Calculate the mass of 6.9 moles of nitrous acid (HNO2). Explain the process or show your work by including all values used to de

I think the state change illustrated in the diagram is deposition. 
Deposition is the transformation of gases into solids without transitioning through a liquid phase. It is the reverse process of sublimation.
A key distinction between gases and solids lies in the spacing of molecules; gases have large spaces between molecules, whereas solids have very minimal spacing, resulting in solids being more densely packed. This is illustrated in the diagram showing the transition from gases to solids.

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

Based on Archimedes' principle, the mass of fresh water and the mass of the cup are equal to the mass of the same volume of seawater.

The mass of freshwater can be calculated using density times volume.

1 cm³ is equivalent to 1 mL.

The mass of freshwater is 0.999 g/cm³ multiplied by 735 cm³, which results in 734.265 g.

The total mass of the freshwater and cup combined is 734.265 g plus 25 g, equating to 759.265 g.

This means the mass for an equal volume of seawater is 759.265 g.

The volume of the seawater displaced is 735 mL, which is 0.735 L (assuming the cup's volume can be disregarded).

We know that 1 liter equals 1000 cm³ or 1000 mL.

The density of seawater can be determined as mass divided by volume.

The density of seawater becomes 759.265 g divided by 0.735 L, yielding 1033.01 g/L.

Conversely, the density of freshwater in g/L is calculated as 0.999 g/(1/1000) L, equating to 999 g/L.

The mass of salt dissolved in 1 liter of seawater is calculated as 1033.01 g - 999 g, which equals 34.01 g.

Thus, the amount of salt in 1 L of seawater is 34 g.

Answer:

The correct choice is: option A.

Justification:

To address this inquiry, we need to evaluate the total number of electrons each orbital can accommodate.

  Orbital                         Number of electrons

   s                                   2

   p                                  6

   d                                 10

   f                                  14

Provided options:

A. 1s² 2s² 2p⁶ 3s²                 This configuration is valid as it aligns with the permitted number of electrons in each orbital and follows the correct sequence.

B. 1s² 2s² 2p⁶ 3s² 3d⁴          This configuration is not accurate because

                                         3d⁴ should follow 3p.

C. 1s² 2s² 2d¹⁰ 2p³                This is incorrect since 2d¹⁰ is not a valid orbital.

D. 1s² 2s^s 2p³ 2d¹⁰            This option contains two errors; s as an exponent does not exist, and 2d¹⁰ is also an invalid description.

I hope this is useful......