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.
The answer is. B hope it helps plz mark me a brainless plz
NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
<span>Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27% </span>
<span>3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
</span><span>0.8211 grams Na + 1.266 grams Cl = 2.087 grams</span>
File:///Users/user/Downloads/Chemistry%20Exam%20Study%20Guide%20Fall%202012.pdf
Idk if this will help you, cuz it says what question are going to be in ur final
The greatest aqueous freezing point is (D) 0.10 KCI
