Answer:
The chemical equation by putting, a 2 on C₅H₁₂O, 15 on O₂, 10 on CO₂ , and 12 on H₂O in the equation;
2C₅H₁₂O + 15O₂ → 10CO₂ + 12H₂O
Explanation:
- Chemical equations are balanced by putting coefficients on the reactants and products to ensure the total number of atoms on the left side equal to those on the right side.
- Balancing chemical equations is done to make chemical equations obey the law of conservation of mass.
- According to the law of conservation of mass, the mass of the reactants should always be equal to the mass of products.
- This is done by balancing chemical equations to ensure the total number of atoms on the left side is equal to that on the right side.
- Therefore, the balanced equation is;
2C₅H₁₂O + 15O₂ → 10CO₂ + 12H₂O
Answer:
C3 H6 Cl 3
Explanation:
C -24.2%
H - 4.0%
Cl - (100-24.2 - 4.0)=73.8 %
We can take 100g of the substance, then we have
C -24.2 g
H - 4.0 g
Cl - 73.8 g
Find the moles of these elements
C -24.2 g/12.0 g/mol =2.0 mol
H - 4.0 g/1.0 g/mol = 4. 0 mol
Cl - 73.8 g/ 35.5 g/mol = 2.1 mol
Ratio of these elements gives simplest formula of the substance
C : H : Cl = 2 : 4 : 2 = 1 : 2 : 1
CH2Cl
Molar mass (CH2Cl) = 1*12.0 +2*1.0 + 1*35.5 = 49.5 g/mol
Real molar mass = 150 g/mol
real molar mass/ Molar mass (CH2Cl) = 150 /49.5=3
So, Real formula should be C3 H6 Cl 3.
Answer:
Dispersion Forces are found between n-Pentane (CH₃-CH₂-CH₂-CH₂-CH₃) and n-Hexane (CH₃-CH₂-CH₂-CH₂-CH₂-CH₃).
Explanation:
Dispersion Forces are present and developed by those compounds which are non-polar in nature. In given statement n-Pentane and n-Hexane both are non-polar in nature as the electronegativity difference between Hydrogen atoms and Carbon atoms is less than 0.4.
When non-polar molecules approaches each other, a Dipole is induced in one of them, this step is known as Instantaneous Dipole, This generated Dipole on approaching another non-polar molecule induces dipole in it and the process propagates. Hence, creating intermolecular interactions.
1. H₂SO₄ + 2NH₄OH ⟶ (NH₄)₂SO₄ + 2H₂O
2. 2NaOH + H₂CO₃ ⟶ Na₂CO₃ + 2H₂O
3. HNO₃ + KOH ⟶ KNO₃ + H₂O
<em>Explanation</em>:
Acid + base ⟶ salt + water
Take the H from the acid and the OH from the base to get water.
Then, join what’s left to get the salt. Write the symbol for the metal first.
For example, in equation 3, take the H from HNO₃ and the OH from KOH.
Combining the remaining parts (NO₃ and K) to get the salt, KNO₃.
Answer: the percentage of acetic acid will be low.
Explanation: The major aim during titration of acids and bases is to determine the endpoint , that is exact point where the acid in the beaker changes colour, (in this case, pink )with an additional drop from the burette containing the base, since it is usually difficult to mark the equivalence point that tells us when all the substrate in the beaker has been neutralized completely with the buretted substance.
Overshooting the end point is an error which can occur when the person involved in the the titration accidently goes beyond this endpoint by adding too much of the substance(base) from the burette into the beaker missing the exact endpoint.
This implies that the person has added too much of the burreted liquid, ie the base than required , making the acid in the beaker to continue to react resulting to a lower concentration of the acid (acetic acid) with excess base.(NaOH)
