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Klio2033 [76]
1 year ago
7

3.the three major types of radioactive decay of an unstable nucleus are alpha particles, beta particles, and gamma rays.

Chemistry
1 answer:
Alenkinab [10]1 year ago
5 0

a) Alpha decay occurs when an unstable nuclide emits an alpha particle, (which is just a helium-4 atom), which contains 2 protons and 4 neutrons. This means that transmutation has occurred (since a nuclide of a different element will be produced), and this newly produced nuclide will have 2 fewer protons and 4 fewer neutrons than the original nuclide.

b) ^{234}_{92} \text{U} \longrightarrow ^{4}_{2} \text{He}+^{230}_{90} \text{Th}

c) Alpha decay occurs when an unstable nuclide emits a beta particle, (also known as an electron). This causes transmutation (since a nuclide of a different element will be produced), and this newly produced nuclide will have 1 more proton than the original nuclide.

d) ^{131}_{53} \text{I} \longrightarrow ^{0}_{-1} \beta+^{131}_{54} \text{Xe}

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You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col
miss Akunina [59]

Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

c) X2Y, YZ15, X6Y

d) hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y  as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

mass of X = 2.0g

mass of Y = 7.0g

amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

5 0
3 years ago
What is an substance
lora16 [44]
<span>a particular kind of matter with uniform properties.

</span>
8 0
3 years ago
Read 2 more answers
Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction
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Answer:

<u>Balanced equation:</u>

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

Explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate

Pb(NO_{3})_{2}(aq)+K_{2}CO_{3}(aq)\rightarrow PbCO_{3}(s)+2KNO_{3}(aq)

<u>Ionic equation:</u>

Pb^{2+}(aq)+2NO_{3}^{-}(aq)+2K^{+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)+K^{+}(aq)+2NO_{3}^{-}

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

Pb^{2+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)

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2 years ago
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