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1 year ago

Beta particles are identical to Multiple Choice protons. helium atoms. hydrogen atoms. helium nuclei.

Chemistry

1 answer:

Vesna [10]1 year ago

4 0

Correct option:

Beta particles are identical to "electrons".

What are particle beams:

There are various types of particle beams, such as (alpha) and (beta) particles, neutron beams, etc. While β-particles are electrons released from a nucleus, helium nuclei made up of two protons and two neutrons are known as α-particles.

When certain radionuclides undergo a process known as beta-decay, high-energy, high-speed electrons (e-) or positrons (p+) are released from the nucleus.

An electron and a beta particle share the same mass and charge.

Beta radiation in high doses can burn the skin, and beta emitters are dangerous if they penetrate the body. Thin sheets of metal or plastic may be able to stop beta particles.

Note: Your question is incomplete, but most probably your full question also includes one more option "Electron".

Beta particles are identical to

Multiple Choice

1. protons. 2. helium atoms. 3. hydrogen atoms. 4. helium nuclei. 5. electrons.

Learn more about the beta particles here,

brainly.com/question/3580345

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3 years ago

What percentage of the mass of hydrofluoric acid (HF) is contributed by hydrogen?

Morgarella [4.7K]

This problem is asking for the percent by mass of hydrogen in hydrofluoric acid. At the end, the answer turns out to be D. 5% as shown below:

<h3>Percent compositions:</h3>

In chemistry, percent compositions are used for us to know the relative amount of a specific element in a compound. In order to do so for hydrogen, we use the following formula, which can also be applied to any other element in a given compound:

$\% H=\frac{m_H*1}{M_{HF}}*100\%$

Where $m_H$ stands for the atomic mass of hydrogen and $M_{HF}$ for the molar mass of hydrofluoric acid. In such a way, we plug in the atomic masses of hydrogen (1.01 g/mol) and fluorine (19.0 g/mol) to obtain:

$\% H=\frac{1.01g/mol*1}{1.01g/mol+19.0g/mol}*100\%\\\\\% H=5\%$

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2 years ago

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vazorg [7]

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Explanation:

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2 years ago

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Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M

Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to identify the limiting reactant:

We have:

0.20 M * 50.0 mL = 10 mmol of AgNO₃

0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. That makes K₂CrO₄ the limiting reactant.

Now we calculate the mass of Ag₂CrO₄ formed, using the limiting reactant:

4 mmol K₂CrO₄ * $\frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4}$ = 1326.92 mg Ag₂CrO₄

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