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3 years ago

What is radioactivity

Chemistry

2 answers:

Mumz [18]3 years ago

7 0

Answer:

Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation. A material containing unstable nuclei is considered radioactive. Three of the most common types of decay are alpha decay, beta decay, and gamma decay, all of which involve emitting one or more particles or photons.

Andrei [34K]3 years ago

3 0

The emission of ionizing radiation or particles caused by the spontaneous disintegration of atomic nuclei.

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What do radio waves and microwaves have in common?

Anastasy [175]

1) A
2)A
3)C
4)A
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3 years ago

The atomic number and the mass number of 18 and 40. Write the information conveyed by this statement.

VashaNatasha [74]

There are 18 protons and electrons and 22 neutrons in the atom

8 0

3 years ago

The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti

lyudmila [28]

$NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}$

Here the base is a benzoate ion, which is a weak base and reacts with water.

$C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)$

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or $[OH^{-}] = 10^{^{-pOH}}$

$[OH^{-}] = 10^{^{-4.96}}$

$[OH^{-}] = 1.1\times 10^{-5}$

The base dissociation equation kb = $\frac{Product}{Reactant}$

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = $1.6\times 10^{-10}$

And value of [OH-] we have calculated as $1.1\times 10^{-5}$ and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

$kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}$

$1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}$

$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

$[C6H5COO^{-}] = 0.76 M$ or $0.76\frac{mol}{L}$

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

8 0

3 years ago

Read 2 more answers

Balance each of the following redox reactions occurring in acidic aqueous solution.

dmitriy555 [2]

Answer:

Part A : Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).

Part B : 3Mg(s) + 2Cr³⁺(aq) → 3Mg²⁺(aq) + 2Cr(s).

Part C: 3MnO₄⁻ + 24H⁺ + 5Al → 5Al³⁺ + 3Mn²⁺ + 12H₂O.

Explanation:

Part A : Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s), Express your answer as a chemical equation. Identify all of the phases in your answer.

It is balanced as written: Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).

The two half reactions are:

The oxidation reaction: Zn(s) → Zn²⁺(aq) + 2e.

The reduction reaction: Sn²⁺(aq) + 2e → Sn(s).

To obtain the net redox reaction, we add the two-half reactions as the no. of electrons in the two-half reactions is equal.

So, the net chemical equation is:

Zn(s) + Sn²⁺(aq) → Zn²⁺(aq) + Sn(s).

Part B: Mg(s) + Cr³⁺(aq) → Mg²⁺(aq) + Cr(s), Express your answer as a chemical equation. Identify all of the phases in your answer.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The oxidation reaction: Mg(s) → Mg²⁺(aq) + 2e.

The reduction reaction: Cr³⁺(aq) + 3e → Cr(s).

To obtain the net redox reaction, we multiply the oxidation reaction by 3 (3Mg(s) → 3Mg²⁺(aq) + 6e) and the reduction reaction by 2 (2Cr³⁺(aq) + 6e → 2Cr(s)) to equalize the no. of electrons in the two-half reactions.

So, the net redox reaction will be:

3Mg(s) + 2Cr³⁺(aq) → 3Mg²⁺(aq) + 2Cr(s).

Part C : MnO⁴⁻(aq) + Al(s) → Mn²⁺(aq) + Al³⁺(aq), Express your answer as a chemical equation. Identify all of the phases in your answer.

To balance and write the net chemical equation, we should write the two-half reactions:

The two half reactions are:

The oxidation reaction: Al → Al³⁺ + 3e.

The reduction reaction: MnO₄⁻ + 8H⁺ + 5e → Mn²⁺ + 4H₂O.

To obtain the net redox reaction, we multiply the oxidation reaction by 5 (5Al → 5Al³⁺ + 15e) and the reduction reaction by 3 (3MnO₄⁻ + 24H⁺ + 15e → 3Mn²⁺ + 12H₂O) to equalize the no. of electrons in the two-half reactions.

So, the net redox reaction will be:

3MnO₄⁻ + 24H⁺ + 5Al → 5Al³⁺ + 3Mn²⁺ + 12H₂O.

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3 years ago

How does water return from atmosphere to land?

vlabodo [156]

Answer:

it falls from the sky

Explanation:

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3 years ago

What is radioactivity​

What is radioactivity