Would an alkali metal make a good replacement for tin in a tin can? Explain.

Answer:

1.2×10²³ atoms.

Explanation:

In the problem, we see the data:

Mole of propanone = 0.20 mole

Calculating the number of atoms in propanone =?

According to Avogadro's principle, one mole of a substance contains 6.022×10²³ atoms.

This means that one mole of propanone also holds 6.022×10²³ atoms.

Thus, we can determine the atom count in 0.20 mole of propanone as:

1 mole of propanone contains 6.022×10²³ atoms.

Accordingly, 0.20 mole of propanone will have = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.

Therefore, 0.20 mole of propanone contains

1.2×10²³ atoms.

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.

The compound is acetone ( CH₃-CO-CH₃)

Explanation:

1) Acetone is represented as CH₃-CO-CH₃.

2) This is a molecule formed by covalent bonds.

3) When it dissolves, compounds with covalent bonds remain as individual molecules, indicating that the primary species in the solution are the molecules themselves, which are surrounded (solvated) by water molecules.

In contrast, ionic compounds ionize. For example, when NaCl dissolves in water, it completely breaks down into ions, hence the predominant species are the ions Na⁺ and Cl⁻, rather than the NaCl formula.

This leads to the conclusion that: when acetone dissolves in water, the primary components are the acetone molecules (there is no need to mention that water molecules are in the solution, as that isn't the question's focus).

Answer:

The concentration of P in the pond at equilibrium is 0.034 g/m³

Explanation:

Given the total mass = 49.9 g

1 day = 24 hours

mass per hour;

Incoming mass = (49.9 g / day) * (1 day /24 hr )

            = 2.079 g/hr

Outgoing mass = 0

Given the half-life = 3.4 hours

For a first-order reaction; k, the rate constant = ln2/t, where t is the half-time

                     ln 2= 0.693, V= volume

                     k = 0.693 / t_half = 0.693 / 3.4 = 0.2038 hr⁻¹

Mass lost to sunlight = k V  

= Incoming mass per hour / kV

= 2.079 g/hr / (0.2038 hr⁻¹ x 300 m³)

0.034 g/m³