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3 years ago

Please help

Chemistry

2 answers:

svlad2 [7]3 years ago

6 0

The heavy particles all passed straight through the foil, because the atoms are mostly empty space.

Answer is third choice

eimsori [14]3 years ago

6 0

Answer: Some of the heavy particles bounced off the foil, because there is a dense, positive area in the atom.

Explanation:

In Rutherford's experiment, he took a gold foil and bombarded it with alpha particles which carry positive charge. He thought that the alpha particles will pass straight through the foil, but to his surprise, many of them passed through, some of them deflected their path and a few of them bounced back.

From this he concluded that in an atom, there exist a small positive charge in the center. Due to this positive charge, the alpha particles deflected their path and some of them bounced straight back their path.

Thus he concluded that there is a dense, positive area in the atom.

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A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to

Snezhnost [94]

Explanation:

It is known that $pK_{a}$ of propionic acid = 4.87

And, initial concentration of propionic acid = $\frac{0.19}{1.20}$

= 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

= 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence, pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.216}{0.158}$

= 4.87 + log (1.36)

= 5.00

Therefore, when 0.02 mol NaOH is added then,

Moles of propionic acid = 0.19 - 0.02

= 0.17 mol

Hence, concentration of propionic acid = $\frac{0.17}{1.20 L}$

= 0.14 M

and, moles of sodium propionic acid = (0.26 + 0.02) mol

= 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.28 mol}{1.20 L}$

= 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.23}{0.14}$

= 4.87 + log (1.64)

= 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

Therefore, when 0.02 mol HI is added then,

Moles of propionic acid = 0.19 + 0.02

= 0.21 mol

Hence, concentration of propionic acid = $\frac{0.21}{1.20 L}$

= 0.175 M

and, moles of sodium propionic acid = (0.26 - 0.02) mol

= 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

$\frac{0.24 mol}{1.20 L}$

= 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

pH = $pK_{a} + log(\frac{[salt]}{[acid]})$

= $4.87 + log \frac{0.2}{0.175}$

= 4.87 + log (0.114)

= 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

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3 years ago

What is NOT a cause of an oil spill?

katrin [286]

Answer:

The answer is #1

Explanation:

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3 years ago

How many lithium ions are present in 30.0 ml of 0.600 m li2co3 solution?

Nikolay [14]

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3 years ago

A certain compound is made up of one phosphorus (P) atom, three fluorine (F) atoms, and one oxygen (O) atom. What is the chemica

poizon [28]

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