1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
SIZIF [17.4K]
3 years ago
15

The radius of an atom of gold (Au) is about 1.35 Å. How many gold atoms would have to be lined up to span 9.0 mm ?

Chemistry
2 answers:
Georgia [21]3 years ago
8 0

We know that the relationship between angstrom and mm is 1:1x10^{-7}.

Knowing the radius of an atom of gold (1.35 angstroms), we can solve for how many atoms fit into 9.0mm:

\frac{1.35angstrom}{1} *\frac{1x10^{-7}mm}{1angstrom} =1.35x10^{-7}mm

\frac{9.0mm}{1} ÷ \frac{1.35x10^{-7}mm}{1}

\frac{9.0mm}{1} *\frac{1}{1.35x10^{-7}mm}=\frac{9.0x10^{0}}{1.35x10^{-7}}= 6.7x10^{7}

Therefore, we now know that 6.7x10^{7} atoms of gold (Au) will line up to span 9.0mm.

atroni [7]3 years ago
3 0

The radius of an atom of gold (Au) is about 1.35 Å. Gold atoms would have to be lined up to span 9.0 mm in the amount of 3.333 * 10^7

<h3>Further explanation </h3>

Gold is a chemical element with the symbol Au and atomic number 79. Its one of the higher atomic number elements that occured naturally. In its purest form, it is a bright, slightly reddish yellow, dense, soft, malleable, and ductile metal

The radius of an atom of gold (Au) is about 1.35 Å. How many gold atoms would have to be lined up to span 9.0 mm ?

Its known that the diameter of an atom is twice the radius = 2*1.35 angstrom = 2.70 angstrom

Then we convert the diameter of an atom in mm

2.70 angstrom * \frac{10^-10 m}{1 angstrom} cot \frac{10^3 mm}{1m} = 2.70 * 10^-7 mm

Then to calculate the number of the gold atoms, we have to divide 1 mm by the diameter of an atom

\frac{9 mm}{2.7*10^-7 mm} = 3.333 * 10^7

<h3>Learn more</h3>
  1. Learn more about gold brainly.com/question/1462511
  2. Learn more about atom brainly.com/question/11829675
  3. Learn more about The radius of an atom brainly.com/question/10452161

<h3>Answer details</h3>

Grade:  9

Subject:  chemistry

Chapter:  atom

Keywords:  span, The radius of an atom, atom, gold, mm

You might be interested in
He distance from Earth to the North Star is about 430 light-years. Which of the following statements describes why scientists us
garri49 [273]

Answer:

D

Explanation:

because the rest don’t make any sense

8 0
2 years ago
2 KClO_2 produces 2 KCl+ 3O_2 when heated. If this reaction produces 82.8 g of KCl how many grams of O2 were produced?
Ad libitum [116K]

Answer:

37.046 grams of oxygen gas were produced.

Explanation:

2KClO_2\rightarrow 2KCl+ 3O_2

Moles of potassium chlorite = \frac{82.2 g}{106.5 g/mol}=0.7718 mol

According to reaction 2 moles of potassium chlorite gives 3 moles of oxygen gas.

Then 0.7718 moles of potassium chlorite will give:

\frac{3}{2}\times 0.7718 mol=1.1577 mol of oxygen gas.

Mass of 1.1577 moles of oxygen gas:

1.1577 mol × 32 g/mol = 37.046 g

37.046 grams of oxygen gas were produced.

4 0
3 years ago
What element has the same number of valence electrons as sulfur?
FinnZ [79.3K]
Oxygen. Is the correct answer


Oxygen has the same number of valence electrons as sulfur. An ion can be an element that gained or lost an electron.
6 0
2 years ago
Use standard enthalpies of formation to calculate ΔH∘rxn for the following reaction______.2H2S(g)+3O2(g)→2H2O(l)+2SO2(g) ΔH∘rxn
Ahat [919]

Answer:

-1,103.39KJ/mol

Explanation:

We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.

In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.

The standard enthalpies of the molecules above are as follows:

H2S = -20.63KJ/mol

H2O = -285.8KJ/mol

SO2 = -296.84KJ/mol

O2 = 0KJ/mol

ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3

 ΔfH⦵(O2)]

ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]

-[ 3 × -20.63)]

= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol

6 0
3 years ago
Which gas law would apply to the following scenario.
chubhunter [2.5K]
The ansewer is Boyle’s law
4 0
3 years ago
Other questions:
  • Tim and Jim finished their lab experiment earlier than their classmates. What should they do while they wait for the others to f
    5·1 answer
  • Bleaching powder was added to excess concentrated hydrochloric acid state whar was observed​
    10·1 answer
  • With a manual transmission, the speed of the vehicle determines
    8·1 answer
  • How many significant figures are in 11.005 g​
    9·1 answer
  • The chart seen here shows the decay of a sample of carbon-14 over a period of 33,000 years. What is the approximate half-life of
    8·2 answers
  • Earthvhas several ideal features which enable it to support life. MOST important is the
    8·1 answer
  • Based on this reaction, how many moles of H2 can be obtained starting with 3 mol CH4?
    13·1 answer
  • Describe the phases/states of matter, excluding plasma
    6·1 answer
  • Circulatory Quiz - 1 chance
    7·1 answer
  • Uranium is classified as what type of element?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!