The molarity of a solution prepared by dissolving 141.6g of citric acid in water is calculated as below
find the the number of moles
moles= mass/molar mass
= 141.6 g/ 192 g/mol = 0.738 moles
molarity= moles/molar mass
= 0.738/3500 x1000 = 0.21 M
The answer is an ENDOTHERMIC REACTION.
HOPE THIS HELPS!!!!!
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:
MEANS:
1 = 97.7
2 = 74.3
3 = 50
4 = 30
5 = 13
UNCERTAINTY:
gimme a sec, ill put it in the comments under this
Answer:
The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.
Explanation:
Mass of compound A in a mixture = 119 mg
Mass of compound A after re-crystallization = 83 mg
Percent recovery from re-crystallization :
Percent recovery of compound A:
Mass of compound B in a mixture = 97 mg
Mass of compound B after re-crystallization = 79 mg
Percent recovery of compound B:
