Energy transformations that occur when a ball is thrown in the air

You want to know how many moles of gas your lungs can hold. You start off with a balloon that has 1.4 moles of gas and occupies

Mrrafil [7]

Number of moles : n₂ = 1.775 moles

<h3>Further explanation</h3>

Given

Moles = n₁ = 1.4

Volume = V₁=22.4 L

V₂=28.4 L

Required

Moles-n₂

Solution

Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles

The ratio of gas volume will be equal to the ratio of gas moles

$\tt \dfrac{V_1}{V_2}=\dfrac{n_1}{n_2}$

Input the values :

n₂ = (V₂ x n₁)/V₁

n₂ = (28.4 x 1.4)/22.4

n₂ = 1.775 moles

4 0

2 years ago

Which method is suitable for collection a a carbon dioxide.Explain

mel-nik [20]

Answer:

Carbon dioxide can be collected over water. Carbon dioxide is slightly soluble in water and denser than air, so another way to collect it is in a dry, upright gas jar.

Explanation:

8 0

2 years ago

What method will you use for your campaign?

Allushta [10]

Sorry I’m only answering so I could upload

6 0

2 years ago

What volume of 0.585 m ca(oh)2 would be needed to neutralize 15.8 l of 1.51 m hcl?

SpyIntel [72]

When the balanced reaction equation is:

2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)

from the balanced equation, we can get the molar ratio between HCl & Ca(OH)2

2:1

∴ the volume of Ca(OH)2 = 15.8 L HCl * 1.51 m HCl * (1mol Ca(OH)2/ 2mol HCl) * (1L ca(OH)2/0.585 mol Ca(OH)2

= 20.4 L

8 0

2 years ago

Consider the following reaction at equilibrium. 2CO2 (g) 2CO (g) + O2 (g) H° = -514 kJ Le Châtelier's principle predicts that th

tiny-mole [99]

Answer:

C. at low temperature and low pressure.

Explanation:

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

For the reaction:

2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.

Effect of pressure:

When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.

The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.

So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), so the equilibrium partial pressure of CO (g) can be maximized at low pressure.

Effect of temperature:

The reaction is exothermic because the sign of ΔH is (negative).

So, we can write the reaction as:

2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.

Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, so the equilibrium partial pressure of CO (g) can be maximized at low temperature.

So, the right choice is:

C. at low temperature and low pressure.

7 0

3 years ago