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
Pepsi [2]
4 years ago
15

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

Chemistry
2 answers:
boyakko [2]4 years ago
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.

You might be interested in
A 25 kg rock is placed in a graduated cylinder with water.the volume of the fluid is 18.3ml.calculate the density of the rock in
ipn [44]

Answer:

=> 1366.120 g/mL.

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m)     = 25 Kg

Volume (v) = 18.3 mL.

From our question, we are to determine the density (rho) of the rock.

The formula:

p = \frac{m}{v}

First let's convert 25 Kg to g;

1 Kg    = 1000 g

25 Kg = ?

= \frac{25 × 1000}{1}

= 25000 g

Substitute the values into the formula:

p =  \frac{25000 g}{18.3 ml}

= 1366.120 g/mL.

Therefore, the density (rho) of the rock is  1366.120 g/mL.

8 0
2 years ago
What is an orbital?
juin [17]

Answer:

Relating to orbit or orbits

Explanation:

6 0
3 years ago
Read 2 more answers
Explain how research from the glomar challenger helped scientist support the theory of seafloor spreading?
Anna71 [15]

Drill cores from the ocean floor were dated and found to be very young compared to the age of the earth. This means the crust had to be formed recently, which can be explained by creation of crust at a spreading center.

3 0
4 years ago
Read 2 more answers
Read the hypotheses below, then answer the questions.
stiks02 [169]
A the type of soil the tulip receive- with or without fertilizer
5 0
3 years ago
Read 2 more answers
What is the ratio of lactic acid (Ka = 1.37x^10-4) to lactate in a solution with pH =4.29
hram777 [196]

Henderson–Hasselbalch equation is given as,

                                         pH  =  pKa  +  log [A⁻] / [HA]   -------- (1)

Solution:

Convert Ka into pKa,

                                         pKa  =  -log Ka

                                         pKa  =  -log 1.37 × 10⁻⁴

                                         pKa  =  3.863

Putting value of pKa and pH in eq.1,

                                         4.29  =  3.863 + log [lactate] / [lactic acid]

Or,

                   log [lactate] / [lactic acid]  =  4.29 - 3.863

                   log [lactate] / [lactic acid]  =  0.427

Taking Anti log,

                             [lactate] / [lactic acid]  =  2.673

Result:

           2.673 M  lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.

6 0
3 years ago
Other questions:
  • Energy is transferred thru heat by each of the following methods EXCEPT?
    5·2 answers
  • Magnesium has 2 valence electrons, and oxygen has 6 valence electrons. Which type of bonding is likely to occur between a magnes
    8·2 answers
  • 8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a
    10·1 answer
  • When the pH of 0.10 m HClO2(aq) was measured, it was found to be 1.2. a. What are the values of Ka and pKa of chlorous acid? b.
    7·1 answer
  • Please help, will mark brainliest
    6·1 answer
  • Which methods could be used to dilute a solution of sodium chloride
    13·1 answer
  • What is another factor that can influence osmosis?
    11·2 answers
  • Ean rule for [Fe(Co) 2(No)2]​
    8·1 answer
  • (GIVING BRAINIEST!)
    5·2 answers
  • What is the empirical formula of CoH1803? (4 points)
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!