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SCORPION-xisa [38]
3 years ago
6

Imagine that you have a 7.00 L gas tank and a 3.00 L gas tank. You need to fill one tank with oxygen and the other with acetylen

e to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 145 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
Chemistry
1 answer:
KatRina [158]3 years ago
3 0

Answer:

The tank filled with acetylene should have a pressure of 135.3 atm

Explanation:

<u>Step 1</u>: Data given

Tank 1 = 7.00 L → filled with oxygen to a pressure of 145 atm

Tank 2 = 3.00 L → filled with acetylene

<u>Step 2:</u> The balanced equation

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

<u> Step 2:</u> Calculate moles of oxygen

p*V = n*R*T

⇒ with p = the pressure = 145 atm

⇒ with V = the volume of the gas = 7.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ T = the temperature = unknown, so we will just use T

n = (p*V)/(R*T)

n( 145*7.00)/(0.08206*T)

n = 12369 T

<u>Step 3:</u> Calculate moles of acetylene

For 2 moles of acetylene we need 5 moles of oxygen to produce 4 moles of CO2 and 2 moles of H2O

For 12369* T moles of oxygen, we have 4947.6*T moles of acetylene

<u>Step 4</u>: Calculate pressure of acetylene

p = (nRT)/V

p = (4947.6* T*0.08206) / 3.00 L

p = 135.3 atm

The tank filled with acetylene should have a pressure of 135.3 atm

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Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

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