If you were given a sample of a cotton ball and a glass stirring rod with identical mass (ex: 5.0 g), which sample would contain

Answer:

2.5 g of platinum

Explanation:

A catalyst is a substance added to a reaction to enhance the reaction speed. It does not undergo any change during the reaction, meaning it remains unchanged after the reaction concludes. The role of a catalyst is to provide an alternative pathway for the reaction by reducing the activation energy required. Therefore, a catalyzed reaction occurs more rapidly and requires less energy compared to an uncatalyzed one.

Since catalysts do not get involved in reactions and retain their mass post-reaction, the amount of platinum will stay the same (2.5g). The mass can only alter if a substance participates in the chemical process. Thus, this is the response.

(c) Cu + S → CuS is classified as a redox reaction

Explanation:

The following reactions are presented:

(a) K₂CrO₄ + BaCl₂ → BaCrO₄ + 2 KCl

(b) Pb²⁺ + 2 Br⁻ → PbBr₂

(c) Cu + S → CuS

Reaction (c) represents a redox reaction, as the oxidation states of the elements are changing. In this case:

Cu + S → CuS

In its elemental form, Cu has an oxidation state of 0, while in CuS (copper sulfide), its oxidation state changes to +2.

Similarly, S in its elemental form has an oxidation state of 0 and is -2 in CuS (copper sulfide).

Learn more about:

redox reactions

Answer:

Ir(NO2)3

Explanation:

The molar mass is 330.2335, in case that's also required.

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.

A. 1.01 is the accurate result

Because

The formula used is Pv= nRT

P=1 atm

V= 22.4 L

N= x

R= 0.0821

T= 273 K (since it’s standard temperature)

Thus, (1)(22.4)=(x)(0.0821)(273)

X= 1.001