When 200 grams of water cools from 50.°C to 25°C, the total amount of heat energy released by the water is?

Response:

The mass percentage of a solution comprising 7.6 grams of sucrose and 83.4 grams of water equals 8.351 %.

Details:

Provided data:

Sucrose mass = 7.6 grams

Water mass = 83.4 grams

In this scenario, sucrose acts as the solute, while water is the solvent.

The calculation for mass percent of a solution is done using the following formula:

Mass percent = (Mass of Solute/Mass of Solution)(100)

As sucrose is the solute, the mass equals 7.6 grams.

The total mass of the solution, which includes both sucrose and water, comes out to:

Total mass = 7.6 grams + 83.4 grams = 91 grams

Therefore, applying the values gives mass percent = (7.6/91)(100) = 8.351 %.

Response:

a) Representation - (in attachment)

b) The geometry is tetrahedral

c) hybrid orbitals involve sigma bonding.

orbitals arise from the overlap of sulfur's d-orbitals with the p-orbitals of oxygen.

Clarification:

a)

Representation in attachment.

The total charge in both structures is -2. Structure (b) is preferred according to the resonance forms since the formal charges within the species are preserved.

Thus, structure (b) represents the sulfate ion more accurately.

b) In the sulfate ion, the sulfur atom forms connections with four distinct oxygen atoms. Using VSEPR theory, the sulfate ion is tetrahedral in shape.

There are four sigma bonds and no lone pairs surrounding the central atom.

Consequently, the sulfur atom's hybridization is

c)

The s and p orbitals of sulfur undergo hybridization to form sigma bonds. The orbitals involved in bonding are

Therefore, bonds result from the overlapping of sulfur's d-orbitals with the p-orbitals of oxygen.

Answer:

Explanation:

AgNO3 + NaCl --> AgCl + NaNO3

Moles

of AgNO3

= molarity * volume

= 1 * 0.01

= 0.01 mol

for NaCl

= 0.01 * 1

= 0.01 mol.

According to stoichiometry, one mole of silver nitrate corresponds to one mole of NaCl reacted. Hence,

Moles of AgCl generated = 0.01 × 1

= 0.01 mol AgCl produced.

Heat gained by the solution as precipitation occurs:

Solution mass = density × volume

= 1 × 20

= 20 g.

Using q = m * Cp * (T2 - T1)

= 20 * 4.18 * (32.6 - 25.0)

= 635 J

The absorbed heat of 635 J indicates the reaction released -635 J

Thus, Delta H = -635 J/0.01 mol

= -63500 J/mol

= -63.5 kJ/mol.