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SVEN [57.7K]
3 years ago
11

Write a nuclear equation to represent radium-226 decaying to radon-222 plus an alpha particle.

Chemistry
1 answer:
valentina_108 [34]3 years ago
4 0
Answer:

<span><span>22688</span>Ra→<span>l<span>22286</span></span>Rn+<span>l<span>42</span></span>He</span>

Explanation:

An α-particle is a helium nucleus. It contains 2 protons and 2 neutrons, for a mass number of 4.

During α-decay, an atomic nucleus emits an alpha particle. It transforms (or decays) into an atom with an atomic number 2 less and a mass number 4 less.

Thus, radium-226 decays through α-particle emission to form radon-222 according to the equation:

<span><span>22688</span>Ra→<span>l<span>22286</span></span>Rn+<span>l<span>42</span></span>He</span>

Note that the sum of the subscripts (atomic numbers or charges) is the same on each side of the equation.

Also, the sum of the superscripts (masses) is the same on each side of the equation.

Now try figuring out Am-241

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Katyanochek1 [597]

Answer:

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2 years ago
An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterize
oee [108]

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s =  M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r =  M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

8 0
3 years ago
Explain how impurities are selected for doping in group 14 semiconductors
aleksklad [387]

Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

<h3>Meaning of Semiconductors</h3>

Semiconductors can be defined as any material that has the ability to exhibit some properties of a conductor and some properties of an insulator.

A semiconductor can be used as either a conductor or an insulator when worked upon.

In conclusion, Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

Learn more about semiconductors: brainly.com/question/1918629

#SPJ1

6 0
1 year ago
Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
mamaluj [8]

Answer:

Explanation:

From the given information:

The equation for the reaction can be represented as:

2SO_2 + O_2 \to 2SO_3

The I.C.E table can be represented as:

                     2SO₂              O₂                   2SO₃

Initial:             14                  2.6                     0

Change:        -2x                -x                      +2x

Equilibrium:   14 - 2x          2.6 - x                2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

SO₃ = 2x,

then x = 1.6/2

x = 0.8 mol

For 2SO₂; we have 14 - 2x

= 14 - 2(0.8)

= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,  

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction 2SO_2 + O_2 \to 2SO_3 = 0.8325;

If we want to find:

SO_2 + \dfrac{1}{2}O_2 \to SO_3

Then:

K_c = (0.8325)^{1/2}

\mathbf{K_c = 0.912}

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

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