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evablogger [386]
2 years ago
14

Rutherford and his colleagues fired a beam of small, positively charged alpha particles at thin gold foil. Some of these alpha p

articles were deflected at angles as they passed through the foil.
What conclusion did Rutherford and his colleagues draw from their results?
A) The atom must contain some neutral charge spread throughout the atom. He called these particles neutrons.
B) The atom must contain some centrally located negative charge. He called this core of negative charge the nucleus.
C) The atom must contain some centrally located positive charge. He called this core of positive charge the nucleus.
D) Thomson's plum pudding model of the atom was correct. The atom contains positive charge spread throughout its structure.
Chemistry
1 answer:
horsena [70]2 years ago
7 0

C) The atom must contain some centrally located positive charge. He called this core of positive charge the nucleus.

Explanation:

Rutherford concluded from his experiment that the atom contains some centrally located positive charge. He called the core the nucleus.

Rutherford developed the nuclear model of the atom.

  • In a bid to explain his observation, Rutherford suggested the nuclear model.
  • The model consists of small positively charged center called the nucleus.
  • Nearly all the mass is concentrated in the nucleus.
  • The extranuclear space is made up of electrons orbiting the nucleus.

Learn more:

Rutherford brainly.com/question/1859083

#learnwithBrainly

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Calculate the pH of 1.00 L of a buffer that contains 0.105 M HNO2 and 0.170 M NaNO2. What is the pH of the same buffer after the
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Answer:

1.- pH =3.61

2.-pH =3.53

Explanation:

In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].

In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.

First Part

pH  = 3.40+ log {0.170 /0.105}

pH =  3.61

Second Part

# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012

# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)

# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol

# mol NaNO₂ remaining = (0.170 - 0.012) mol  = 0.158

# mol HNO₂ produced = 0.012 mol

# mol HNO₂ initial = 0.105

# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol

Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.

pH = 3.40 + log {0.158/.0117}

pH = 3.53

Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.

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