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Describe an experiment to find how the volume of gold can be measured

Md Ariful Islam

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Md Ariful Islam

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  1. Partially fill a graduated cylindrical container with liquid.
  2. Mark the initial liquid level on the outside of the container with a marker.
  3. Add the gold.
  4. Mark the new liquid level.
  5. Measure the distance between the original and final liquid levels.

Since the cylinder is marked in milliliters (ml), the difference between the final and initial volume will give the volume of gold in cm3.

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Throwing some scrap iron in a gold nitrate solution causes the gold metal to precipitate. How many liters of 0.5 M gold nitrate solution would react with 2.8x10<sup>23</sup> particles of iron metal?

Reaction:Au3+ + Fe = Fe3+ + AuV(Au(NO3)3 solution) = 2.8x1023/(6.02x1023)/0.5 = 0.93 LAnswer: 0.93 L

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1 Answers 1 views
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What is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3 g/cm3?

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1 Answers 1 views
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Gold cannot displace copper from copper sulphate solution. This is because gold is very much less reactive than that of copper hence it cannot replace copper from copper sulphate solution...

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Q = c*m*ΔT = 0.128*3100*(69-25) = 17459 (J) = 17.5 (kJ)

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Q = m*s*dT m(metal) = 1.6 g dT = 18 C s(metal) = Q/(m*dT) = 5.8 J / (1.6 g * 18 C) = 0.201 J/(g*C) The specific heat of...

1 Answers 1 views

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