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andriy [413]
2 years ago
11

A helium balloon can hold about 6.8 mol of the gas. Assuming the balloon is kept at room temperature (25°C) and had a maximum of

1.2L, what is the pressure inside teh balloon
Chemistry
2 answers:
Kipish [7]2 years ago
6 0

Answer:

P= 138.6 atm or 1.4x10²

Explanation:

You will need to use PV=nRT to solve for P, pressure

Where the R constant is  0.08206 L atm mol^-1 K^-1

V is 1.2 L

T is temperature in kelvin, 273.15+25= 298.15

n is 6.8 mol

P*1.2=6.8*0.08206*298.15 ---> P*1.2=166.3700852 ---> P=166.3700852/1.2

P=138.6 atm

or 1.4x10²

Elza [17]2 years ago
3 0

Answer:

138.47atm

Explanation:

Data obtained from the question include:

n (number of mole) = 6.8 mol

T (temperature) = 25°C = 25 + 273 = 298K

V (volume) = 1.2L

R (gas constant) = 0.082atm.L/Kmol

P (pressure) =?

Using the ideal gas equation PV = nRT, we can confidently calculate the pressure inside the balloon as shown below:

PV = nRT

P = nRT/V

P = (6.8 x 0.082 x 298)/1.2

P = 138.47atm

Therefore, the pressure inside the balloon is 138.47atm

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A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu
bogdanovich [222]

Answer:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

Explanation:

The temperature of an object can be written using different temperature scales.

The two most important scales are:

- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).

- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

T(K)=T(^{\circ}C)+273.15

Therefore  in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

6 0
3 years ago
Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g o
stich3 [128]

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of butane

\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles

b) moles of oxygen

\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

According to stoichiometry :

2 moles of butane require 13 moles of O_2

Thus 0.09 moles of butane will require =\frac{13}{2}\times 0.09=0.585moles  of O_2

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.

7 0
2 years ago
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