What is the correct ratio of the atoms of the two elements present in a sample of the compound copper chloride, CuCl2?
1 answer:
The correct ratio of the atoms of the two elements in the sample of copper chloride compound would be 1:2.
<h3>Atoms of elements</h3>Copper chloride has a chemical formula, .
Thus, an atom of copper, Cu, combines with 2 atoms of chlorine, Cl.
The 2 chlorine atoms are recipients of an electron each from the copper atom which happens to have 2 electrons in its outermost valence shell.
Therefore, the ratio of copper to chlorine atoms in the compound is 1:2.
More on atoms of elements can be found here: brainly.com/question/15173445
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This is an incomplete question, the complete question is attached below.
Answer : The coefficient of silver in the final, balanced equation for this reaction is, 3
Explanation :
Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.
Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.
Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.
The given redox reaction is,
The oxidation-reduction half reaction will be :
Oxidation :
Reduction :
In order to balance the electrons, we multiply the reduction reaction by 3 and then added both equation, we get the balanced redox reaction.
The balanced redox reaction will be,
From the balanced redox reaction we conclude that, the coefficient of silver in the final balanced equation for this reaction is 3.
Hence, the correct option is 3.
<h3>Answer:</h3>
458 g H₂SO₄
<h3>General Formulas and Concepts:</h3><u>Math</u>
<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
<u>Stoichiometry</u>
- Using Dimensional Analysis
<u>Step 1: Define</u>
4.67 mol H₂SO₄
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of H - 1.01 g/mol
[PT] Molar Mass of S - 32.07 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of H₂SO₄ - 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol
<u>Step 3: Convert</u>
- Set up:
- Multiply/Divide:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
458.08 g H₂SO₄ ≈ 458 g H₂SO₄