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3 years ago

Identify the product of radioactive decay and classify the given nuclear reactions accordingly.

2 answers:

8 0

Radioactive decay is the means by which unstable nuclei move towards stability by emitting energy and subatomic particles. The types are:

Alpha particle decay:
Alpha particles, which are identical to helium nuclei, are released. The atomic mass of the substance decreases by 4 units and the atomic number decreases by 2.

Beta particle decay:
These particles are similar to mass as an electron, and during this decay, the atomic mass remains the same but atomic number increases by 1.

Gamma ray decay:
Gamma rays are released, which are high energy electromagnetic waves, by some radioactive nuclei to rid the excess energy they have.

Eva8 [605]3 years ago

8 0

The product of radioactive decay :

₂He⁴
₋₁e⁰
γ
₁e⁰

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

The main particles are emitted by radioactive elements so that they generally decay are alpha (α), beta (β) and gamma (γ) particles

General formulas used in decay:

$\large {\boxed {Nt = No. (\frac {1} {2}) ^ {\frac {T} {t1 / 2}}}}$

Information:

T = duration of decay

t1 / 2 = elemental half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

At the core, the reaction applies the law of eternity

• 1. energy

the energy before and after the reaction is the same

• 2. atomic number

the number of atomic numbers before and after the same

• 3. mass number

the number of mass numbers before and after the same

Particles that play a role in core reactions include

• alpha α particles ₂He⁴

atomic number decreases by 2, mass number decreases by 4

• beta β ₋₁e⁰ particles

atomic number remains the same , mass number decreases by -1

• gamma particles γ

Core reactions that occur usually release high energy

• positron particles ₁e⁰

atomic number remains the same , mass number decreases by +1

In general, the core reaction equation can be written:

X = target core

a = particle fired

Y = new core

b = the particle produced

Q = heat energy

or can be written simply

X (a, b) Y

<h3>Learn more </h3>

the identity of the missing nucleus

brainly.com/question/4207569

the process of radioactive decay

brainly.com/question/1770619

nuclear fusion and nuclear fission reactions

brainly.com/question/1949468

the product of radioactive decay

brainly.com/question/6029197

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What is the molar solubility of nickel(II) sulfide in 0.091 M KCN? For NiS, Ksp = 3.0 × 10 –19; for Ni(CN) 4 2–, Kf = 1.0 × 10 3

Marta_Voda [28]

Answer:

The value is $x = 0.0227 \ M$

Explanation:

From the question we are told that

The concentration of $KCN \ \ i.e \ \ CN^{-}$ is $M_1 = 0.091 \ M$

The solubility product constant for $NiS$ is $K_{sp} = 3.0 *10^{-19}$

The stability constant for $Ni(CN)_4 ^{2-}$ is $K_f = 1.0 *10^{31}$

Generally the dissociation reaction for NiS is

$Ni S \underset{}{\stackrel{}{\rightleftharpoons}} Ni^{2+} + S^{2-}$

Generally the formation reaction for $Ni(CN)_4 ^{2-}$ is

$4CN^- + N_i ^{2+} \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_{4}$

Combining both reaction we have

$4CN^ - + NiS \ \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_4 + S^{2-}$

Gnerally the equilibrium constant for this reaction is

$K_c = K_{sp} * K_f$

=> $K_c = 3.0 *10^{-19 } * 1.0 *10^{31}$

=> $K_c = 3.0*10^{12}$

Generally the I C E table for the above reaction is

$4CN^ - \ \ \ + \ \ \ NiS \ \ \ \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ Ni(CN)^{2-}_4 \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ S^{2-}$

initial [ I] 0.091 0 0

Change [C] -4x +x + x

Equilibrium [E ] 0.091 - 4x x x

Here is x is the amount in term of concentration that is lost by $CN^-$ and gained by $Ni(CN)_4 ^{2-}$ and $S^{2-}$

Gnerally the equilibrium constant for this reaction is mathematically represented as

$K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}$

=> $3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}$

=> $3.0*10^{12}* [0.091 - 4x ]^4 = x^2$

=> $[0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}$

=> $[0.091 - 4x ] = \sqrt[4]{ \frac{x^2}{3.0*10^{12}}}$

=> $[0.091 - 4x ] = \frac{\sqrt{x} }{1316}$

=> $119.8 - 5264x =\sqrt{x}$

Square both sides

$(119.8 - 5264x)^2 =x$

=> $14352.04 - 1261255 x + 27709696x^2 = 0$

=> $27709696x^2 - 1261255 x + 14352.04 = 0$

Solving using quadratic equation

The value of x is $x = 0.0227 \ M$

Hence the amount in terms of molarity (concentration) of $Ni(CN)_4 ^{2-}$ and $S^{2-}$ produced at equilibrium is $x = 0.0227 \ M$ it then means that the amount of NiS (nickel(II) sulfide) lost at equilibrium is $x = 0.0227 \ M$

So the molar solubility of nickel(II) sulfide at equilibrium is

$x = 0.0227 \ M$

3 0

3 years ago

For the decomposition of hydrogen peroxide in dilute sodium hydroxide at 20 °C 2 H2O2(aq)2 H2O(l) + O2(g) the following data hav

saul85 [17]

Answer:

K= 0.06611

Explanation: The rate of reaction is defined as the change in concentration of any of reactant or products per unit time. From the given reaction, the rate of reaction may be equal to the rate of disappearance of reactant which is equal to the rate of appearance of products.

The average rate of disappearance of H2O2 over the time period from t=0 min at 8.92×10^-2 to t=9.63min at 4.72×10^-2 is given as -4.36×10^-3Mmin-1.

We can say:

•The initial concentration [H2O2]o is 8.92×10^-2M

•The concentration at time t. [H2O2]t is 4.72×10^-2

•The time (t) is 9.63 min

The expression of rate constant for a first order reaction is shown as

K=2.303/t log[H2O2]o/ [H2O2]t

Substitute the values of t, [H2O2]o and [H2O2]t in the equation of rate constant.

K=2.303/9.63 log [8.92×10^-2]/ [4.72×10^-2]

K= 0.2391 (log 8.92×10^-2 - log 4.72×10^-2)

K= 0.2391 [-1.0496-(-1.3261)]

K= 0.2391 (-1.0496+1.3261)

K= 0.2391 (0.2765)

K= 0.06611

Since the value of k is almost constant, the decomposition of H2O2 is a first order reaction.

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3 years ago

What is the formula for phosphorus trihydride

Usimov [2.4K]

PH3
Also known as phosphine

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