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A thorium-232 nucleus decays by a series of alpha and beta emissions until it reaches astatine-212. how many alpha emissions and

Arturiano [62]

There are 5 alpha and 5 beta particle emit emission of thorium-232 nucleus into astatine-212.

A beta ray would be released from such an atomic nucleus during a radioactive decay process known as beta decay. The proton in the nucleus changes from a proton to a neutron throughout beta decay, and vice versa.

It is given that, The atomic mass of Th = 232

Atomic mass of astatine = 212

Atomic number of Th = 90.

Atomic number of astatine = 85

Number of alpha particle = difference between atomic mass / 4

Number of alpha particle = 232-212 /4

Number of alpha particle = 20/4

Number of alpha particle = 5

Thus, the number of alpha particle will be 5.

Number of beta particle = difference between atomic number

Number of beta particle = 90-85

Number of beta particle = 5

Thus, the number of beta particle will be 5

Therefore, the number of alpha and beta particle will be 5 and 5.

To know more about beta particle

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How chemicals in chemical reactions are an important part of our life

pashok25 [27]

Because if we didnt have chemical reactions we would have life because we use chemical reactions to live

7 0

3 years ago

The ionization constant for acetic acid is 1.8 × 10-5, and that for hydrocyanic acid is 4 × 10-10.

mihalych1998 [28]

Answer:

D. [H+] of the acetic acid solution is greater than that of the hydrocyanic acid solution.

Explanation:

$CH_3 COOH(aq)+ H_2 O(l) CH_3 COO^- (aq)+H_3 O^+ (aq)$

Initial 0.1M 0 0

Change -x +x +x

Equilibrium 0.1M-x +x +x

$Ka =\frac {((x)(x))}{(0.1M-x)}$

$1.8\times10^{-5}= \frac {x^2}{(0.1M-x)}$

(-x is neglected) so we get

$1.8\times10^{-5}\times0.1=x^2\\\\x^2=1.8\times10^{-6}$

$x=\sqrt{x^2}=1.34\times10^{-3} M=H^3 O^{+}$ is the $[H^+ ]$ concnetration of acetic acid

$HCN(aq)+ H_2 O(l) CN^- (aq)+H_3 O^+ (aq)$

Initial 0.1M 0 0

Change -x +x +x

Equilibrium 0.1M-x +x +x

$Ka =\frac {((x)(x))}{(0.1M-x)}$

$4.0\times10^{-10}= \frac {x^2}{(0.1M-x)}$

(-x is neglected) so we get

$4.0\times10^{-10}\times0.1=x^2\\\\x^2=4.0\times10^{-11}$

$x=\sqrt{x^2}=6.32\times10^{-6} M=H^3 O^{+}$ is the $[H^+ ]$ concnetration of Hydrocyanic acid

Thus we see here, [H+] of the acetic acid solution $1.34 \times 10^{-3} M$ is greater than that of the hydrocyanic acid solution $6.32 \times10^{-6} M$