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pochemuha
2 years ago
10

Determine the unknown mineral using the information given below. Be sure to use significant figure rules when doing calculations

. Mass is shown on the triple beam. Scale Test Mineral Mineral H. Density, g/cc 10 diamond tellurium 2 2.07 9 corundum galena 2.5 7.58 8 topaz anglesite 2.5 - 3 6.4 7 quartz chalcocite 2.5 - 3 5.6 6 feldspar copper 2.5 - 3 9.0 5 apatite gold 2.5 - 3 19.3 4 fluorite silver 2.5 - 3 10.5 3 calcite arsenic 3.5 5.7 2 gypsum barite 3 - 3.5 4.4 1 talc dolomite 3.5 - 4 2.9 platinum 4.5 21.5 willemite 5.5 4.0 magnetite 6 5.18 pyrite 6 - 6.5 5.02 pyrolusite 6 - 6.5 5.0 cassiterite 6.5 6.9 diamond 10 3.52 Volume of water displaced = 0.051 L
Mass: 981.0, 908, 908.1, 981 g
Density: 192.35, 19,235, 19, 19.2 g/cc
Hardness 4, 2.5, 1, 3, 5.5 - 5, 3, 2, 4.5, 1.5
Determine the unknown mineral.
Platinum, Gold
Chemistry
1 answer:
Ivenika [448]2 years ago
6 0

The unknown mineral is gold because of the similar properties of gold to that unknown mineral.

<h3>What is unknown mineral?</h3>

The unknown mineral is Gold because it has similar properties to the gold. Gold has density of 19.3 grams per cubic centimeter which is same to this unknown mineral. The hardness and mass of the mineral are also similar to the Gold.

So we can conclude that the unknown mineral is gold because of the similar properties of gold to that unknown mineral.

Learn more about mineral here: brainly.com/question/15844293

#SPJ1

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4 0
1 year ago
A stover-to-ethanol plant produces 40,000 tonne/yr of ethanol and contains 8functionalunits:feedstockhandling,pretreatment,simul
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Answer:

FCI=458000*8*40000^{0.3}\\FCI=880818398.52 M

FCI=88.0818 MM≅88 MM

Explanation:

Empirical correlation based on the work of Bridgwater and Mumford (1979):

For Liquid or solid phase Plants:

FCI=458000*N*F^{0.3}                       F<60,000 tonne/yr     Eq (1)

FCI=7000*N*F^{0.68}                         F≥60,000 tonnes/yr   Eq (2)

Where:

N is the number of functional units

F is the process throughput tonnes/yr

In our case F=40,000 tonne/yr <60,000 tonne/yr, We are going to use Eq (1)

FCI=458000*N*F^{0.3}                       F<60,000 tonne/yr

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FCI=458000*8*40000^{0.3}\\FCI=880818398.52 M

FCI=88.0818 MM≅88 MM

8 0
3 years ago
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moles (mol) = mass (g) / molar mass (g/mol)

According to the given data,
     moles = 4.60 mol
     mass = ?
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By substitution,
   4.60 mol = mass / (28.02 g/mol)
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