 323.15

P2 = 674.37mmHg

The final pressure is 674.37mmHg

P1 = 0.470atm

T1 = 60°C = (60 + 273.15)K = 333.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2,

P2 = (P1 × T2) / T1

P2 = (0.470 × 273.15) / 333.15

P2 = 0.385atm

The final pressure is 0.385atm

mestny [16]2 years ago

3 0

Answer:

1. $T_2=307.23K=34.23\°C$

2. $P_2=0.385atm$

Explanation:

Hello,

1. In this case, the standard pressure is 1 atm or 760 mmHg, therefore, using the Gay-Lussac's law we can compute the corresponding temperature considering the initial 50.0 °C in absolute Kelvins:

$\frac{T_1}{P_1}= \frac{T_2}{P_2}\\\\T_2=\frac{T_1P_2}{P_1}=\frac{(50+273.15)K*760mmHg}{799mmHg} \\\\T_2=307.23K=34.23\°C$

2. As well as in the previous case, we now compute the pressure at 273 K which is the standard temperature:

$\frac{P_1}{T_1}= \frac{P_2}{T_2}\\\\P_2=\frac{P_1T_2}{T_1}=\frac{273K*0.47atm}{(60+273)K} \\\\P_2=0.385atm$

Best regards.

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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.

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2 years ago

Read 2 more answers

1. Which class of compounds contains at least one element from Group 17 of the Periodic Table ? A) aldehyde B ) amine C) ester D

blondinia [14]

Halide

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2 years ago

an element x has two naturally occurring isotopes: X-79 (abundance+50.69%, mass +78.918amu) and X-81 (abundance+49.31% mass+80.9

Lana71 [14]

The answer would be B. Brimone. I had the same question before, but let me know if it is not right. Cause certain schools have the same questions but different answers for them.

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2 years ago

Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bis

almond37 [142]

Answer: a) $BaCl_2$

b) $BiO_3H_3$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

a) Mass of Ba= 66.06 g

Mass of Cl = 34.0 g

Step 1 : convert given masses into moles.

Moles of Ba = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{66.06g}{137g/mole}=0.48moles$

Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ba = $\frac{0.48}{0.48}=1$

For O = $\frac{0.96}{0.48}=2$

The ratio of Ba: Cl= 1:2

Hence the empirical formula is $BaCl_2$

b) Mass of Bi= 80.38 g

Mass of O= 18.46 g

Mass of H = 1.16 g

Step 1 : convert given masses into moles.

Moles of Bi = $\frac{\text{ given mass of ba}}{\text{ molar mass of Ba}}= \frac{80.38g}{209g/mole}=0.38moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{18.46g}{16g/mole}=1.15moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.16g}{1g/mole}=1.16moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Bi= $\frac{0.38}{0.38}=1$

For O = $\frac{1.15}{0.38}=3$

For H= $\frac{1.16}{0.38}=3$

The ratio of Bi: O: H= 1:3: 3

Hence the empirical formula is $BiO_3H_3$

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2 years ago

Write an equation for the formation of bf3(g) from its elements in their standard states.

blagie [28]

In nature, boron is monoatomic. Therefore, its formula is B.
On the other hand, fluorine is diatomic. Therefore, its formula is F2

Now, the basic unbalanced equation is:
B + F2 .........> BF3

Now, we need to balance this equation. As you can see, we have two fluorine moles entering the reaction and 3 formed in the products.
Balancing the equation, we will reach the following balanced reaction:
2B + 3F2 .......> 2BF3

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2 years ago