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slavikrds [6]
3 years ago
14

On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i

s the new volume of the balloon? Selected Answer:
Chemistry
1 answer:
ddd [48]3 years ago
8 0

Answer:

1.6 L

Explanation:

Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (12 + 273.15) K = 285.15 K  

T₂ = (32 + 273.15) K = 305.15 K  

Using above equation as:

\frac{1.5}{285.15}=\frac{V_2}{305.15}

V_2=\frac{1.5\cdot \:305.15}{285.15}

New volume = 1.6 L

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Answer:

90

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Explanation:

90 - atomic number

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3 years ago
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Answer:

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Explanation:

4 0
2 years ago
Read 2 more answers
What volume (in L) will a 32 g sample of butane gas, C4H10(g), occupy at a temperature of 45.0 oC and a pressure of 728 mm Hg?
larisa86 [58]

Answer:

15.0 L

Explanation:

To find the volume, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (mmHg)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)

-----> T = temperature (K)

To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).

Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)

Molar Mass (C₄H₁₀): 58.124 g/mol

32 grams C₄H₁₀              1 moles
-------------------------  x  -----------------------  = 0.551 moles C₄H₁₀
                                    58.124 grams

P = 728 mmHg                      R = 62.36 L*mmHg/mol*K

V = ? L                                    T = 45.0 °C + 273.15 = 318.15 K

n = 0.551 moles

PV = nRT

(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)

(728 mmHg)V = 10922.7632

V = 15.0 L

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1 year ago
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