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4 years ago

In a heterogeneous mixture the materials are mixed together and

1 answer:

5 0

In a heterogeneous mixture the materials are mixed together and will be easily separated. The correct option among all the options that are given in the question is the second option or option "b". An example of a heterogeneous mixture is a bowl of colored candies. I hope the answer helps you.

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Given 4.80g of ammonium carbonate, find:

V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

1) Number of moles of the compound:

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

2) Number of moles of ammonium ions :

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ = (2.0)(0.05 mol) = 0.1 mol.

3) Number of moles of carbonate ions :

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

4) Number of moles of hydrogen atoms:

Every 1.0 mol of (NH₄)₂CO₃ contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO₃ = (8.0)(0.05 mol) = 0.4 mol.

5) Number of hydrogen atoms:

It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

Using cross multiplication:

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

∴ The no. of atoms in 0.4 mol of H atoms = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = 2.4 x 10²³ molecules.

8 0

4 years ago

How many moles of methane are in 7.31*10^25 molecules?

GrogVix [38]

<h3>Answer:</h3>

121 mol CH₄

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Organic

Writing chemical compounds
Writing organic structures
Prefixes
Alkanes, Alkenes, Alkynes

Atomic Structure

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

Step 1: Define

7.31 × 10²⁵ molecules CH₄

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$\displaystyle 7.31 \cdot 10^{25} \ molecules \ CH_4(\frac{1 \ mol \ CH_4}{6.022 \cdot 10^{23} \ molecules \ CH_4} ) = 121.388 \ mol \ CH_4$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

121.388 mol CH₄ ≈ 121 mol CH₄

7 0

3 years ago

identify the element in period 3 that requires the least amount of energy to remove the most loosely held electrons from a mole

OLga [1]

Answer:

The element is Na

Explanation:

Ionization energy is the energy needed to release the last electron from an atom in its ground state to the gaseous state. It is a periodic property that increases as we go through the periods of the periodic table, but decreases if we move in groups. Sodium has thr ionic radius (another periodic property) that is too large, making it easier to release the electron away, since it is too far from the nucleus.

3 0

3 years ago

1. _ LiBr2 + __K2CO3 -->___ LiCO3 +_ KBr Reaction Type:

Arada [10]

Answer:

Double displacement chemical reaction.

Explanation:

Hello,

In this case, for the given reaction we firstly balance it:

$LiBr_2 +K_2CO3 \rightarrow LiCO3 +2KBr$

Then, since the lithium and potassium cations are being exchanged to each other from bromide and carbonate, we are talking about a double displacement chemical reaction.

Best regards.

5 0

4 years ago

Ca(ClO4)2 Ionic compound name

qwelly [4]

Answer:

I thing Calcium Perchlorate

7 0

3 years ago