I really need help with this!

tamaranim1 [39]

I read and said it's a weak base

3 0

2 years ago

Which of the following statements about Avogadro’s Law is false? a. One mole of He occupies 22.4 L at STP. b. At constant P and

Ronch [10]

Answer:

The correct answer to the question which statements about Avogadro’s Law is false is

c. At constant T and P, doubling the moles of gas decreases the volume by half.

Explanation:

Avogadro's law describes the relationship between the volume of a mass of gas and the number of moles present. Avogadro's law states that at standard (or the same) temperature and pressure, equal volumes od all gases contain equal number of molecules

That is mathematically

$\frac{V_{1} }{n_{1} } = \frac{V_{2} }{n_{2} }$ where

V₁ = volume of first sample

V₂ = volume of second sample

n₁ = number of moles in first sample

n₂ = number of moles in second sample

6 0

2 years ago

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

5 0

2 years ago

What does an emission spectrum look like?

aniked [119]

Answer:

It should be B

8 0

2 years ago

Read 2 more answers

Salt is dissolved in a flask of tap water. Distilling the mixture causes the salt to separate from the water. Which type of ener

Alik [6]

Heat energy is required.
In distillation, the solution is first heated, where heat energy is required, such as using a bunsen burner.
When the solution is heated, the water may reach its boiling point and evaporate. However, salt does not. When water molecules evaporates, it travels through a condenser that cools it down into liquid again. Therefore we get pure water. Salt is also obtained in the original beaker.
Therefore to first start this process, heat energy is required.

6 0

2 years ago

Read 2 more answers