How many molecules are 0.5000 miles of H2?

determine the percent error if the density of mercury is 13.534g/mL and you measured the density to be 13.000g/mL

9966 [12]

percentage error=difference/actual value x 100

=0.534/13.0 x 100

=4.11%

3 0

3 years ago

If a single gold atom has a diameter of 2.9×x10−8 cm, how many atoms thick was rutherford's foil

ale4655 [162]

27,586

<h3>Further explanation</h3>

<u>Given:</u>

A single gold atom has a diameter of $\boxed{ \ 2.9 \times 10^{-8} \ cm. \ }$

From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately $\boxed{ \ 8 \times 10^{-3} \ mm. \ }$

<u>Question:</u>

How many atoms thick were Rutherford's foil?

<u>The Process:</u>

Convert thickness from mm to cm.

$\boxed{ \ 8 \times 10^{-3} \ mm = 8 \times 10^{-3} \times 10{-1} \ cm \ } \rightarrow \boxed{ \ 8 \times 10^{-4} \ cm \}$

The number of atoms is calculated from gold foil thickness divided by the atomic diameter.

$\boxed{ \ = \frac{8 \times 10^{-4} \ cm}{2.9 \times 10^{-8} \ cm} \ }$

$\boxed{ \ =2.7586 \times 10^4 \ atoms \ }$

Therefore, we get an atomic thickness of 27,586 atoms.

<u>Notes:</u>

In 1909-1910, Ernest Rutherford with two of his assistants, namely Hans Geiger and Ernest Marsden, conducted a series of experiments to find out more about the arrangement of atoms. They fired at a very thin gold plate with high-energy alpha particles.
One of their observations is that a small portion of alpha particles are reflected. This greatly surprised Rutherford. The reflected alpha particle must have hit something very dense in the atom. This fact is incompatible with the atomic model proposed by J.J. Thomson where the atoms are described as homogeneous in all parts with electrons and protons evenly distributed.
In 1911, Rutherford was able to explain the scattering of alpha rays by proposing ideas about atomic nuclei. According to him, most of the mass and positive charge of atoms are concentrated at the center of the atom, hereinafter referred to as the nucleus.

<h3>Learn more</h3>

The energy density of the stored energy brainly.com/question/9617400
The theoretical density of platinum which has the FCC crystal structure. brainly.com/question/5048216
Compound microscope brainly.com/question/4000241

Keywords: if a single gold atom, has a diameter of 2.9 x 10⁻⁸ cm, how many, atoms thick, Rutherford's foil, his scattering experiment

7 0

3 years ago

Read 2 more answers

Predict the products of the following reaction: <br> RbNO2 +BaCO3->

sukhopar [10]

Answer:

2RbNO2 + BaCO3 → Rb2CO3 + Ba(NO2)2

Explanation:

The balanced reaction equation is shown below;

2RbNO2 + BaCO3 → Rb2CO3 + Ba(NO2)2

This reaction is possible because the reduction potential of Rb is -2.98V while that of Ba is –2.92 V. Hence Rb can displace Ba from its salt solution.

The equation is balanced since the number of atoms of each element on the left and right hand sides of the reaction equation are equal.

6 0

3 years ago

Which is generally true of enzymes in cells?

Wewaii [24]

The answer is A, enzymes allow reactions to proceed at body temperature.

4 0

4 years ago

Read 2 more answers

If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone

GREYUIT [131]

Answer:

m of NH3 = 6.46 g

Explanation:

First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:

NH₃ + O₂ ---------> NO + H₂O

Now, let's balance the equation:

4NH₃ + 5O₂ ---------> 4NO + 6H₂O

Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:

MM NH3 = 17 g/mol

MM O2 = 32 g/mol

moles NH3 = 18.1 / 17 = 1.06 moles

moles O2 = 27.2 / 32 = 0.85 moles

Now, let's compare these moles with the theorical moles that the balanced equation gave:

4 moles NH3 --------> 5 moles O2

1.06 moles ----------> X

X = 1.06 * 5 / 4 = 1.325 moles of O2

These means in order to NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.

Now, let's see how many grams in excess we have left after the reaction is complete.

4 moles NH3 --------> 5 moles O2

X moles NH3 ----------> 0.85 moles

X = 0.85 * 4 / 5 = 0.68 moles of NH3

This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:

moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles

Finally the mass:

m = 0.38 * 17

8 0

3 years ago