Overuse of the same chemicals can result in the pest becoming immune to the pesticides.
Answer:
T₂ = 721 k
Explanation:
Given data:
Initial volume = 285 mL
Initial pressure = 1.88 atm
Initial temperature = 355 K
Final temperature = ?
Final volume = 435 mL
Final pressure = 2.50 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁ / P₁V₁
T₂ = 2.50 atm × 435 mL × 355 K / 1.88 atm × 285 mL
T₂ = 386062.5 atm. mL. K /535.8 atm. mL
T₂ = 721 k
The amount of energy in kilocalories released from 49 g of glucose given the data is -4.4 Kcal
How to determine the mole of glucose
Mass of glucose = 49 g
Molar mass of glucose = 180.2 g/mol
Mole of glucose = ?
Mole = mass / molar mass
Mole of glucose = 49 / 180.2
Mole of glucose = 0.272 mole
How to determine the energy released
C₆H₁₂O₆ →2C₂H₆O + 2CO₂ ΔH = -16 kcal/mol
From the balanced equation above,
1 mole of glucose released -16 kcal of energy
Therefore,
0.272 mole of glucose will release = 0.272 × -16 = -4.4 Kcal
Thus, -4.4 Kcal were released from the reaction
Learn more about stoichiometry:
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Answer: 16 g/mol
Explanation: Molarity is Moles/Liters or Mol/L
(0.50 mol/L)(1.30L) = 0.65mol
(cancels out L)
Molar mass is g/mol so...
10.5g/0.65mol = 16.1 g/mol
Rounded to 2 sig figs is 16 g/mol