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Nikitich [7]
3 years ago
11

What did Ernest Rutherford’s gold foil experiment demonstrate about atoms? Their positive charge is located in a small region th

at is called the nucleus. Their negative charge is located in small particles that are called electrons. Their nucleus makes up the majority of the volume of the atom. Their electrons are floating in a sea of positive charges.
Chemistry
1 answer:
sergey [27]3 years ago
5 0

Their positive charge is located in a small region that is called the nucleus.

Explanation:

Ernest Rutherford in his gold foil experiment was able to demonstrate that the nucleus is made up of positive charges which occupies a small and tiny nucleus.

Rutherford bombarded a thin gold foil with alpha particles from a radioactive source.

  1. He observed that all the particles passed through but a small portion was deflected back.
  2. This led to his proposition of the nuclear model the atom.

learn more:

Ernest Rutherford brainly.com/question/1859083

#learnwithBrainly

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Qué clase de alimentos podemos encontrar que contengan oligoelementos como Cromo (Cr), Litio (Li), Selenio (Se), Manganeso (Mn),
Slav-nsk [51]

Answer: There are many foods who have trace elements like seafood, legume, vegetables, tomatoes, bananas, eggs, fishes, meat.

Their five mainly functions are accept or receive electrons in redox reactions, act like cathalytic centers which are necessaries to live, control biologics process like the hormon activation and gen expression, provide stability, 3d structures in important molecules.

Explanation: Trace elements participate in redox reactions because they have acid-base components which alkalize o acidify the way, protecting the organism from free radicals.

Biological sources which are controlled by trace elements are:

  • assimilation of ATP
  • methabolic sources relationed with insuline
  • oxygen transport in the organism
  • stomach acid formation
  • 3d structures from important moleculess are mantained by balance of electrons which make up the atoms from the molecule.
8 0
3 years ago
A. 14.8
Scilla [17]

Answer:

Explanation:

1)

Given data:

Initial volume  = 2.5 L

Initial temperature = 300 k

Final temperature = 80 k

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 2.5 L × 80 K / 300 k

V₂ = 200 L.K / 300 K

V₂ = 0.67 L

2)

Given data:

Initial volume  = 752 mL

Initial temperature = 25.0°C (25+273 = 298 K)

Final temperature = 50.0°C(50+273 = 323 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 752 mL × 323 K / 298 k

V₂ = 242896 mL.K / 298 K

V₂ = 815.1  mL

3)

Given data:

Initial volume  = 2.75 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = ?

Final volume = 2.46 L

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂  = V₂T₁/V₁

T₂ = 2.46 L × 293 K / 2.75  L

T₂ = 720.78 L.K /  2.75 L

T₂ = 262.1 K

4)

Given data:

Initial volume  = 1500 L

Initial temperature = 5°C (5+273 = 278 K)

Final temperature = 30 °C(30+273 = 303 K)

Final volume of heated air = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1500 L × 303 K / 278 k

V₂ = 454500 L.K / 278 K

V₂ = 1634.89  L

5)

Given data:

Initial volume = 15.5 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = 7.0 °C(7.0+273 = 280 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 15.5 L × 280 K / 293 k

V₂ = 4340 L.K / 293 K

V₂ = 14.8  L

6)

Given data:

Initial volume = 150 mL

Initial temperature = 23.5°C (23.5+273 = 296.5 K)

Final temperature = 72.5 °C(72.5+273 = 345.5 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 150 mL × 345.5 K / 296.5 k

V₂ = 51825 mL.K / 296.5 K

V₂ = 174.79  mL

6 0
3 years ago
The melting point of a substance occurs at the same temperature as its _____ point. A. boiling B. freezing C. condensing
shusha [124]
A) Boiling point. :)
8 0
3 years ago
Read 2 more answers
2FePO4+3 Na2so4--->Fe2(so4) + 2 Na3Po4 if i ise 25grams of iron iii phosphate 18.5 g iton what is my percent yeild?
Strike441 [17]

Answer:

56%

Explanation:

<em>If I use 25 grams of iron (III) phosphate and obtain 18.5 g of iron (III) sulfate, what is my percent yield?</em>

Step 1: Write the balanced equation

2 FePO₄ + 3 Na₂SO₄ ⇒ Fe₂(SO₄)₃ + 2 Na₃PO₄

Step 2: Calculate the theoretical yield of Fe₂(SO₄)₃

The mass ratio of FePO₄ to Fe₂(SO₄)₃ is 301.64:399.88.

25 g FePO₄ × 399.88 g Fe₂(SO₄)₃/301.64 g FePO₄ = 33 g Fe₂(SO₄)₃

Step 3: Calculate the percent yield of Fe₂(SO₄)₃

We will use the following expression.

%yield = (experimental yield/theoretical yield) × 100%

%yield = (18.5 g/33 g) × 100% = 56%

4 0
3 years ago
How does altitude affect climate patterns in a region? [Hint: relationship of altitude and temperature]
schepotkina [342]

Answer:

Higher in the atmosphere temperature changes

Explanation: The tops of tall mountain might have ice or snow because it’s cooler in higher parts of the atmosphere.

4 0
3 years ago
Read 2 more answers
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