How many grams of aluminum oxide can be made if you start with 108 g of aluminum reacting completely with oxygen?

A helium balloon is filled on the ground where the atmospheric pressure is 768 torr. The volume of the balloon is 8.00 m3. When

Anvisha [2.4K]

Answer:

The correct option is: B. 366 torr

Explanation:

Given: On the ground- Initial Volume: V₁ = 8.00 m³, Initial Atmospheric Pressure: P₁= 768 torr;

At 4200 m height- Final Volume: V₂ = 16.80 m³, Final Atmospheric Pressure: P₂ = ?

Amount of gas: n, and Temperature: T = constant

According to the Boyle's Law, for a given amount of gas at constant temperature: P₁ V₁ = P₂ V₂

⇒ P₂ = P₁ V₁ ÷ V₂

⇒ P₂ = [(768 torr) × (8.00 m³)] ÷ (16.80 m³)

⇒ P₂ = 365.71 torr ≈ 366 torr

Therefore, the final air pressure at 4200 m height: P₂ = 366 torr.

6 0

3 years ago

The pressure of a sample of helium in a 200. ml. container is 2.0 atm. If the 5 points

just olya [345]

The pressure of the gas = 40 atm

<h3>Further explanation</h3>

Given

200 ml container

P = 2 atm

final volume = 10 ml

Required

Final pressure

Solution

Boyle's Law

At a fixed temperature, the gas volume is inversely proportional to the pressure applied

$\tt \rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}$

Input the value :

P₂ = P₁V₁/V₂

P₂ = 2 x 200 / 10

P₂ = 40 atm

3 0

2 years ago

(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of

Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

$P x V = k$