Answer:
The resulting pressure is 2.81 atm
Explanation:
According to Dalton's Law of Partial Pressure, each of the gases (A and B) will exert their pressure independently. If we use Boyle's Law to calculate the pressure of each of the gases separately we have:
Pressure of gas A:
p1V1 = p2V2
p1 = 2.4 atm
V1 = 722 mL
V2 = 722 + 169 = 891 mL
p2 =?
Clearing p2:
p2 = (p1V1)/V2 = (2.4*722)/891 = 1.94 atm
Pressure of gas B:
p1 = 4.6 atm
V1 = 169 mL
V2 = 169+722 = 891 mL
p2=?
Clearing p:
p2 = (4.6*169)/891 = 0.87 atm
Dalton's expression for total partial pressures is equal to:
ptotal = pA + pB = 1.94+0.87 = 2.81 atm
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:
• One mole of oxygen is equivalent to 16 grams.
→ But at STP, 22.4 dm³ are occupied by 1 mole.
Answer:
32.6%
Explanation:
Equation of reaction
2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)
Molar volume of Oxygen at s.t.p = 22.4L / mol
since the gas was collected over water,
total pressure = pressure of water vapor + pressure of oxygen gas
0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C
pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1
P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p
Using ideal gas equation
= 
V2 = 
V2 = 664.1052 ml
245.2 yielded 67.2 molar volume of oxygen
0.66411 will yield = 
= 2.4232 g
percentage of potassium chlorate in the original mixture = 
= 32.6%
