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Shkiper50 [21]
3 years ago
13

If I have 12.0 volume of gas at pressure of 1.10 and temperature of 200k, what is the number of moles? (R=8314)

Chemistry
1 answer:
arsen [322]3 years ago
5 0

Answer:

7.94 x 10^6 mol

Explanation:

PV=nRT

n=(PV)/(RT)

n=(1.10*12.0)/(8314*200)

n=7.94 x 10^6 mol

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Answer: Charles's law, Avogadro's law andd Boyle's law.

Charles law states the constant ratio of volume to temperature, at constant pressure. Boyle's law states the constat product of pressure and volumen at constant temperature. Avogadro's law states that equal volumes of gases at the same temperature and pressure have equal number of particles.

So, all those three laws combined state the relation of pressure, volume, temperature and number of particles of a gas, which is what the ideal gas law does: PV = n RT.
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3 years ago
A cliff near the water has been worn away. How did erosion change the shape of the land?
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8 0
3 years ago
a 10.99g sample of NaBr contains 22.34% Na by mass. Considering the law of constant composition (define proportions), how many g
leonid [27]

Given :

A 10.99 g sample of NaBr contains 22.34% Na by mass.

To Find :

How many grams of sodium does a 9.77g sample of sodium bromine contain.

Solution :

By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.

Therefore , percentage of Na by mass in NaBr will be same for every amount .

Percentage of Na in 9.77 g NaBr is 22.34 % too .

Gram of Na = 9.77\times \dfrac{22.34}{100}=2.18\ g .

Hence , this is the required solution .

7 0
3 years ago
Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W
alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

4 0
3 years ago
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