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SpyIntel [72]
3 years ago
11

The radius of an atom of gold (Au) is about 1.35 Å.

Chemistry
2 answers:
Degger [83]3 years ago
8 0

Answer : The number of gold atoms will be, 3.52\times 10^7

Explanation :

First we have to determine the diameter of an atom of gold.

Diameter=2\times Radius

Given :

Radius of an atom of gold = 1.35\AA

Diameter=2\times 1.35\AA=2.7\AA

Conversion used : (1\AA=10^{-7}mm)

Diameter=2.7\AA=2.7\times 10^{-7}mm

Now we have to calculate the number of gold atoms.

\text{Number of gold atoms}=\frac{\text{Span length}}{\text{Diameter of an atom of gold}}

\text{Number of gold atoms}=\frac{9.5mm}{2.7\times 10^{-7}mm}=3.52\times 10^7

Therefore, the number of gold atoms will be 3.52\times 10^7

Eduardwww [97]3 years ago
6 0
Two radius of an atom is equal to the diameter. Adding up all the diameter of  the atoms, it should be equal to 9.5 mm. Therefore, we simply convert the units to the same units then divide 1.35 A to 9.5 mm. We calculate as follows:

no. of atoms = 0.0095 m / 1.35x10^-9 m = 7037037 atoms 

Hope this answers the question. Have a nice day.
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How many liters of CO2 gas can be produced at 30.0 °C and 1.50 atm from the reaction of 5.00 mol of C3H8 and an excess of O2 acc
lbvjy [14]

Answer:

249 L

Explanation:

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C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈

The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol

Step 3: Convert "30.0°C" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 30.0°C + 273.15 = 303.2 K

Step 4: Calculate the volume of carbon dioxide

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm

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5 0
3 years ago
Which choice gives the correct oxidation numbers for all three elements in rb2so3 in the order that the elements are shown in th
vlabodo [156]
<h3>Answer:</h3>

             Rb  =  + 1

              S    =  + 4

              O   =  - 2

<h3>Explanation:</h3>

                   Oxidation states of the elements were calculated keeping in mind the basic rules of assigning oxidation states which included assignment of +1 charge to first group elements i.e. Rubidium (Rb) and assignment of -2 charge to Oxygen atom. Then the oxidation state of Sulfur was calculated as follow,

Rb₂ + S + O₃  =  0

Above zero (0) means that the overall molecule is neutral.

Putting values of Rb and O,

                                           (+1)₂ + S + (-2)₃  =  0

                                           (+2) + S + (-6)  =  0

                                           +2 + S - 6  = 0

                                           S - 6  =  -2

                                          S  =  -2 + 6

                                           S  =  + 4

7 0
3 years ago
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