You must burn 1.17 g C to obtain 2.21 L CO2 at
STP.
The balanced chemical equation is
C+02+ CO2.
Step 1. Convert litres of CO, to moles of CO2.
STP is 0 °C and 1 bar. At STP the volume of 1 mol
of an ideal gas is 22.71 L.
Moles of CO2= 2.21 L CO2 × (1 mol CO2/22.71 L
CO2) = 0.097 31 mol CO2
Step 2. Use the molar ratio of C:CO2 to convert
moles of CO to moles of C
Moles of C= 0.097 31mol CO2 × (1 mol C/1 mol
CO2) = 0.097 31mol C
Step 3. Use the molar mass of C to calculate the
mass of C
Mass of C= 0.097 31mol C × (12.01 g C/1 mol C) =
1.17 g C
It looks as if you are using the old (pre-1982)
definition of STP. That definition gives a value of
1.18 g C.
The reaction which shows oxidation and reduction simultaneously is C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l).
<h3>What are redox reactions?</h3>
Those reaction in which oxidation as well as reduction of substances takes place simultaneously will known as redox reactions.
- SO₂(g) + H₂O(l) → H₂SO₃(aq)
- CaCO₃(aq) → CaO(s) + CO₂(g)
- Ca(OH)₂(s) + H₂CO₃(l) CaCO₃(aq) + 2H₂O(l)
Above reaction are not the redox reactions as in these reaction oxidation and reduction simultaneously not takes place.
- C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)
In the above reaction reduction of oxygen takes place as its oxidation state changes from 0 to -2, and at the same time oxidation of carbon takes place as its oxidation state changes from 0 to +4.
Hence correct option is (4).
To know more about redox reactions, visit the below link:
brainly.com/question/7935462
<h3>
Answer:</h3>
2.04 mol CBr₄
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Organic</u>
- Writing Organic Compounds
- Writing Covalent Compounds
- Organic Prefixes
<u>Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
675 g CBr₄
<u>Step 2: Identify Conversions</u>
Molar Mass of C - 12.01 g/mol
Molar Mass of Br - 79.90 g/mol
Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol
<u>Step 3: Convert</u>
<u />
<u />
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<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2.03552 mol CBr₄ ≈ 2.04 mol CBr₄
The average atomic mass of Sn is 118.71 g/mol
the percentage of heaviest Sn is 5.80%
the given mass of Sn is 82g
The total moles of Sn will be = mass / atomic mass = 82/118.71=0.691
Total atoms of Sn in 82g = 
the percentage of heaviest Sn is 5.80%
So the total atoms of 
= 5.80% X 
Total atoms of =
atoms
the mass of will be = 
Ans: Have the same number of electron shells
