A chemistry student in lab needs to fill a temperature-control tank with water. The tank measures 24.0 cm long by 21.0 cm wide by 13.0 cm deep. In addition, as shown in the sketch below, the student needs to allow 2.0 cm between the top of the tank and the top of the water, and a round-bottom flask

with a diameter of 10.5 cm will be just barely submerged in the water.

Calculate the volume of water in liters which the student needs. Round your answer to the nearest 0.1 L. ol.2 cm

M

flask

OL

6

water

The required volume of water the student needs is 4.9 litres of water

Explanation:

From the diagram related to the question, we have;

The dimensions of the tank are;

Length of tank = 24.0 cm = 0.24 m

Width of tank = 21.0 cm = 0.21 m

Depth of tank = 13.0 cm. = 0.13 m

Allowance provided between the top of the tank and the top of the water = 2.0 cm

Diameter of the round bottom flask, D = 10.5 cm = 0.105 m

Therefore, the radius of the round bottom flask, r = 0.105/2 = 0.0525 m

Therefore we have;

Depth of water in the tank = Depth of tank - Allowance provided between the top of the tank and the top of the water

∴ Depth of water in the tank = 13.0 - 2.0 = 11.0 cm = 0.11 m

Given that the flask is immersed in the water contained in the tank to raise the tank water level, we have;

Volume of water + Volume of flask in the tank = Length of tank × Width of tank × Depth of water in the tank

Volume of water + Volume of flask in the tank =  0.24 × 0.21 × 0.11 = 0.005544 m³ = 0.005544 m³× 1000 l/m³ = 5.544 l

The volume of the spherical flask = 4/3·π·r³ = 4/3·π·0.0525³ = 6.06×10⁻⁴ m³ = 6.06×10⁻⁴ m³ × 1000 l/m³ = 0.606 l

The required volume of water the student needs , V = Volume of water + Volume of flask in the tank - The volume of the spherical flask = 5.544 l - 0.606 l = 4.9 l.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

"the periodic table is arranged in order of atomic number."