What is the pressure of 2.50 moles of an ideal gas if it has a volume of 50.0 liters when the temperature is 300.0 K

1 answer:

7 0

Start with the ideal gas equation, PV=nRT PV=nRT and rearrange for pressure to get p=nRTV p=nRTV . You have all the necessary variables in their proper units, so plug em' into the equation to solve for pressure in units of atmospheres.

P=(2.5 mol)(300 K)(0.08206 L atm K −1 mol −1 50.0 L =1.23 atm P=(2.5 mol)(300 K)(0.08206 L atm K−1 mol−150.0 L=1.23 atm

All that needs to be done now is converting atmospheres to mm Hg Hg .

1.23 atm∗760 mm Hg1 atm =935 mm Hg 1.23 atm∗760 mm Hg1 atm=935 mm Hg .

That value makes sense, since the original pressure in atmospheres was above 1, the pressure in mm Hg Hg will be above 760.

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A 0.216 g sample of carbon dioxide, CO2, has a volume of 507 mL and a pressure of 470 mmHg. What is the temperature of the gas i

Gala2k [10]

Answer:

The temperature of the gas is 876.69 Kelvin

Explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

P= 470 mmHg
V= 570 mL= 0.570 L
n= 0.216 g= 0.0049 moles (being the molar mass of carbon dioxide is 44 g/mole)

R= 62.36367 $\frac{mmHg*L}{mol*K}$

Replacing:

470 mmHg*0.570 L= 0.0049 moles* 62.36367 $\frac{mmHg*L}{mol*K}$ *T

Solving:

$T=\frac{470 mmHg*0.570 L}{0.0049 moles* 62.36367\frac{mmHg*L}{mol*K} }$

T= 876.69 K

The temperature of the gas is 876.69 Kelvin

6 0

2 years ago

How many grams of carbonic acid were produced by the 3.00 g sample of NaHCO

stellarik [79]

I think the given is 3 g sample of NaHCO3. then if it will be reacted with an acid, it will produce H2CO3.
so the reaction NaHCO3 + HCl --> NaCl + H2CO3

mas of H2CO3 = 3 g NaHCO3 ( 1 mol NaHCO3 / 84 g ) ( 1 mol H2CO3 / 1 mol NaHCO3) ( 62.03 g / 1 mol )
mass of H2CO3 = 2.22 g H2CO3

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Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted

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When PH + POH = 14
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

7 = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L

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