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2 years ago

4.

Chemistry

1 answer:

Naya [18.7K]2 years ago

5 0

For all the reactions listed, the moles of products and limiting reactants are;

1. Molar cofficient - 6, oxygen is the limiting reactant

2. Molar coefficient - 8 chlorine is the limiting reactant

3. Molar coefficient - 4 and CO2 is the limiting reactant

<h3>What is a chemical reaction?</h3>

A chemical reaction is a combination of reactants that yields products. Recall that the limiting reactant is the reactant that nis present in the least amount.

Let us now complete each reaction;

1. The molar coefficient of the product in reaction 1 is 6 and oxygen is the limiting reactant

2. The molar For all the reactions listed coefficent of the product in reaction 2 is 8 and chlorine is the limiting reactant

3. The molar coefficient of the product in reaction 3 is 4 and the limiting reactant is CO2.

Learn more about limiting reactant: brainly.com/question/14225536

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3 years ago

How many atoms of cobalt are in 4 moles of cobalt?

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<h3>Answer:</h3>

2 × 10²⁴ atoms Co

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

4 mol Co (Cobalt)

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 4 \ mol \ Co(\frac{6.022 \cdot 10^{23} \ atoms \ Co}{1 \ mol \ Co})$
Multiply/Divide: $\displaystyle 2.4088 \cdot 10^{24} \ atoms \ Co$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

2.4088 × 10²⁴ atoms Co ≈ 2 × 10²⁴ atoms Co

4 0

3 years ago

When 231. mg of a certain molecular compound X are dissolved in 65. g of benzene (CH), the freezing point of the solution is mea

Elan Coil [88]

Answer:

Molar mass X = 18.2 g/mol

Explanation:

Step 1: Data given

Mass of compound X = 231 mg = 0.231 grams

Mass of benzene = 65.0 grams

The freezing point of the solution is measured to be 4.5 °C.

Freezing point of pure benzene = 5.5 °C

The freezing point constant for benzene is 5.12 °C/m

Step 2: Calculate molality

ΔT = i*Kf*m

⇒ΔT = the freezing point depression = 5.5 °C - 4.5 °C = 1.0 °C

⇒i = the Van't hoff factor = 1

⇒Kf = the freezing point depression constant for benzene = 5.12 °C/m

⇒m = the molality = moles X / mass benzene

m = 1.0 / 5.12 °C/m

m = 0.1953 molal

Step 3: Calculate moles X

Moles X = molality * mass benzene

Moles X = 0.1953 molal * 0.065 kg

Moles X = 0.0127 moles

Step 4: Calculate molar mass X

Molar mass X = mass / moles

Molar mass X = 0.231 grams / 0.0127 moles

Molar mass X = 18.2 g/mol

5 0

3 years ago