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steposvetlana [31]
2 years ago
5

a gas that exerts a pressure of 215 torr in a container with a volume of 51.0 mL will exert a pressure of ? torr when transferre

d to a container with a volume of 18.5L
Chemistry
1 answer:
zhannawk [14.2K]2 years ago
3 0
To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:

P1V1 = P2V2,

Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:

(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr

The final pressure exerted by the gas would be 0.593 torr.

Hope this helps!
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The flame produced by the burner of a gas (propane) grill is a pale blue color when enough air mixes with the propane (C3H8) to
adelina 88 [10]

Answer:

At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane

Explanation:

The combustion stoichiometry is as follows:

      C₃H₈ + 5O₂  = 4 H₂O + 3CO₂      The molecular weights (g/mol) are:

MW  44    5x32      4x18    3x44

So each gram of propane is 1/44 = 0.02272 mol propane

and will need 5 x 0.02272 = 0.1136 mol oxygen

At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.

At the low pressure in the burner we can use the Ideal Gas Law

PV=nRT, or V = nRT/P

P = 1.1 x 101325 Pa = 111457 Pa

T = 195°C + 273 = 468 K

R = 8.314

and we calculated n = number of moles air = 0.54 mol

So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.

6 0
2 years ago
An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial
andrey2020 [161]

Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state )  P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁  →  P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ =  n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2  = 2.25 atm

P₂= 2.25 atm

8 0
2 years ago
In the reaction MgCl2 + 2KOH Mg(OH)2 + 2KCl, if 3 moles MgCl2 are added to 4 moles KOH, what determines how much Mg(OH)2 is made
LenKa [72]
The correct option is C. The amount of MgCl2. we know this because <span>no matter how much you increase KOH, if you dont increase Mgcl2, the amount of Mg(OH)2 remains the same. Hope this works for you</span>
8 0
2 years 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
Grace [21]

Answer : The volume of the cube submerged in the liquid is, 29.8 mL

Explanation :

First we have to determine the mass of ice.

Formula used :

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

Given:

Density of ice = 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 will float when 40.5 g of liquid is displaced.

Now we have to determine the volume 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

Thus, the volume of the cube submerged in the liquid is, 29.8 mL

5 0
2 years ago
Three moles of helium gas (molar mass MM = 4.00 g/molg/mol) are in a rigid container that keeps the volume of the gas constant.
Anastaziya [24]

Answer:

The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Explanation:

Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg

The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)

where u = initial rms speed of the gas = 850 m/s

Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J

Then, 3600 J is added to the gas,

New kinetic energy of the gas = 4335 + 3600 = 7935 J

New kinetic energy of Helium atoms = (1/2)(m)(v²)

where v = final rms speed of the gas = ?

7935 = (1/2)(0.012)(v²)

v² = (7935×2)/0.012

v² = 1,322,500

v = 1150 m/s

Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Hope this Helps!!!

8 0
1 year ago
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