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3 years ago

Write the equation for the decay of kr-85

Chemistry

2 answers:

iris [78.8K]3 years ago

8 0

Answer:see image attached

Explanation:

Kr-85 turns spontaneously into a Rubidium nucleus by emitting a beta particle and a gamma ray. The rubidium nucleus has the same mass as the krypton nucleus. The gamma rays carries off the excess energy released. It is actually a photon of light. A beta particle is actually an electron. A lot of energy is usually released in this process.

Ray Of Light [21]3 years ago

7 0

<u>Answer</u>:-

Kr-85 changes into Rb-86 by mostly beta decay.

Equation:

85Kr → Rb-86 + β-particle + Energy

Explanation:

1)Link:

http://www.nucleide.org/DDEP_WG/Nuclides/Kr-85_tables.pdf

2) Photo

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Which product is used on the natural nail prior to product application to assist in adhesion and serves to chemically bond the e

ira [324]

The product that is used on the natural nail prior to application to assist in adhesion and serves to chemically bond the enhancement product to the natural nail is known as nail primer.

<h3>What is a nail primer?</h3>

A nail primer is a chemical agent used in esthetic centers before applying a colored polish to the nails and serves as an adhesive product.

The nail primers are also very useful for improving the cleaning efficiency of the product before its application.

Nail care products include different types of chemical formulations such as, for example, creams that reinvigorate the cuticle.

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2 years ago

Consider the following reversible reaction.

andriy [413]

Answer:

No one is correct. The correct expression is:

Keq = [H₂]² . [O₂]² / [H₂O]²

Explanation:

To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.

The balance reaction is:

<u>2</u> H₂O (g) ⇄ <u>2</u> H₂ (g) + O₂ (g)

Keq = [H₂]² . [O₂] / [H₂O]²

In opposite side: <u>2</u> H₂ (g) + O₂ (g) ⇄ <u>2</u> H₂O (g)

Keq = [H₂O]² / [H₂]² . [O₂]

6 0

3 years ago

Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr

Delvig [45]

<h2>Hello!</h2>

The answer is: Charle's Law.

The law that states that the volume and absolute temperature of a fixed quantity of gas (ideal gas) are proportional under constant pressure is the Charle's Law, also known as the law of volumes.

The law describes how a gas kept under constant pressure tends to expand when the temperature increases and it's described by the following equation:

$\frac{V}{T}=k$

Where,

$V=Volume\\T=Temperature\\k=constant$

Also, to describe the relationship between two differents volumes at different temperatures, we have:

$\frac{V_{i}}{T_{i}}=\frac{V_{f}}{T_{f}}$

Where,

$V_{i}=InitialVolume\\T_{i}=InitialTemperature\\V_{f}=FinalVolume\\T_{f}=FinalTemperature$

Have a nice day!

8 0

3 years ago

Read 2 more answers

Indicate which solution in each pair has the lower pH. Your response should be a four letter "word". The first letter should be

JulijaS [17]

Answer:

bcfh

Explanation:

HClO₄ reacts with water thus:

HClO₄ + H₂O → H₃O⁺ + ClO₄⁻

That means HClO₄ produce H₃O⁺ that decreases pH. That means the higher concentration of HClO₄ decreases pH. Thus, lower pH will be:

b) 0.2 M HClO4

The reaction of NaClO₄ is:

NaClO₄ + H₂O → OH⁻ + HClO₄ + Na⁺

The higher concentration of NaClO₄ the higher production of OH⁻ that increase pH, that means the lower concentration of NaClO₄ the lower pH, thus, the answer is:

<em>c) 0.1 M NaClO or</em>

HF reacts with water thus;

HF ⇄ H⁺ + F⁻

The equilibrium constant is:

k = [H⁺] [F⁻] / [HF] = 3,5x10⁻⁴

For HNO₂ equilibrium is:

HNO₂ ⇄ H⁺ + NO₂⁻

k = [H⁺] [NO₂⁻] / [HNO₂] = 4,5x10⁻⁴

As k value is higher for HNO₂, the concentration of H⁺ will be higher in this system doing the HNO₂ with the lower pH.

f) 0.1 M HNO2

NaOH is a strong base that produce OH⁻ that increase pH, pure water is neutral, thus, the lowe pH is:

h) pure water

I hope it helps!

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3 years ago

Why are the oxidation and reduction half-reactions separated in an<br> electrochemical cell?

Mrac [35]

Answer:

The half-cells separate the oxidation half-reaction from the reduction half-reaction and make it possible for current to flow through an external wire.

Explanation:

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3 years ago

Read 2 more answers