The type of _______ and the climate determines the rate at which weathering occurs

Given, 2Ca + O2 -> 2CaO, what is the 2 located in front of calcium called??

EleoNora [17]

Moles used to react

3 0

3 years ago

Select the correct answer. which substance in this redox reaction is the oxidizing agent? cu 2agno3 → 2ag cu(no3)2 a. n b. agno3

lana66690 [7]

AgNO₃ will act as the oxidising agent.

<h3>For the given chemical equation:</h3>

Cu + 2AgNO₃ → 2Ag + Cu(NO₃)₂

Half reactions for the given chemical reaction:

Reducing agent:

Cu → Cu²⁺ + 2e⁻

Copper is a reducing agent because it is losing 2 electrons, which causes an oxidation process.

Oxidising Agent:

Ag⁺ + e⁻ → Ag

The silver ion undergoes a reduction process and is regarded as an oxidizing agent since it is acquiring one electron per atom.

Hence, AgNO₃ is considered as an oxidizing agent and therefore the correct answer is Option B.

<h3>Oxidising and Reducing agents</h3>

An oxidizing agent is a substance that reduces itself after oxidizing another material. It passes through a reduction process in which it obtains electrons and the substance's oxidation state is decreased.
A reducing agent is a chemical that oxidizes after reducing another material. It passes through an oxidation process in which it loses electrons and the substance's oxidation state increases.

To know more about the process of Oxidation and Reduction, refer to:

brainly.com/question/4222605

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8 0

1 year ago

What is the difference in uniformity between heterogeneous and homogeneous mixtures?

Aneli [31]

Answer:

The correct answer according to the uniformity the difference is whether you can distinguish more than one part in the mixture.

Explanation:

A homogeneous mixture is one in which with the naked eye or with a microscope no different parts can be distinguished. Its composition and properties are the same at all points, the substance is uniform.

In a heterogeneous mixture, we can observe different parts. Its composition is variable. It can be broken down into simpler substances by physical processes. The substance, in this case, is not uniform.

Have a nice day!

4 0

3 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago

A. 207 kJ B. 4730 kJ O C. 9460 kJ O D. 414 kJ

slavikrds [6]

Answer:

C = 9460 Kj

Explanation:

Given data:

Mass of copper = 2kg

Latent heat of vaporization = 4730 Kj/Kg

Energy required to vaporize 2kg copper = ?

Solution:

Equation

Q= mLvap

by putting values,

Q= 2kg × 4730 Kj/Kg

Q = 9460 Kj

3 0

3 years ago