The molar heat capacity of solid aluminium is 24.4\text{ J K}^{-1}\text{ mol}^{-1} \text{ at } 25^{\circ}\text{C}24.4 J K −1 mol

<span>According to crystal field strength, the Cl ligand results in the longest d-d transition when coordinated with Ti(III) due to its classification as a weak field ligand that causes minimal d orbital splitting.</span>

Answer: The mole fraction of hydrogen gas at 20°C is 0.975

Explanation:

The information provided includes:

Water vapor pressure at 20°C is 17.5 torr

Total pressure at 20°C = 700.0 torr

Hydrogen gas vapor pressure at 20°C = (700.0 - 17.5) torr = 682.5 torr

To find hydrogen gas's mole fraction at 20°C, we utilize Raoult's law, represented by:

where,

= pressure of hydrogen gas = 682.5 torr

= total pressure = 700.0 torr

= mole fraction of hydrogen gas =?

Substituting the values into the equation yields:

Thus, the mole fraction of hydrogen gas at 20°C equals 0.975

Response: Option A) The lattice energy rises as cations become smaller, as demonstrated by LiF and KF.

Clarification: It has been observed that the lattice energy is largely determined by two primary factors regarding ionic solids:

i) The ionic charges - An increase in the charge of the ions corresponds to an increase in lattice energy.

and

ii) The size or radius of the ions - As the ionic size grows, the lattice energy diminishes accordingly.

Therefore, in this context, the latter factor is evident. Thus, it can be concluded that as cation sizes decrease among ionic solids, the lattice energy increases.

Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

Answer:

The ratios arranged in ascending order are; The ratio of the mass of Y to X in XY2 divided by the mass of Y to X in XY, The ratio of the mass of Y to X in XY3 divided by the mass of Y to X in XY, The ratio of the mass of Y to X in XY4 divided by the mass of Y to X in XY

1) Mass ratio = 3

2) Mass ratio = 2

3) Mass ratio = 4

Explanation:

Comprehensive calculations are displayed in the attachment.