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Reika [66]
3 years ago
12

Which of elements would you expect to form the following with copper? Substitutional solid solution with complete solubility Sub

stitutional solid solution with incomplete solubility Interstitial solid solution.
Chemistry
1 answer:
Kipish [7]3 years ago
3 0

Answer:

Element Atomic Radius (nm) Crystal Structure Electronegativity Valence

Cu 0.1278 FCC 1.9 +2

C 0.071

H 0.046

O 0.060

Ag 0.1445 FCC 1.9 +1

Al 0.1431 FCC 1.5 +3

Co 0.1253 HCP 1.8 +2

Cr 0.1249 BCC 1.6 +3

Fe 0.1241 BCC 1.8 +2

Ni 0.1246 FCC 1.8 +2

Pd 0.1376 FCC 2.2 +2

Pt 0.1387 FCC 2.2 +2

Zn 0.1332 HCP 1.6 +2

The answers to the question are

(a) Copper will form substitutional solid solution with complete solubility with

Ni due to their very similar size, equal valency and similar electronegativities as well as Pd, and Pt but less likely with aluminium due to size and valency difference

(b) Copper will form substitutional solid solution of incomplete solubility with Ag, Al, Co, Cr, Fe, Zn

(c) Copper will form interstitial solid solution with

C, H, O due to the large difference between the size of copper atomic radius and the smaller atomic radii of C, H and O

Explanation:

Substitutional solid solution rules

the following are the Hume-Rothery rules for substitutional solid solutions:

1. Less than 15%difference between the atomic radius of the solute and solvent:

2. Similarity in the crystal structures of solute and solvent;

3. The valency of the solvent and solute must be similar before they can be said to be completely soluble in each other.

A lower valency metal is more likely to dissolve in one higher valency.

4. The electronegativities of the solute and solvent should be similar a wide variation in electronegativity will lead to the formation of intermetallic compounds rather than a solid solutions.

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The experimental density of CO2 at STP is 0.10/0.056=1.78 g/L. The percent error equals to (1.96-1.78)/1.96*100%=9.18%. So the answer is 9.18%.
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3 years ago
What does the atomic weight (mass number) of an element represent? ____________________________ .
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The atomic weight of an element represents the ratio of the average mass of atoms of a chemical element.

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Compound A and Compound B are binary compounds containing only elements X and Y. Compound A contains 1.000 g of X for every 2.10
ycow [4]

Answer:

Explanation:

a) for 1.000 g X: 0.1621 g Y

ratio of mass of element Y = 2.100g : 0.1621g

= 1 : 0.07

b) 1.000 g X: 0.7391 g Y

ratio of mass of element Y = 2.100g : 0.7391g

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c) 1.000 g X: 0.2579 g Y

ratio of mass of element Y = 2.100g : 0.2579g

= 1 : 0.12

d) 1.000 g X: 0.2376 g Y

ratio of mass of element Y = 2.100g : 0.2376g

= 1: 0.11

e) 1.000 g X: 0.2733 g Y

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From the values obtained , the closest that is in compliance with the law f multiple proportions is option B

7 0
3 years ago
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What components of an atom has no charge
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Answer:

Neutrons

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B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS

Explanation:

Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.

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