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Sidana [21]
3 years ago
7

Calculate the density of nitrogen gas, in grams per liter, at stp.

Chemistry
1 answer:
lesantik [10]3 years ago
4 0

Standard temperature is 273 K

Standard pressure is 1 atm

We use the ideal gas equation to find out density of nitrogen gas in g/L

Ideal gas equation:

PV = nRT\\  PV = (\frac{Mass}{Molar mass)}RT\\   P(Molar mass) = (\frac{Mass}{Volume})RT\\  \frac{Mass}{Volume}=\frac{P(molar mass)}{RT} \\  Density = \frac{P(Molar mass)}{RT}

Molar mass of N_{2} = 28 g/mol

Pressure = 1 atm

Temperature = 273 K

Density = \frac{(1atm)(28 g/mol)}{(0.08206 \frac{L.atm}{mol.K})(273 K)}

= 1.25 g/L

Therefore, density of nitrogen gas at STP is 1.25 g/L

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Lyrx [107]
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4 0
3 years ago
☝️☝️☝️help☝️☝️☝️!!!!
nataly862011 [7]
8. A. Density is independent of size (an intensive property).
9. B. The highest density is 1786 g/20cm^3, or 89.3 g/cm^3. The least is 512 g/20cm^3, or 25.6 g/cm^3. Therefore, the difference is 89.3 - 25.6, or 63.7 g/cm^3.
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7 0
3 years ago
Can you store cuso4 in an aluminum container? explain your reasoning
mash [69]

We can store the copper sulphate solution in alumiun container, if cover on alumiun is present.

<h3>Can you store cuso4 in an aluminum container?</h3>

Aluminium is more reactive than copper so the Aluminium will displace copper sulphate from its solution by reacting with it but if there is cover on the aluminium then the alumium can't react with copper.

So we can store the copper sulphate solution in alumiun container.

Learn more about container here: brainly.com/question/11459708

3 0
2 years ago
How many atoms are in 12 moles of silver?
lord [1]

Answer:

NA=72.264x1023

Explanation:

I got you

7 0
3 years ago
When the potential difference across a lamp is 3.3V the current is 0.15 calculate the resistance of the lamp and give the unit
marissa [1.9K]

Answer:

22Ω

Explanation:

Given parameters:

Potential difference  = 3.3V

Current  = 0.15A

Unknown:

Resistance  = ?

Solution:

According to ohm's law, potential difference, current and resistance are related by the expression below;

             V = I R

where V is the voltage

           I is the current

          R is the resistance

          3.3 = 0.15 x R

             R = \frac{3.3}{0.15}   = 22Ω

6 0
3 years ago
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