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2 years ago

What is the mass of butane gas, C4H10, that can be held in a 3.00 l container at STP?

Chemistry

1 answer:

ipn [44]2 years ago

7 0

The mass of butane gas at the given volume and Standard temperature and pressure is 0.0022g.

<h3>What is Ideal Gas Law?</h3>

The Ideal gas law or general gas equation emphasizes on the state or behavior of a hypothetical ideal gas. It states that "the pressure multiplied by volume is equal to moles multiply by the universal gas constant multiply by temperature.

It is expressed as;

PV = nRT

Where P is pressure, V is volume, n is the amount of substance, T is temperature and R is the ideal gas constant ( 0.08206 Latm/molK )

Given the data in the question;

Volume of the gas V = 3L
At STP; P = 1.0atm, T = 273.15K
Amount of gas n = ?

PV = nRT

n = PV / RT

n = ( 1.0atm × 3.00L ) / ( 0.08206 Latm/molK × 273.15K )

n = 3.00Latm / 22.41 Latm/mol

n = 0.13mol

Now, since Molar mass of brutane gas is 58.12 g/mol.

Mass = n × Molar mass

Mass = 0.13mol × 58.12 g/mol

Mass = 0.0022g

Therefore, the mass of butane gas at the given volume and Standard temperature and pressure is 0.0022g.

Learn more about Ideal Gas Law here: brainly.com/question/4147359

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Formic acid (HCO2H) has a dissociation constant of 1.8 x 10^-4 M. The acid dissociates 1:1. What is the [H+] if 0.35 mole of for

Tju [1.3M]

Initial [ HCO2H] = moles * volume
=0.35 moles * 1 L = 0.35 M

by using ICE table:

HCO2H ↔ H+ + HCO2-
initial 0.35 M 0 0
change - X +X +X
Equ (0.35 - X) X X

∴ Ka = [H+][HCO2-] / [HCO2H]

by substitution:

1.8 x 10^-4 = X^2 / (0.35-X) by solving for X

∴ X = 0.0079 or 7.9 x 10^-3

∴ [H+] = X = 7.9 x 10^-3 M

7 0

3 years ago

If you are given an ideal gas with pressure (p)259,392.00 pa and temperature (T)=200°c of 1 mole Argon gas in a volume 8.8dm3,ca

GuDViN [60]

Answer: $R=4.82436 \frac{Pa. m^{3}}{mol. K}$

Explanation:

The Ideal Gas equation is:

$P.V=n.R.T$ (1)

Where:

$P$ is the pressure of the gas

$n$ the number of moles of gas

$R=8.3144598 \frac{Pa. m^{3}}{mol. K}$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin.

$V$ is the volume

It is important to note that the behavior of a real gas is far from that of an ideal gas, taking into account that <u>an ideal gas is a single hypothetical gas</u>. However, under specific conditions of standard temperature and pressure (T=0\°C=273.15 K and P=1 atm=101,3 kPa) one mole of real gas (especially in noble gases such as Argon) will behave like an ideal gas and the constant R will be $8.3144598 \frac{Pa. m^{3}}{mol. K}$ .

However, in this case we are not working with standard temperature and pressure, therefore, even if we are working with Argon, the value of R will be far from the constant of the ideal gases.

Having this clarified, let's isolate $R$ from (1):

$R=\frac{PV}{nT}$ (2)

Where:

$P=259392 Pa$

$n=1 mole$

$T=200\°C=473.15 K$ is the absolute temperature of the gas in Kelvin.

$V=8.8 dm^{3}=0.0088 m^{3}$

$R=\frac{(259392 Pa)(0.0088 m^{3})}{(1 mole)(473.15 K)}$ (3)

Finally:

$R=4.82436 \frac{Pa. m^{3}}{mol. K}$

4 0

3 years ago

Why do scientists think the Grand Canyon was once covered by an ocean? (1 point)

PIT_PIT [208]

Seawater found to be seeping up through the canyon floor is why

scientists think Grand Canyon was once covered by an ocean.

Grand cayon is located in Arizona, United States and comprises of a lot of

fossil records and rocks which makes it one of the best sites for geological

study.

The Grand cayon was found to be seeping through its floor which made

scientists think it was once covered by an ocean years ago.

Read more about Grand cayon on brainly.com/question/20629780

4 0

2 years ago

Which is the strongest acid listed in the table

denpristay [2]

Answer:

Carborane

Explanation:

4 0

3 years ago

You have 17 liters of gas at STP. If the temperature rises to 94C and while the volume decreases to 12 liters, what will the ne

fenix001 [56]

Answer:

$P_2=1.90atm$

Explanation:

Hello!

In this case, according to the ideal gas equation ratio for two states:

$\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}$

Whereas both n and R are cancelled out as they don't change, we obtain:

$\frac{P_1V_1}{P_2V_2} =\frac{T_1}{T_2}$

Thus, by solving for the final pressure, we obtain:

$\frac{P_2V_2}{P_1V_1} =\frac{T_2}{T_1}\\\\P_2=\frac{T_2P_1V_1}{V_2T_1}$

Now, since initial conditions are 1.00 atm, 273.15 K and 17 L and final temperature and volume are 94 + 273 = 367 K and 12 L respectively, the resulting pressure turns out to be:

$P_2=\frac{367K*1.00atm*17L}{12L*273.15K}\\\\P_2=1.90atm$

Best regards!

7 0

2 years ago