1. Adakah benar bahawa anda tidak boleh membasuh rambut, meminum air sejuk dan

C: 91 kJ/mol; no.

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

<span>Some solutions demonstrate colligative properties, which rely on the quantity of solute in a solvent. To find the elevation in boiling point, we use the formula:

</span><span>ΔT(boiling point)  = (Kb)mi

where Kb represents a constant, m is the solution's molality, and i is the van't Hoff factor.

From the provided information, we can easily determine i as follows:

</span>ΔT(boiling point)  = (Kb)mi
103.45 - 100  = (0.512)3.90i
i = 1.73 <-------van't Hoff factor

Monomers combine through electron sharing during the polymerization process. This leads to the formation of a polymer, which consists of repeating units. The resulting substance has various applications.

Answer: The mole fraction of nitrogen is calculated to be 0.4615.

Explanation: In the mixture of nitrogen () and hydrogen (), the mole ratio established is 1: 1.5.

At this juncture, we identify that () serves as the limiting reagent.

When 0.4 moles of () are produced, it will require 0.4 moles of () along with 3.4 moles of ().

The total amount of remaining () ends up being 0.6, while the amount of () left in the mixture is 0.3 moles.

The mole fraction of () is computed as 0.6 divided by the sum of 0.6, 0.4, and 0.3, resulting in 0.4615.