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Bogdan [553]
3 years ago
9

In alpha decay, alpha particles are ejected from the nucleus. Which equation represents alpha decay?

Chemistry
2 answers:
Paladinen [302]3 years ago
6 0

Answer:

₉₅Am²²⁴   →    ₉₃Np²³⁷ + ₂He⁴ is correct on edge

Alexxandr [17]3 years ago
4 0

Answer:

The first equation shows the alpha decay.

₉₅Am²²⁴   →    ₉₃Np²³⁷ + ₂He⁴

Explanation:

The americium with atomic wight  224 undergo alpha decay and produce

₉₃Np²³⁷ . The alpha particle emitted is also called helium nuclei. During this decay some gamma radiations also produce as a byproduct.

₉₅Am²²⁴   →    ₉₃Np²³⁷ + ₂He⁴

Properties of alpha radiation:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number 4 less than and atomic number 2 less than the starting atom.

Alpha radiations can travel in a short distance.

These radiations can not penetrate into the skin or clothes.

These radiations can be harmful for the human if these are inhaled.

These radiations can be stopped by a piece of paper.

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True or false elements in the same colum have simular properties
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Why can molar mass be used as a conversion factor
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From the relative atomic mass of each element, it is possible to determine each element's molar mass by multiplying the molar mass constant (1 g/mol) by the atomic weight of that particular element. The molar mass value can be used as a conversion factor to facilitate mass-to-mole and mole-to-mass conversions.

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Read 2 more answers
Which atom has a mass of 12 u
Andrews [41]

Answer:

Carbon-12

Explanation:

4 0
3 years ago
Read 2 more answers
A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0
3 years ago
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