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ivolga24 [154]
3 years ago
9

Which of the following observations indicates that an atom has neutrons?

Chemistry
2 answers:
olganol [36]3 years ago
8 0

Answer:

The correct answer is C: A radiation consisting of uncharged particles is emitted when alpha particles strike beryllium atoms.

Explanation:

Chadwick discovered neutron while experimenting with a gold foil. A stream of alpha particles produced from a polonium source was directed at beryllium target. It was noticed that some penetrating radiations were produced. These uncharged particles were called neutrons because on the charge detector these particles showed no deflection.

Their nuclear reaction is as follow.

                             ₂He + ₄Be ----------> ₀n + ₆C

Chadwick noticed that neutrons cannot ionize gases and are highly penetrating particles.

-Dominant- [34]3 years ago
4 0

Answer: C. A radiation consisting of uncharged particles is emitted when alpha particles strike beryllium atoms.

A. Some uncharged particles are scattered by a beryllium atom when it hits a gold foil.

B. Some uncharged particles bounce back from a gold foil when it is bombarded with alpha particles.

C. A radiation consisting of uncharged particles is emitted when alpha particles strike beryllium atoms.

D. A radiation which attracts electrons is produced when a beryllium atom is bombarded with alpha particles.

Explanation:

I just took the test and the answer is C. A radiation consisting of uncharged particles is emitted when alpha particles strike beryllium atoms.

Good Luck!!

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What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

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3 years ago
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