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Anvisha [2.4K]
3 years ago
13

A gaseous mixture contains 434.0 Torr H 2 ( g ) , 434.0 Torr H2(g), 389.9 Torr N 2 ( g ) , 389.9 Torr N2(g), and 77.9 Torr Ar (

g ) . 77.9 Torr Ar(g). Calculate the mole fraction, χ , χ, of each of these gases.
Chemistry
1 answer:
weqwewe [10]3 years ago
8 0

Answer:

H₂: 0.48,  N₂: 0.43;  Ar: 0.09

Explanation:

First of all, sum all the pressures to know the total pressure in the mixture.

434 Torr + 389.9 Torr + 77.9 Torr = 901.8 Torr

Mole fraction = Pressure gas / Total Pressure

Mole Fraction H₂: 434 Torr /901.8 Torr = 0.48

Mole Fraction N₂: 389.9 /901.8 Torr =0.43

Mole Fraction Ar:  77.9 /901.8 Torr = 0.09

Remember: <u>SUM OF MOLE FRACTION = 1</u>

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Can you please help me and can you show your work please
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Answer:

1. 25 moles water.

2. 41.2 grams of sodium hydroxide.

3. 0.25 grams of sugar.

4. 340.6 grams of ammonia.

5. 4.5x10²³ molecules of sulfur dioxide.

Explanation:

Hello!

In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:

1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:

1.5x10^{25}molecules*\frac{1mol}{6.022x10^{23}molecules} =25 molH_2O

2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:

1.2mol*\frac{40.00g}{1mol} =41.2g

3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:

45g*\frac{1mol}{180.15g}=0.25g

4. Here, since the molar mass of ammonia is 17.03 g/mol:

20mol*\frac{17.03g}{1mol}=340.6g

5. Here, since the molar mass of SO2 is 64.06 g/mol:

48g*\frac{1mol}{64.06g} *\frac{6.022x10^{23}molecules}{1mol} =4.5x10^{23}molecules

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