Decay of Uranium-238 produces gamma radiaton whereas decay of thorium-234 releases beta radiation.
<h3>What type of radiation is produced?</h3>
In the decay of U-238, two gamma rays of different energies are emitted in addition to the alpha particle while on the other hand, in the decay of thorium-234 , beta rays are emitted.
So we can conclude that decay of Uranium-238 produces gamma radiaton whereas decay of thorium-234 releases beta radiation.
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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
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The answer is D . I hope this help you :) .
According to the balanced equation of the reaction:
2C2H2 + 5O2 → 4CO2 + 2H2O
So we can mention all as liters,
A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O
So, when we have 75L of CO2
and when we have 2 L of C2H2 reacts and gives 4 L of CO2
2C2H2 → 4CO2
∴ The volume of C2H2 required is:
= 75L / 2
= 37.5 L
B) and, when we have 75 L of CO2
and 4CO2 → 2H2O
∴ the volume of H2O required is:
= 75 L /2
= 37.5 L
C) and from the balanced equation and by the same way:
when 5 liters O2 reacts to give 4 liters of CO2
and we have 75 L of CO2:
5 O2 → 4 CO2
?? ← 75 L
∴ the volume of O2 required is:
= 75 *(5/4)
= 93.75 L
D) about the using of the number of moles the answer is:
no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.
