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Artemon [7]
3 years ago
9

1. A weather balloon is filled with helium on a cool morning when the temperature is 13.0°C. The temperature is predicted to rea

ch 35.0°C around 4 PM. When first inflated, the balloon has a volume of 12.9 L and is at 238 kPa – these are the standard volume and pressure for this type of weather balloon. Based on the balloon design, it can only withstand pressures up to 252 kPa.
How many moles of helium are in this hot air balloon?
Will the balloon burst when the temperature reaches 35.0 C?

Please someone help, Im mostly stumped by R. I know it is 8.314 L kPa/mole K but i dont know how to calculate that please help
Chemistry
1 answer:
azamat3 years ago
4 0

Answer:

Number of moles: n = 1.29 moles

If the temperature reaches 35°C which is 308.15 K the balloon will burst

Explanation:

Using the formula: PV = nRT  

Where P is pressure in atm, V is volume in L, n is the number of moles, R is the gas constant which can be 8.314 J/mol K or 0.08206 atm L/mol K, and T is temperature in K.

In order to know which R to use, just take a look at the units. If you are given the pressure in atm, then use 0.08206 atm L/mol K. If you are given energy in Joules instead, then use 8.314 J/mol K.

Problem solution:

1 atm = 101.325 kPa, so 238 kPa is equal to 2.349 atm

1°C is 274.15 K, so 13°C is 286.15 k

Using the formula PV = nRT

2.349 atm * 12.9 L = n * 0.08206 atm L/mol K * 286.15 k  

n = (30.302 atm L) / (23.481 atm L/mol)

n = 1.29 moles

If the temperature reaches 35°C which is 308.15 K the balloon will burst:

P = nRT/V

P = (1.29 moles * 0.08206 atm L/mol K * 308.15 k) / 12.9 L

P = 2.529 atm = 256.21 kPa

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Which elements are out of order in terms of atomic mass. co,Ni or Li,be or I,xe​
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7 0
3 years ago
Calculate the percent by volume of a solution that has 75 mL of solute dissolved in 375 mL of solution. Show your work and round
Mandarinka [93]

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The answer is 20 % V/V

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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
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<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

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