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2 years ago

A 50.5 L flask contains 3.25 mol of an unknown gas at 288.6 K, what is the pressure of this gas?

Chemistry

1 answer:

Ivahew [28]2 years ago

3 0

Answer:

1.52atm is the pressure of the gas

Explanation:

To solve this question we must use the general gas law:

PV = nRT

Where P is pressure in atm = Our incognite

V is volume = 50.5L

n are moles of gas = 3.25moles

R is gas constat = 0.082atmL/molK

And T is absolute temperature = 288.6K

To solve pressure:

P = nRT / V

P = 3.25mol*0.082atmL/molK*288.6K / 50.5L

P = 1.52atm is the pressure of the gas

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Which equation correctly relates kinetic energy mass and velocity

jolli1 [7]

Answer: The equation for kinetic energy is $\frac{1}{2}mv^2$

Explanation:

Kinetic energy is the energy possessed by the virtue of object's motion. It is defined as the work needed to move a body of a given mass from rest to its velocity.

Mathematically,

$K.E.=\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body.

Hence, above equation relates kinetic energy to the mass and velocity of the body.

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3 years ago

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What occurs during a chemical reaction? *

Romashka [77]

Answer:

c. Atoms of two or more elements bond together

Explanation:

a chemical reactions forms products ( new substances); where it rearranges themselves to form new bonds.

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2 years ago

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determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)

sasho [114]

The mass of water formed is

calculation

Use the ideal gas equation to calculate the moles of NH3 and O2

that is Pv= n RT

where; P= pressure,

V= volume,

n = number of moles,

R=gas constant = 0.0821 l .atm/ mol.K

make n the formula of the subject by diving both side by RT

n = PV /RT

The moles of NH3

n= (1.50 atm x 12.5 L) /( 0.0821 L. atm /mol.k x 298 K) =0.766 moles

The moles of O2

=(1.1 atm x 18.9 L) / ( 0.0821 L. atm/ mol.k x 323 K) = 0.784 moles

write the reaction between NH3 and O2

4 NH3 + 5 O2 →4 No +6H2O

from equation above 0.766 moles of NH3 reacted to produce

0.766 x 6/4 =1.149 moles of H2O

0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20

since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced= 0.9408 moles

mass of H2O = moles x molar mass

from periodic table the molar mass of H2O = (1 x2)+16= 18 g/mol

mass = 18 g/mol x 0.9408 moles= 16.93 grams

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3 years ago

45) George is making spaghetti for dinner. He places 4.01 kg of water in a pan and brings it to a boil.

lesya692 [45]

On adding salt.....The boiling temperature increases.....

So ∆t= KB * molality
=O.52*(58/58)/4
= O.52*1/4
= 0.13
So increase is 100+.13=100.13°c

5 0

3 years ago

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

( $1 atm = 760 mmHg$ )

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

$(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole$

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

$\frac{49.4}{100}\times 0.0438 mol=0.02164 mol$

$2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2$

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

$\frac{2}{1}\times 0.02164 mol=0.04328 mol$

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0

3 years ago