An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g

Answer: The correct option is 3.

Explanation: Radioisotopes that emit alpha-particles are termed alpha-emitters. These isotopes undergo alpha-decay.

Those radioisotopes that emit beta-particles are called beta-emitters. They undergo beta-minus decay, in which a neutron converts to a proton and an electron.

Isotopes that emit positrons are known as positron-emitters, undergoing beta-plus decay where a proton becomes a neutron.

From the options given,

Option 1: All three isotopes undergo beta-minus decay.

Option 2: Cs-137 and Tc-99 undergo beta-minus decay.

Fr-220 undergoes alpha-decay.

Option 3: Kr-85 undergoes beta-minus decay.

Ne-19 undergoes positron decay.

Rn-222 undergoes alpha decay.

Option 4: All three isotopes undergo beta-minus decay processes.

Therefore, the correct choice is 3.

The mixture’s density is 1.57 g/cm³.

Step 1: Determine the mass of the butter.

Step 2: Determine the mass of the sand.

Step 3: Determine the density of the mixture.

Total mass = 0.860 g + 2.28 g = 3.14 g.

Total volume = 1 cm³ + 1 cm³ = 2 cm³

Answer:

Angle ABE measures 27°.

Explanation:

Refer to the attached diagram related to this question.

The given values are ∠ABE=2n+7 and ∠EBF=4n-13.

Clearly seen in the diagram, ∠ABE and ∠EBF are equal in measure.

Move variable components to one side of the equation.

Split both sides by 2.

The solution for n arrives at 10.

The next step is to calculate ∠ABE.

Consequently, the measurement of angle ABE is 27°.

Answer:

Quantity of generated will be reduced to fifty percent of its initial amount.

Explanation:

Equilibrium reaction:

In accordance with the balanced equation, 1 mol of HCl interacts with 1 mol of NaOH leading to the formation of 1 mol of

0.5 mol of HCl interacting with 0.5 mol of NaOH yielding 0.5 mol of .

Consequently, it is clear that the total of produced will be halved if the quantities of the reactants are halved.

Answer:

C. connecting an active metal to designate the pipe as the cathode in an electrochemical cell.

Explanation:

Cathodic protection involves a method to manage the accelerated corrosion of a metal surface by designating it as the cathode within an electrochemical cell. This is accomplished by attaching the protected metal to a more sacrificial metal, which acts as the anode.

This method helps to preserve the metal by introducing a highly reactive metal that serves as the anode, supplying free electrons. By adding these free electrons, the active metal gives up its ions, protecting the less reactive steel from corrosion.