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3 years ago

1. A given mass of air has a volume of 6.00 L at 80.0°C. At constant pressure, the temperature is

Chemistry

1 answer:

earnstyle [38]3 years ago

4 0

3 L will be the final volume for the gas as per Charle's law.

Answer:

Explanation:

The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.

V∝ T

Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be

$\frac{V_{1} }{T_{1} }= \frac{V_{2} }{T_{2} }$

So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.

$\frac{6}{80}=\frac{V_{2} }{40}$

$V_{2}=\frac{6*40}{80}=3 L$

So, 3 L will be the final volume for the gas as per Charle's law.

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⦁answer Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg. ⦁ Determine the molar mass of a gas that has a

STALIN [3.7K]

Answer:

The correct answers are:

- Krypton: density= 2.8 g/L

- Molar Mass= 63.99 g/mol

- Mass of O₂= 15.29 g

Explanation:

The general equation of an ideal gas is the folllowing:

P x V = n x R x T

Where: P= pressure (in atm), V= volume; n= number of moles, R= gas constant (0,082 L.atm/K.mol) and T= temperature (in K).

For krypton:

P= 671 mmHg = 0,882 atm

V= 478 ml x 1000 ml/1 L= 0,478 L

T= 47ºC= 320 K

MM= 83.8 g/mol (from Periodic Table, Kr is an inert gas so it is a monoatomic gas)

P x V = n x R x T

Since the number of moles of a compound can be calculated by dividing the mass of compound (m) into its molar mass (MM):

n= m/MM

We can replace the expression in the first equation to obtain:

$P x V= \frac{m}{MM} x R x T$

m/V= $\frac{P x MM}{R x T}$

Density (d) is equal to the mass per volume (m/V), so we can directly calculate the density:

d= m/V= \frac{P x MM}{R x T}=

= (0.882 atm x 83.8 g/mol)/(0.082 L.atm/K.mol x 320 K)

= 2.81 g/L

For the gas:

d= 2.18 g/L

T= 66ºC= 339 K

P= 720 mmHg= 0.947 atm

d= \frac{P x MM}{R x T}

⇒MM = $\frac{dx R x T}{P}$

= (2.18 g/L x 0.082 L.atm/K.mol x 339 K)/(0.947 atm)

= 63.99 g/mol ≅ 64 g/mol

For the O₂:

V= 5.60 L

P= 1.75 atm

T= 250 K

MM(O₂) = 2 x Atomic Mass O= 2 x 16 g/mol= 32 g/mol

We can use the second equation:

P x V= \frac{m}{MM} x R x T

⇒ m = $\frac{P x V x MM}{R x T}$ = (1.75 atm x 5.6 L x 32 g/mol)/(0.082 L.atm/K.mol x 250 K)

= 15.29 g≅ 16 g

4 0

3 years ago

Read 2 more answers

Do viruses have locomotion

Dvinal [7]

Answer:

Explanation:

They do not have any sensing organs. They can't see, smell or move on their own. They rely on random motions from their host to move around.

5 0

3 years ago

Read 2 more answers

First person with the right answer gets brainliest thanks (btw the numbers on the right are the answers choose the right one)

kap26 [50]

Molar mass of FE2O3=2(55.85)+3(16)=159.7

2.56g*1mol/159.7*2mol/1mol*55.85g/1mol=1.79g

6 0

3 years ago

How many grams of H2 will be produced, given 27.4 g of Al to start with?

pishuonlain [190]

3.07g H2

27.4/26.98/2x3x1.01x2=3.07

6 0

3 years ago

Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:

kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g = 0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3 = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂ I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0

3 years ago