Answer:
The half-life varies depending on the isotope.
Half-lives range from fractions of a second to billions of years.
The half-life of a particular isotope is constant.
The reaction of Phosphorous Pentaoxide with water yield Phosphoric Acid as shown below,
P₄O₁₀ + 6 H₂O → 4 H₃PO₄
According to balance equation,
283.88 g (1 mole) P₄O₁₀ requires = 108 g (6 mole) of H₂O
So,
100 g P₄O₁₀ will require = X g of H₂O
Solving for X,
X = (100 g × 108 g) ÷ 283.88 g
X = 38.04 g of H₂O
So, 100 g P₄O₁₀ requires 38.04 g of H₂O, while we are provided with 200 g of H₂O which means that water is in excess and P₄O₁₀ is limiting reagent. Therefore, P₄O₁₀ will control the yield of H₃PO₄. So,
As,
283.88 g (1 mole) P₄O₁₀ produced = 391.96 g (4 mole) of H₃PO₄
So,
100 g P₄O₁₀ will produce = X g of H₃PO₄
Solving for X,
X = (100 g × 391.96 g) ÷ 283.88 g
X = 138.07 g of H₃PO₄
Result:
Theoretical Yield of this reaction is 138.07 g.
Answer:
The volume is 1.2L
Explanation:
Initial volume (V1) = 700mL = 0.7L
Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K
Initial pressure = 106.6kPa = 106600Pa
Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K
Final pressure (P2) = 66.6kPa = 66600Pa
Final volume (V2) = ?
To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.
(P1 × V1) / T1 = (P2 × V2) / T2
solve for V2 by making it the subject of formula,
P1 × V1 × T2 = P2 × V2 × T1
V2 = (P1 × V1 × T2) / (P2 × T1)
V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)
V2 = 22397193 / 18657990
V2 = 1.2L
The final volume of the gas is 1.2L
Colligative properties include lowering the freezing point, lowering the vapor pressure, and raising the boiling point.
Answer:
28.25 F per hour
Explanation:
It has a rather very easy solution. First find the temperature difference between the the initial and final internal temperatures which is 145 - 32 = 113 F. The time span is about 4 hours, divide 113 F by 4 , which is 28.25 F per hour.
