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OleMash
11 days ago
15

a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill

imeters of mercury
Chemistry
1 answer:
lions [2.9K]11 days ago
5 0
1223.38 mmHg Using the ideal gas equation: where, P is the pressure V is the volume n is the number of moles T is the temperature R is the gas constant valued at Also, Moles = mass (m) / Molar mass (M) Density (d) = Mass (m) / Volume (V) Thus, the ideal gas equation can be expressed as: Given that: d = 1.80 g/L Temperature = 32 °C To convert T(°C) to T(K): T(K) = T(°C) + 273.15 So, T = (32 + 273.15) K = 305.15 K Molar mass of nitrogen gas = 28 g/mol Applying the equation: P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K Thus, P = 1223.38 mmHg. Therefore, the pressure of the nitrogen gas must be 1223.38 mmHg.
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Calculate the ratio of the velocity of helium atoms to the velocity of neon atoms at the same temperature.
Tems11 [2777]

Answer:

vHe / vNe = 2.24

Explanation:

To determine the velocity of an ideal gas, one should apply the formula:

v = √3RT / √M

In this equation, R represents the gas constant (8.314 kgm²/s²molK); T refers to temperature, and M indicates the molar mass of the gas (4x10⁻³kg/mol for helium and 20.18x10⁻³ kg/mol for neon). Hence:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio simplifies to:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

vHe / vNe = 2.24

I hope it assists you!

8 0
1 month ago
The metallic radius of a potassium atom is 231 pm. What is the volume of a potassium atom in cubic meters?
Anarel [2989]
1 pm equals 10^{-10} cm. Therefore, 230 pm can be converted to 2.3 × 10^{-8} cm. Atoms are spherical in shape, and the volume of a sphere can be calculated using the formula 4/3πr³. Consequently, the atom's volume is calculated as 4/3π(2.3 × 10^{-8})³. Evaluating this gives 4/3 × (3.142 × 12.167 × 10^{-24}), which simplifies to approximately 5.096 × 10^{-23} cm³. Since 1 m³ is equal to 1,000,000 cm³, we find that the volume of the atom is 5.096 × 10^{-29} m³.
8 0
1 month ago
1. Adakah benar bahawa anda tidak boleh membasuh rambut, meminum air sejuk dan
KiRa [2933]

Answer:

Is it true that you shouldn't wash your hair, drink cold water, eat ice cream, or exercise during your period? Please explain your answer.

No, this is not accurate; doing any of these activities is perfectly fine. None of them affects us because they are not connected to our bodily systems. Also, I apologize for any language errors as I utilized Google Translate.

I hope this is helpful :)

7 0
2 months ago
The melting of an ice sculpture of BEVO at room temperature requires 10 kJ of energy. Calculate the change in entropy of the sur
KiRa [2933]

Answer:

The entropy change of the surroundings is -33.5 J/K

Explanation:

The heat gained by the ice is 10 kJ, equating to 10000 J

The heat released by the surroundings is the negative of the heat gained by the ice.

Thus,

The heat lost by the surroundings = -(10000 J)

Given the room temperature = 25^{o}C = 273 + 25 = 298 K

Change\,\,in\,\,entropy\,\,surrounding =\frac{Heat\,\,lost\,\,by\,\,surrounding}{Room\,\,temperature}= \frac{-10000}{298}= 33.5\,J/k

Thus, the entropy change of the surroundings is -33.5 J/K

7 0
1 month ago
An ice cube with a volume of 45.0ml and a density of 0.9000g/cm3 floats in a liquid with a density of 1.36g/ml. what volume of t
VMariaS [2998]

Response: The volume of the cube that is underwater is 29.8 mL

Clarification:

Initially, we need to find the mass of the ice.

Equation applied:

\text{Mass of ice}=\text{Density of ice}\times \text{Volume of ice}

Provided information:

Ice density = 0.9000g/cm^3=0.9000g/mL

Volume of ice = 45.0 mL

\text{Mass of ice}=0.9000g/mL\times 45.0mL

\text{Mass of ice}=40.5g

The cube displaces 40.5 g of liquid to float.

Next, we will ascertain how much of the cube is submerged in the liquid.

\text{Volume of ice}=\frac{\text{Mass of liquid}}{\text{Density of liquid}}

\text{Volume of ice}=\frac{40.5g}{1.36g/mL}

\text{Volume of ice}=29.8mL

Therefore, the submerged portion of the cube is 29.8 mL

5 0
24 days ago
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