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Maurinko [17]
3 years ago
15

A geologist determines that a sample of a mineral can't be scratched by a steel nail but can be scratched by a masonry drill bit

. Based on this information, the sample mineral has to be softer than A. orthoclase. B. fluorite. C. apatite. D. corundum.
Chemistry
2 answers:
MatroZZZ [7]3 years ago
7 0

Answer:

D. corundum.

Explanation:

Corundum is the hardest mineral in nature after diamond. For this reason it can only be scratched and worked by it. Mohs' hardness scale proves that corundum is indeed a very hard mineral and almost impossible to scratch. This scale states that the corundum has a hardness of 9.0, which indicates that it cannot be scratched by a drill.

Using the same scale, we can find out the hardness of the mineral exposed in the question. The mineral can be scratched by a drill, which means it has a hardness lower than 8.5, and the mineral cannot be scratched by a nail, which indicates that it has a hardness greater than 6.5. Based on this data, we can guarantee that this meniral is softer than the corundum.

KIM [24]3 years ago
4 0
D. Corundum

The mineral has a hardness greater than 6.5 (steel nail) and less than 8.5 (drill bit). All other choices have a hardness less than 8.5 but corundum has a hardness of 9.0. *Mohs hardness scale
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7 0
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zysi [14]

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A)Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

B)Ksp=1.75×10^−10

C)Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=​Δ​G​f​,​p​r​o​d​u​c​t​s​∘​−​Δ​G​f​,​r​e​a​c​t​a​n​t​s​∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

A)

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​C​l​(​a​q​)​−​∘​]​−​Δ​G​f​,

A​g​C​l​(​s​)​⇌​A​g​(​a​q​)​+​+​C​l​(​a​q​)​−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​B​r​(​a​q​)​−​∘​]​−​Δ​G​f​,

Δ​G​r​x​n​∘​=​[​77.1​k​J​/​m​o​l​+​(​−​104.0​k​J​/​m​o​l​)​]​−​(​−96.90kj/mol

Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K

Ksp=5.45×10^−13

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3 years ago
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