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2 years ago

What do the following have in common: MgCl2, AlF3, CaI2, KCl

Chemistry

2 answers:

arsen [322]2 years ago

8 0

Answer:

They are salts, ionic compounds

Explanation:

To know if they are covalent or ionic we need to check their difference in electronegativity. As an example mgcl2, mg has electronegativity of 1.2 and cl 3 so 3-1.2=1.8 which is bigger than 1.7 means it's ionic therefore they are salts

Kay [80]2 years ago

5 0

They are salts and are all ionic compounds because in the formula one of the element has to have a positive and the other a negative charge.

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1 year ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Mama L [17]

<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

<u>Explanation:</u>

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

$N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}$

where,

$N_{Au}$ = number of gold atoms per cubic centimeters

$N_A$ = Avogadro's number = $6.022\times 10^{23}atoms/mol$

$C_{Au}$ = Mass percent of gold in the alloy = 42 %

$\rho_{Au}$ = Density of pure gold = $19.32g/cm^3$

$\rho_{Ag}$ = Density of pure silver = $10.49g/cm^3$

$M_{Au$ = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

$N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3$

Hence, the number of gold atoms per cubic centimeters in the given alloy is $1.83\times 10^{22}$

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2 years ago