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3 years ago

Which of the following methods of fossil formation is formed in dry areas due to a lack of moisture?

Chemistry

1 answer:

KengaRu [80]3 years ago

5 0

I don’t know the answer but my guess would be mummification

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Consider a solution that is 2.5×10−2 M in Fe2+ and 1.1×10−2 M in Mg2+. (Ksp for FeCO3 is 3.07×10−11 and Ksp for MgCO3 is 6.82×10

Bumek [7]

Answer:

This question is incomplete, here's the complete question:

Consider a solution that is 2.1×10−2 M in Fe2+ and 1.6×10−2 M in Mg2+.

Part A

If potassium carbonate is used to selectively precipitate one of the cations while leaving the other cation in solution, which cation will precipitate first? ANSWER: Fe 2+

Part B

What minimum concentration of K2CO3 is required to cause the precipitation of the cation that precipitates first? ANSWER: [K2CO3] = 1.5×10−9 M

Part C

What is the remaining concentration of the cation that precipitates first, when the other cation just begins to precipitate? (ANSWER IS NOT .021 or 2.0E-6)

Explanation:

Part A

Fe2+ will precipitate first as solubility product of FeCO3 is lesser than solubility product of MgCO3.

Part B

FeCO3\= Fe2+ + CO32-

Ksp = [Fe2+][CO32-] = 3.07 x 10-11

3.07 x 10-11 = (2.1 x 10-2)[CO32-]

[CO32-] = 1.5 x 10-9 M to precipitate the Fe2+ ions

Part C

MgCO3\= Mg2+ + CO32-

Ksp = [Mg2+][CO32-] = 3.07 x 10-11

6.82 x 10-6 = (1.6 x 10-2)[CO32-]

[CO32-] = 4.3 x 10-4 M to precipitate the Mg2+ ions

Ksp = [Fe2+][CO32-] = 3.07 x 10-11

3.07 x 10-11 = [Fe2+](4.3 x 10-4)

[Fe2+] = 7.1 x 10-8 M when the Mg2+ ions precipitates

5 0

3 years ago

Low melting point and poor conductor of electricity

WINSTONCH [101]

i THINK THE ANSWER IS A

5 0

3 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$