All categories

2 years ago

Place the following in order of increasing standard molar entropy H2O(l) H2O(g) H2O(s) a)H2O(S)< H2O(l)

Chemistry

1 answer:

Black_prince [1.1K]2 years ago

7 0

Answer : The order of increasing standard molar entropy is:

$H_2O(s)$

Explanation :

Entropy : It is defined as the measurement of randomness or disorderedness in a system.

The order of entropy will be,

As we are moving from solid state to liquid state to gaseous state, the entropy will be increases due to the increase in the disorderedness.

As we are moving from gaseous state to liquid state to solid state, the entropy will be decreases due to the decrease in the disorderedness.

From this we conclude that the solids have less disorder in the system than liquids because the molecules in the solid are closely or tightly packed as compared to liquid.

The liquids have less disorder in the system than gases because the molecules in the liquid are more close as compared to gases.

That means the more disorderedness in a system, the more will be the entropy.

So, the order of increasing standard molar entropy is:

$H_2O(s)$

You might be interested in

One molecule of a compound weighs 2.03x10-22 g. what is the molar mass of that compound?

stealth61 [152]

The molar mass of the compound is 122 g.

6 0

2 years ago

Read 2 more answers

The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

2 years ago

Read 2 more answers

Please help!!

Sati [7]

In a chemical reaction, the limiting reagent is the chemical being used up while the excess reactant is the chemical left after the reaction process.

Before calculating the limiting and excess reactant, it is important to balance the equation first by stoichiometry.

C25N3H30Cl + NaOH = C25N3H30OH + NaCl

Since the reaction is already balanced, we can now identify which is the limiting and excess reagent.

First, we need to determine the number of moles of each chemical in the equation. This is crucial for determining the limiting and excess reagent.

Assuming that there is the same amount of solution X for each reactant

1.0 M NaOH ( X ) = 1.0 moles NaOH

1.00 x 10-5 M C25N3H30Cl ( X ) = 1.00 x 10-5 moles C25N3H30Cl

The result showed that the crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this chemical reaction is crystal violet and the excess reactant is NaOH.

3 0

2 years ago

84. Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass = 18.01), as a result o

ipn [44]

Answer:

relative rate of diffusion is 1.05

Explanation:

According to Graham's law of difussion:

Rate of diffusion is inversely proportional to the square root of molecular weight of a molecule.

For two given molecules:

$\frac{(rate)_{1}}{(rate)_{2}}=\frac{\sqrt{M_{2}} }{\sqrt{M_{1}}}$