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FromTheMoon [43]
3 years ago
7

Recall all the models you described in task 1. Think about the results each model would predict for the experiment with hydrogen

gas. Which models of the atom does the experimental evidence support? Explain why these models are compatible with the experimental results.
Chemistry
1 answer:
SpyIntel [72]3 years ago
3 0

Answer:

Rutherford's theory laid the foundation upon which Bohr's model is founded. Rutherford established the fact that at the center of the atom, there is a nucleus whose radius is smaller than the radius of the atom. This nucleus is positively charged and most of the mass of the atom is concentrated there. Electrons move round this nucleus in orbits.

The experimental evidences of the Bohr's model shows that Rutherford's model was fundamentally correct. However, Bohr's model introduced the idea of quantization of the energy of electrons in an atom. The model went further to explain the spectra lines of the hydrogen atom.

Explanation:

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How many liters of gas will be in the closed reaction flask when 36.0L of ethane (C2H6) is allowed to react with 105.0L of oxyge
Ivan

Answer:- Volume of the gas in the flask after the reaction is 156.0 L.

Solution:-  The balanced equation for the combustion of ethane is:

2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.

Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

36.0LC_2H_6(\frac{7LO_2}{2LC_2H_6})

= 126 L O_2

126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.

let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

105.0LO_2(\frac{4LCO_2}{7L O_2})

= 60.0 L CO_2

Similarly, let's calculate the volume of water vapors formed:

105.0L O_2(\frac{6L H_2O}{7L O_2})

= 90.0 L H_2O

Since ethane is present in excess, the remaining volume of it would also be present in the flask.

Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

105.0LO_2(\frac{2LC_2H_6}{7LO_2})

= 30.0 L C_2H_6

Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L

Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L

Hence. the answer is 156.0 L.

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