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if liquid N2 has a boling point of 77.36K and desnity of 0.807g/ml. 1.00mL of liquid nieigen is placed into a deflated baloon, WHat will be the volume of nitorgen gas in the balloon at STP3

Sagot :

Answer:

2.47

Explanation:

Here's how to find the volume of nitrogen gas in the balloon at STP (Standard Temperature and Pressure):

**1. Convert liquid nitrogen volume to mass:**

- We know the volume of liquid nitrogen (V_liquid) = 1.00 mL

- We know the density of liquid nitrogen (ρ_liquid) = 0.807 g/mL

Mass (m_liquid) = Volume (V_liquid) x Density (ρ_liquid)

m_liquid = 1.00 mL * 0.807 g/mL

m_liquid = 0.807 g

**2. Assuming all liquid nitrogen vaporizes:**

The entire mass of liquid nitrogen (0.807 g) will convert to gaseous nitrogen.

**3. Calculate the number of moles of nitrogen gas:**

- Molar mass of nitrogen (M_nitrogen) = 28.02 g/mol (nitrogen molecule has two nitrogen atoms)

Number of moles (n_gas) = Mass (m_liquid) / Molar mass (M_nitrogen)

n_gas = 0.807 g / 28.02 g/mol

n_gas = 0.0288 mol (rounded to four decimals)

**4. Apply the ideal gas law at STP:**

At STP, temperature (T) = 273.15 K and pressure (P) = 1 atm (101.325 kPa).

The ideal gas law equation is: PV = nRT

where:

 - P = Pressure

 - V = Volume

 - n = Number of moles

 - R = Ideal gas constant (approximately 8.314 J/mol*K)

 - T = Temperature

We want to solve for V (volume of gas). Since we know all other values at STP, we can rearrange the equation:

V = nRT / P

V = 0.0288 mol * 8.314 J/mol*K * 273.15 K / (101.325 kPa)

**Note:** To convert kPa to atm, divide by 101.325.

**5. Convert Joules to Liters*atmosphere (L*atm) for volume:**

1 J/mol*K = 0.00008205 L*atm/mol*K (conversion factor)

V = 0.0288 mol * 8.314 J/mol*K * 273.15 K / (101.325 kPa * 0.00008205 L*atm/mol*K)

V ≈ 2.47 L (rounded to two decimals)

Therefore, the volume of nitrogen gas in the balloon at STP will be approximately 2.47 liters.

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