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4 years ago

Order the theories from oldest to most recent.

Chemistry

1 answer:

rjkz [21]4 years ago

5 0

Answer:

Explanation:

The following are the order of point from oldest to most recent

1. Atoms are tiny indivisible particles that make up all mater.

it is Postulate of Dalton's atomic theory. this theory was put forward in 1803

2. Negatively charged electrons are embedded in a mass of positive charge.

In 1897 J.J Thomson discovered that atom consist of tiny negatively charged particles called electrons that are uniformly spread in +vely charged matrix this model called as plum pudding model.

3. Atoms consist of mostly empty space with a dense nucleus of positive charge.

While doing experiment on gold foil Rutherford presented a model in 1909 and stated that atom consist mostly empty space with dense nucleus.

4. Electrons occupy specific energy levels surrounding a positively charged nucleus.

Niels Bohr in 1913 put forward a model to explain atomic orbitals/energy level. This is a postulate of Bohr model.

5. Electrons move about a positively charged nucleus in clouds that are defined by probabilities.

In 1926 Erwin Schrödinger explain the wave function of electrons and its probability.

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(I). The gases are present at the end of the experiment are O₂ and NO₂

(II). The pressure of O₂ is 0.158 atm.

The pressure of NO₂ is 0.681 atm

The Pressure of NO is zero.

Explanation:

Given that,

Volume of large bulb = 6.00 L

Pressure = 0.850 atm

Volume of small bulb = 1.50 L

Pressure = 2.50 atm

Temperature = 22°C = 295 K

We need to calculate the moles in NO

Using formula of moles

$n=\dfrac{PV}{RT}$

Put the value into the formula

$n=\dfrac{0.850\times6.00}{0.0821\times295}$

$n=0.211\ moles$

We need to calculate the moles in O

Using formula of moles

$n=\dfrac{PV}{RT}$

Put the value into the formula

$n=\dfrac{2.50\times1.50}{0.0821\times295}$

$n=0.155\ moles$

The balance equation for the reaction is

$2NO+O_{2}\Rightarrow 2NO_{2}$

So, The gases are present at the end of the experiment are O₂ and NO₂

We need to calculate the remaining moles of O₂

Using formula for remaining moles

Moles of O₂ remaining = $0.155-\dfrac{0.211}{2}$

$Moles\ of \ O_{2}\ remaining =0.049\ moles$

Moles of NO₂ = 0.211 moles

Total volume $V=V_{l}+V_{s}$

Put the value into the formula

$V=6.00+1.50$

$V=7.5\ V$

(2). If the gas was consumed completely

We need to calculate the pressure of O₂

Using formula of pressure

$P=\dfrac{moles\times R\times T}{V}$

Put the value into the formula

$P=\dfrac{0.049\times0.0821\times295}{7.5}$

$P=0.158\ atm$

We need to calculate the pressure of NO₂

Using formula of pressure

$P=\dfrac{moles\times R\times T}{V}$

Put the value into the formula

$P=\dfrac{0.211\times0.0821\times295}{7.5}$

$P=0.681\ atm$

If the gas was consumed completely

Then, Pressure of NO is zero.

Hence, (I). The gases are present at the end of the experiment are O₂ and NO₂

(II). The pressure of O₂ is 0.158 atm.

The pressure of NO₂ is 0.681 atm

The Pressure of NO is zero.

8 0

3 years ago