An electron can be added to halogen atom to force a halide ion with 8 valence electrons
<h3>What is an atom?</h3>
An atom can be defined as the smallest part of an element which can take part in a chemical reaction.
However whenever, an electron is added to halogen atom to force a halide ion with 8 different valence electrons
So therefore; an electron can be added to halogen atom to force a halide ion with 8 valence electrons
Learn more about halogens:
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Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
Answer:
hope this help !
Explanation:
Use the given functions to set up and simplify 173 ° C .
1.5 =
CH4 = CH4
4.4 = CH4
173 ° C = CH4
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Answer
b. 95%
Explanation
Given:
Mass of K₂O produced (actual yield) = 28.56 g
Mass of K that reacted = 25.00 g
Equation: 4K(s) + O₂(g) → 2K₂0(s)
What to find:
The percent yield of K₂O.
Step-by-step solution:
The first step is to calculate the theoretical yield of K₂O produced.
From the balanced equation, 4 mol K produced 2 mol K₂O
Molar mass of K₂O = 94.20 g/mol)
Molar mass of K = 39.10 g/mol)
This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O
So 25.00 g K will produce:

Actual yield of K₂O = 28.56 g
Theoretical yield of k₂O = 30.1151 g
The percent yield for the reaction can now be calculated using the formula below:

Therefore, the percent yield for the reaction is 95%.