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2 years ago

Which substance is the oxidizing agent in this reaction? 2CuO+C→2Cu+CO2 Express your answer as a chemical formula.

1 answer:

3 0

The oxidizing agent is the one that is reduced in the reaction. In this reaction, the charge of Cu falls from +2 to zero charge (neutral atom in the right side). Hence, CuO is the oxidizing agent. The reducing agent, the one being oxidized is carbon from zero charge to +4. The answer is CuO.

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a.) The diameter of a uranium atom is 3.50 Å . Express the radius of a uranium atom in both meters and nanometers. b.) How many

Svetradugi [14.3K]

Answer:

Th answer to your question is:

a) 3.5 x10⁻¹⁰ meters; 0.35 nm

b) 6857142.86 atoms

c) Volume = 2.06 x 10⁻²³ cm³

Explanation:

a) data

Uranium atoms = 3.5A°

meters

1 A° ---------------- 1 x 10 ⁻¹⁰ m

3.5A° --------------- x

x = 3.5(1 x10⁻¹⁰)/ 1 = 3.5 x10⁻¹⁰ meters

1 A° ------------------ 0.1 nm

3.5 A° ---------------- 0.35 nm

b) 2.4 mm

Divide 2,40 mm / uranium diameter

But, first convert 3,5A° to mm = 3.5 x 10⁻⁷ mm

# of uranium atoms = 2.4 / 3.5 x 10⁻⁷ = 6857142.86

c) volume in cubic cm

Convert 3.5A° to cm = 3.5 x 10⁻⁸

Volume = 4/3 πr³ = (4/3) (3.14)(1.7 x10⁻⁸)³

Volume = 2.06 x 10⁻²³ cm³

6 0

2 years ago

Karen has discovered a new organism in the Amazon Rainforest. It moves by itself, is a consumer, and its cells do not have a cel

EastWind [94]

The correct answer is option B, that is, Animalia.

The animals refer to the multicellular eukaryotic species, which forms the biological kingdom known as Animalia. The majority of the animals breathe oxygen, consume organic substances, and possess the tendency to move, has the tendency to reproduce, and develop from a hollow sphere of cells, the blastula at the time of embryonic development. In comparison to Kingdom Plantae, the cells in animals do not exhibit cell wall. Thus, on the basis of the characteristics, the newly discovered species must be categorized under Kingdom Animalia.

3 0

2 years ago

A sample of SO3 is introduced into an evacuated sealed container and heated to 600 K. The following equilibrium is established:

Andreas93 [3]

Answer: The value of $K_p$ is 0.050.

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_x=x\times P$

As we know the mole fraction of $O_2$ is 0.12

The partial pressure of $O_2=0.12\times 3.0atm=0.36atm$

The partial pressure of $SO_2=2\times 0.36atm=0.72atm$ Thus the partial pressure of $SO_3$ is = [3 - (0.36+0.720)] atm = 1.92 atm

$p_{SO3}$ = 1.92 atm

$2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)$

$K_p=\frac{p_{O_2}\times (p_{SO_}2)^2}{(p_{SO_3})^2}$

$K_p=\frac{0.36\times (0.72)^2}{(1.92)^2}$

$K_p=0.050$

The value of $K_p$ is 0.050.

7 0

2 years ago

A geologist took this aerial photo of a location where a river flows out of a mountain range.

rodikova [14]

Answer:

A<u> alluvial fan</u>

Explanation:

3 0

2 years ago

Read 2 more answers

Mass in grams of 6.25 mol of copper (II) nitrate?

podryga [215]

Cu = 63.546
N= 14.001 g/mol
O= 15.999 g/mol * 3 = 47.997

Copper (II) Nitrate has a MW of 125.544 g/mol

6.25 x 125.544

= 784.65 <--- is your answer, if there were was a multiple choice or not :)

8 0

2 years ago

Read 2 more answers