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3 years ago

A

Chemistry

1 answer:

meriva3 years ago

6 0

Answer:

Option B is correct. A nuclear alpha decay

Explanation:

Step 1

This equation is a nuclear reaction. So it can be an alpha decay or a beta decay

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 that is showed.

A Beta decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.

Option B is correct. A nuclear alpha decay

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A 100-watt light bulb radiates energy at a rate of 100 J/s. (The watt, a unit of power or energy over time, is defined as 1 J/s.

Semmy [17]

Answer

2.7956 * 10^19 photons

Givens

Wavelength = λ = 525 * 10^-9 meters [1 nmeter = 1*10^-9 meters]
c = 3 * 10^8 meters
E = ???
W = 100 watts
t = 1 second
h= plank's Constant = 6.26 * 10^-34 J*s

Formula

E = h * c / λ

W = E / t

Solution

E = 6.26 * 10^-34 j*s * 3 * 10^8 m/s /525 * 10^-9 (m)

The meters cancel out. So do the seconds. You are left with Joules as you should be.

E = 3.577 * 10^-18 Joules

What you have found is the energy of 1 photon.

Now you have to find the Joules from the watts.

W = E/t

100 * 1 second = 100 joules

1 photon contains 3.577 * 10 ^ - 18 Joules

x photon = 100 joules

1/x = 3.577 * 10^-18 / 100 Cross multiply

100 = 3.577 * 10 ^ - 18 * x Divide both sides by 3.577 * 10 ^ - 18

100/3.577 * 10 ^ - 18 = 3.577 * 10 ^ - 18x / 3.577 * 10 ^ - 18

2.7956 * 10^19 photons = x

7 0

3 years ago

A reaction occurs in which hydrogen-1 and hydrogen-2 form helium-3. Is this a chemical reaction or a nuclear reaction, and how d

olga nikolaevna [1]

Cause -3 is -3sjsnsnsnsnsnnsnsnsnnsnsnsnsn

7 0

3 years ago

Match the following chemical reactions:

S_A_V [24]

Answer:

Synthesis - 4

reversible- 2

exchange- 1

decomposition-3

Explanation:

In synthesis reaction two or more components combines to form a single product. example 2H2+O2⇒2H2O

In reversible reaction two reactants combine to form two products . The products then reacts and forms back the reactants. example N2 +3H2 ⇒2NH3

In exchange reaction there is an alternation of ions of reactants to form new products. AB+CD ⇒AC + BD

In decomposition reaction, molecules of a compound break down by the action of heat or light or catalyst. example CaCO3 ⇒CaO +CO2

3 0

2 years ago

In the following reaction, 451.4 g of lead reacts with excess oxygen forming 365.0 g of lead(II) oxide. Calculate the percent yi

user100 [1]

Let MM(x) be the molar mass of x.

MM(Pb) : MM(PbO)
=207.21 : 223.20 = 451.4 g : x g

cross multiply and solve for x
x=223.2/207.21*451.4
= 486.23 g

Percentage yield = 365.0/486.23= 0.75067 = 75.07% (rounded to 4 sign. fig.)

4 0

3 years ago

You are given a solid that is a mixture of na2so4 and k2so4.

Murljashka [212]

Here we have to calculate the amount of $SO_{4}^{2-}$ ion present in the sample.

In the sample solution 0.122g of $SO_{4}^{2-}$ ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ + $SO_{4}^{2-}$

(Na)₂SO₄=2Na⁺ + $SO_{4}^{2-}$

Thus, BaCl₂+ $SO_{4}^{2-}$ = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of $SO_{4}^{2-}$ ion is precipitated in this reaction.

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion ( $SO_{4}^{2-}$ ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present. So in 1 g of BaSO₄ $\frac{96.06}{233.3}=0.411$ g of $SO_{4}^{2-}$ ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of $SO_{4}^{2-}$ ion is present.

5 0

3 years ago