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Basile [38]
2 years ago
14

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

Chemistry
1 answer:
Vesna [10]2 years 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

#SPJ4

 

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Katena32 [7]
An acid dissociation constant, Ka<span>, (also known as acidity </span>constant<span>, or </span>acid<span>-ionization </span>constant<span>) is a quantitative measure of the strength of an </span>acid<span> in solution. It is the equilibrium </span>constant<span> for a chemical reaction known as </span>dissociation<span> in the context of </span>acid<span>–base reactions.

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sergejj [24]

The scientists responsible for testing the hypothesis are already famous

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Which energy source has been developed and improved by chemists?
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2 years ago
Hydrogen is diffusing through solid nickel. At a temperature of 358 K, the diffusivity is 1.16 x 10-8 cm2/s. At a temperature of
Lelu [443]

Answer:

Q_d=35881 J/mol

So the activation energy is 35881 J/mol.

Explanation:

Consider the following equations:

lnD_{1} =lnD_{o} -\frac{Q_{d} }{R*T_{1} }

lnD_{2} =lnD_{o} -\frac{Q_{d} }{R*T_{2} }

Solving the above two equation to find the Q_d in term of diffusivity and temperature we will get:

Q_{d}=-R*\frac{lnD_{1} -lnD_{2} }{\frac{1}{T_{1} }-\frac{1}{T_{2} }  }

where:

Q_d is the activation energy

D_1 is the diffusivity at T_1

D_2 is the diffusivity at T_2

Q_{d}=-8.31*\frac{ln1.16*10^-12 -ln1.05*10^-11 }{\frac{1}{358 }-\frac{1}{438 }  }

Q_d=35881 J/mol

So the activation energy is 35881 J/mol.

3 0
4 years ago
Which of the following releases chemical energy?
GaryK [48]

Answer:

D. burning wood

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7 0
3 years ago
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