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3 years ago

Select three of the following compounds that are ionic

Chemistry

1 answer:

Orlov [11]3 years ago

5 0

Explanation:

Ionic compounds are compounds formed by the bonding of a metal with a non metal.

A. NO2

This is not an ionic compound because N and O are non metals.

B. CaCl2

This is an ionic compound because Ca is a metal and Cl is a non metal.

C. KI

This is an ionic compound because K is a metal and I is a non metal.

D. CO2

This is not an ionic compound because C and O are non metals.

E. H2O

This is not an ionic compound because H and O are non metals.

F. NaCl

This is an ionic compound because Na is a metal and Cl is a non metal.

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Answer:

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At 517 mm Hg and 24 °C, a sample of gas occuples a volume of 95 ml. The gas is transferred to a 225-ml flask and the temperature

vodomira [7]

Answer:

$P_2=194.78mmHg$

Explanation:

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In this case, we employ the combined ideal gas law in order to understand the volume-gas-pressure behavior as shown below:

$\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}$

Hence, solving for the final pressure P2, we obtain (do not forget temperature must be absolute):

$P_2=\frac{P_1V_1T_2}{V_2T_1}=\frac{517mmHg*95mL*(-8.0+273.15)K}{(24+273.15)K*225mL}\\ \\P_2=194.78mmHg$

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They are the molecules that are formed when four atoms chemically bond together.

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The force of a gas's outward push divided by the area of the walls of the container is the gas's

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

Sonja [21]

Answer:

The Number of gold atoms are = $N_{Au}=1.81718*10^{22}\ atoms/cm^3$

Explanation:

The formula we are going to use is:

$N_{Au}=\frac{N_A*C_{Au}}{{\frac{C_{Au}A_{Au} }{\rho_{Au}}}+\frac{A_{Au}}{\rho_{Ag}}(100-C_{Au})}$

Where:

$N_{Au}$ are number of gold atoms.

$N_A$ is Avogadro Number.

$C_{Au}$ is the amount of gold.

$A_{Au}$ is the atomic weight of gold.

$\rho_{Au}$ is the density of gold.

$\rho_{Ag}$ is the density of silver.

$C_{Ag}$ is the amount of silver.

$N_{Au}=\frac{6.023*10^{23}*45\%wt}{\frac{45\%wt*196.97}{19.32}+\frac{196.97}{10.49}(100-45\%wt)}\\ N_{Au}=1.81718*10^{22}\ atoms/cm^3$

The Number of gold atoms are = $N_{Au}=1.81718*10^{22}\ atoms/cm^3$

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4 years ago