Which conclusion could be made from Ernest Rutherford’s gold foil experiment?

2 answers:

5 0

The conclusion was that there must be space between the particles of an atom. The way he knew this is because while some of the radiation reflected off of the gold foil, other radiation went right through it.

Katarina [22]4 years ago

5 0

Answer:

Mass of an atom is concentrated in a positively charged core called the nucleus.

Explanation:

Rutherford gold foil experiment expanded our understanding of the structure of an atom. When positively charged alpha particles were bombarded against a thin gold foil, Rutherford observed that majority of particles passed through the foil undisturbed whereas a small percentage retraced their path.

There were two main conclusions from the experiment:

-Since most alpha particles simply passed through undeviated this suggested that majority of the space inside the atom is empty

- Also, since some of them completely bounced back this suggested that there is a positively charged core in which the mass of the nucleus is concentrated.

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The acetylene tank contains 35.0 mol C2H2, and the oxygen tank contains 84.0 mol O2.

harkovskaia [24]

Answer:- As per the question is asked, 35.0 moles of acetylene gives 70 moles of carbon dioxide but if we solve the problem using the limiting reactant which is oxygen then 67.2 moles of carbon dioxide will form.

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

$2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)$

From the balanced equation, two moles of acetylene gives four moles of carbon dioxide. Using dimensional analysis we could show the calculations for the formation of carbon dioxide by the combustion of 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{4molCO_2}{2molC_2H_2})$

= $70molCO_2$

The next part is, how we choose 35.0 moles of acetylene and not 84.0 moles of oxygen.

From balanced equation, there is 2:5 mol ratio between acetylene and oxygen. Let's calculate the moles of oxygen required to react completely with 35.0 moles of acetylene.

$35.0molC_2H_2(\frac{5molO_2}{2molC_2H_2})$

= $87.5molO_2$

Calculations shows that 87.5 moles of oxygen are required to react completely with 35.0 moles of acetylene. Since only 84.0 moles of oxygen are available, the limiting reactant is oxygen, so 35.0 moles of acetylene will not react completely as it is excess reactant.

So, the theoretical yield should be calculated using 84.0 moles of oxygen as:

$84.0molO_2(\frac{4molO_2}{5molO_2})$

= $67.2molCO_2$

7 0

4 years ago

Read 2 more answers

At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

algol [13]

Answer:

820 K

Explanation:

As per Boltzman equation, kinetic energy (KE) is in direct relation to the temperature, measured in absolute scale Kelvin.

KE α T.

Then, the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be 3 times such temperature.

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80