Which of the following types of radiation can be shielded by piece of paper or fabric

1 answer:

8 0

There are 3 major types of radiation. The Alpha, Beta and Gamma radiation. It is the Alpha radiation that can be stopped by a piece of paper or fabric. Beta on the other hand can be stopped by plastic or a thin sheet of metal and Gamma by a piece of aluminum foil or lead.

You might be interested in

Starting with lead (ii) carbonate describe how a solid sample of lead (ii) sulphate can be prepared

tatuchka [14]

The preparation of lead (ii) sulphate from lead (ii) carbonate occurs in two steps:

insoluble lead carbonate is converted to soluble lead (ii) nitrate
soluble lead (ii) nitrate is reacted with sulphuric acid to produce lead (ii) sulphate.

<h3>How can a solid sample of lead (ii) sulphate be prepared from lead (ii) carbonate?</h3>

Lead (ii) carbonate and lead (ii) sulphate are both insoluble salts of lead.

In order to prepare lead (ii) sulphate, a two step process is performed.

In the first step, Lead (ii) carbonate is reacted with dilute trioxonitrate (v) acid to produce lead (ii) nitrate.

PbCO₃ + 2HNO₃ → Pb(NO₃)₂ + CO₂ + H₂O

In the second step, dilute sulfuric acid is reacted with the lead (ii) nitrate to produce insoluble lead (ii) sulphate which is filtered and dried.

Pb(NO₃)₂ + H₂SO₄ → PbSO₄ + 2HNO₃

In conclusion, lead (ii) sulphate is prepared in two steps.

Learn more about lead (ii) sulphate at: brainly.com/question/188055

#SPJ1

4 0

2 years ago

In a single replacement reaction between 2.57 moles Aluminum and 3.59 moles of Hydrochloric acid, how many moles of Hydrogen can

Alenkasestr [34]

Answer:

1.795 mole of H2.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2Al + 6HCl —> 2AlCl3 + 3H2

Step 2:

Determination of the limiting reactant.

From the balanced equation above,

2 moles of Al reacted with 6 moles

Therefore, 2.57 moles of Al will react with = (2.57 x 6)/2 = 7.71 moles of HCl.

From the calculation made above, it will require a higher amount of HCl than what was given to react completely with 2.57 moles of Al. Therefore, HCl is the limiting reactant and Al is the excess reactant.

Step 3:

Determination of the number of mole H2 produced from the reaction.

Here, we shall be using the limiting reactant because it will produce the maximum yield of the reaction since all of it were consumed by the reaction.

The limiting reactant is HCl and the amount of H2 produce can be obtained as follow:

From the balanced equation above,

6 moles of HCl reacted to produce 3 moles of H2.

Therefore, 3.59 moles of HCl will produce = (3.59 x 3)/6 = 1.795 mole of H2.

From the calculations made above, 1.795 mole of H2 is produced from the reaction.

5 0

3 years ago

SOMEONE PLEASE HELP

Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)