Answer:
- <em>Option d. Its empirical formula is CH</em><em>₂</em><em>.</em>
Explanation:
The percent composition of the compound allow you to calculate the empirical formula of the compound but is not enough to calculate either the molar mass or the molecular formula. So, since now you can discard options b. and c.
Telling that it is a hydrocarbon (option e.) is true but very vague compared with finding the empirical formula. So, you can also discard the option e.
The fact that the product has a triple bond cannot be concluded from the percent composition, you should find the molecular formula to assert whether it contains or not a triple bond. So, you could discard option a., which lets you only with choice d.
Let us find the empirical formula to be certain that it is CH₂.
1. <u>First, assume a basis of 100 g of compound</u>:
- H: 14.5% × 100 g = 14.0 g
- C: 85.5% × 100 g = 85.5 g
2. <u>Divide each element by its atomic mass to find number of moles</u>:
- H: 14.0 g / 1.008 g/mol = 14.38 mol
- C: 85.5 g / 12.011 g/mol = 7.12 mol
3. <u>Divide both amounts by the smallest number, to find the mole ratio</u>:
- H: 14.38 mol / 7.12 mol ≈ 2
- C: 7.12 mol / 7.12 mol = 1.
Hence, the ratio is 2:1 and the empirical formula is CH₂.
Answer:
LHS-RHS
1 C - 1 C
4 H-2 H
2 O- 3 O
So on right side 2 hydrogen are less and one oxygen is more..so
Balanced equation is
CH4+2O2==CO2 + 2H2O
Explanation:
2 cm east is the answer because you go east 5.5 cm and then go back in a sense 3.5 so its basically 5.5-3.5 because its backwards while still facing eat
The element that gains electrons, becomes reduced.
While the one which loses electrons, becomes oxidized.
In this equation,
CH₃OH + Cr₂O₇²⁻---- --> CH₂O + Cr³⁺.
By balancing the equation, we will get:
3CH₃OH + Cr₂O₇²⁻ + 8H⁺ --> 3CH₂O + 2Cr³⁺ + 7H₂O
Here the oxidation state of Cr changes from +6 to +3 that is it is being reduced thus serving as a oxidizing agent while other element retain their charges.
Here Cr₂O₇²⁻ is reduced while CH₃OH is oxidized.
So Cr₂O₇²⁻ serves as a oxidizing agent, while CH₃OH serves as reducing agent .
Both figures are mixtures,
Figure II is a heterogenous mixture
Figure I is a homogenous mixture
