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10 months ago

.400 moles of CO2 gas are confined in a 5.00-liter container at 25 °C. Calculate the pressure exerted in atmospheres and mm Hg.

1 answer:

8 0

The pressure exerted by 0.400 moles of carbon dioxide in a 5.00 Liter container at 25 °C would be 1.9563 atm or 1486.788 mm Hg.

<h3>The ideal gas law</h3>

According to the ideal gas law, the product of the pressure and volume of a gas is a constant.

This can be mathematically expressed as:

pv = nRT

Where:

p = pressure of the gas

v = volume

n = number of moles

R = Rydberg constant (0.08206 L•atm•mol-1K)

T = temperature.

In this case:

p is what we are looking for.

v = 5.00 L

n = 0.400 moles

T = 25 + 273

= 298 K

Now, let's make p the subject of the formula of the equation.

p = nRT/v

= 0.400 x 0.08206 x 298/5

= 1.9563 atm

Recall that: 1 atm = 760 mm Hg

Thus:

1.9563 atm = 1.9563 x 760 mm Hg

= 1486.788 mm Hg

In other words, the pressure exerted by the gas in atm is 1.9563 atm and in mm HG is 1486.788 mm Hg.

More on the ideal gas law can be found here: brainly.com/question/28257995

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They are very stable (not reactive).

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How many objects in the solar system have been confirmed to currently support life?

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B. One
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In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.

trasher [3.6K]

Answer:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction

The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction

The second run has twice the surface area - yes, 44 sqcm to 22 sqcm

Explanation:

A catalyst is a material which speeds up a reaction without being consumed in the process. A heterogeneous catalyst is one which is of a different phase than the reactants. The effectiveness of a catalyst is dependent on the available surface area. The first step for this question is to determine the total available surface area of catalyst in both processes.

Step 1: Determine radius of large sphere

$V_{L}=\frac{4}{3}*pi*{r_{L}}^{3}$

$10=\frac{4}{3}*pi*{r_{L}}^{3}$

${r_{L}}^{3}=\frac{7.5}{pi}$

$r_{L}=1.337cm$

Step 2: Determine surface area of large sphere

$A_{L}=4*pi*{r_{L}}^{2}$

$A_{L}=4*pi*1.337^{2}$

$A_{L}=22.463 cm^{2}$

Step 3: Determine radius of small sphere

$V_{s}=\frac{4}{3}*pi*{r_{s}}^{3}$

$1.25=\frac{4}{3}*pi*{r_{s}}^{3}$

${r_{s}}^{3}=\frac{0.9375}{pi}$

$r_{s}=0.668cm$

Step 4: Determine surface area of small sphere

$A_{s}=4*pi*{r_{s}}^{2}$

$A_{s}=4*pi*0.668^{2}$

$A_{s}=5.607 cm^{2}$

Step 5: Determine total surface area of 8 small spheres

$A_{S}=8*A_{s}$

$A_{S}=8*5.607$

$A_{S}=44.856 cm^{2}$

$A_{L}=22.463 cm^{2}$ - Surface area of 1 large sphere

$A_{S}=44.856 cm^{2}$ - Surface area of 8 small spheres

Options:

The second run will be faster - true, the increased surface area of catalyst will increase the rate of reaction
The second run will be slower - false, the increased surface area of catalyst will increase the rate of reaction
The second run will have the same rate as the first - possible, in case there is a factor other than catalyst limiting the reaction
The second run has twice the surface area - yes, 44 sqcm to 22 sqcm
The second run has eight times the surface area - no, 44 sqcm to 22 sqcm
The second run has 10 times the surface area - no, 44 sqcm to 22 sqcm

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3 years ago

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4. A sample of gas has a pressure of 700 mmHg and 30.0°C. Ar what temperature would the pressure be 600 mmHg if the volume remai

shtirl [24]

Answer:

T₂ = 259.84 K

T₂ = -13.31 °C

Explanation:

Given data:

Initial pressure = 700 mmHg

Initial temperature = 30.0°C (30+273.15 K = 303.15 K)

Final temperature = ?

Final pressure = 600 mmHg

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

700 mmHg /303.15 K = 600 mmHg / T₂

T₂ = 600 mmHg × 303.15 K / 700 mmHg

T₂ =181890 mmHg.K /700 mmHg

T₂ = 259.84 K

Temperature in celsius

259.84 K - 273.15 = -13.31 °C

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