In the late 1800s, the British scientist J. J. Thomson discovered that atoms were composed of positively-charged and negatively-charged particles. He proposed that these two types of particles were evenly mixed throughout the structure of the atom. How does this atomic model differ from current atomic models?A. Current atomic models show that atoms do not have positively-charged particles.

B. Current atomic models show an atom's positively-charged particles concentrated in the center of the atom.

C. Current atomic models show that atoms do not have negatively-charged particles.

Answers

Answer 1
Answer: Option B.

Current atomic models show the positive charged particles (protons) concentrated in the center (nucleus) of the atom, along with the neutrons, and the negative charged particles (electrons) distributed around (far from) the nucleus.

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An ion that consists of 7 protons, 6 neutrons, and 10 electrons has a net charge of(I ) 4- (3) 3+(2) 3- (4) 4+

Suppose you are asked to find the area of a rectangle that is 2.1-cm wide by 5.6-cm long. Your calculator answer would be 11.76 cm2 . Now suppose you are asked to enter the answer to two significant figures. (Note that if you do not round your answer to two significant figures, your answer will fall outside of the grading tolerance and be graded as incorrect.)

Answers

1) ALL non-zero numbers (1,2,3,4,5,6,7,8,9) are ALWAYS significant.2) ALL zeroes between non-zero numbers are ALWAYS significant.3) ALL zeroes which are SIMULTANEOUSLY to the right of the decimal point AND at the end of the number are ALWAYS significant.4) ALL zeroes which are to the left of a written decimal point and are in a number >= 10 are ALWAYS significant.

so it will be 11.76 = 12 bc 7 rounds up

As the result of experimental work, a student finds the density of a liquid to be .137 g/ml. the known density of the liquid is .1478 g/ml. find the student's experimental error (ee) and percentage error (pe).

Answers

the known density of the liquid is

d_known = 0.1478 g /ml

the density of a liquid

d_calculated = 0.137 g /mol

experimental error = d_known – d_calculated

= 0.1478 – 0.137 = 0.0108 g /mL

Thus, experimental error of 0.0108 g /mL is present.


Percentage error = [ (d_known – d_calculated) / (d_known) ] * 100

=[ (0.0108) / 0.1478 ] * 100

= 7.31 %

Thus, Percentage error of 7.31 % is present.

What is the pH of a 2.0 M solution of HClO4?

Answers

HClO4 is a strong acid. Strong acids dissociate completely in water. 2.0 M of HClO4 is equal to 2.0 M of the H+ ion component of the substance. To solve the pH, the ff formula is to be used: 
pH=-log[H+]

Which one has the less density?

Answers

Answer:

A

Large mass but not as much volume, so not as dense...

Explanation:

Answer:

C

Explanation:

You take the mass/volume to find density.

Bromothymol blue is an indicator that turns yellow in solutions when the ph is below 6, green when the ph is between 6 and 8, and blue when the ph is above 8. what color would it turn when added to water? a. yellow b. green c. blue d. pink

Answers

b. green

water solutions have neutral pH ( ~7) so the indicator will turn green

Final answer:

Bromothymol blue would become green when added to water, due to water's neutral pH of around 7. This response uses pH levels and the color changes of bromothymol blue as indicators of acidity or basicity.

Explanation:

The color of bromothymol blue in water would be green. That's because the pH level of pure water is around 7, which falls within the range of 6 to 8 where bromothymol blue would turn green. Bromothymol blue is an indicator used in chemistry to identify pH levels by presenting different colors in solutions of different pHs: it turns yellow in solutions under pH 6, green between pH 6 to 8, and blue when the pH is above 8.

Learn more about Bromothymol Blue in Water here:

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Consider the following reaction: 2HCl + Ca(OH)2 CaCl2 + 2H2O A scientist wants to neutralize 25 mL of 0.001 M Ca(OH)2 with a volume of 0.005 M HCl. What is the minimum volume of HCl required?

Answers

The balanced chemical equation for the above reaction is as follows ;
Ca(OH)2 + 2HCl —> CaCl2 + 2H2O
Stoichiometry of Ca(OH)2 to HCl is 1:2
Number of Ca(OH)2 moles present - 0.001 mol/L / 1000 mL/L x 25 mL
Number of Ca(OH)2 moles = 2.5 x 10^(-5) mol
Number of HCl moles needed for neutralisation = 2.5 x 10^(-5) mol x2 = 5 x 10 ^(-5) mol
The molarity of HCl solution is 0.005 M
The solution contains 0.005 mol in 1000 mL
Therefore volume of 5x10^(-5) mol in = 1000/0.005 x 5 x 10^(-5) mol = 10 mL
10 mL of HCl is needed for neutralisation