How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethy
1 answer:
Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether
Explanation:
First we have to calculate the moles of diethyl ether
As, 1 mole of diethyl ether require heat = 26.5 kJ
So, 1.01 moles of diethyl ether require heat =
Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether
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Answer:
E = h v = 6.626 x x 97.20 x
= 6.44 x
J = 4.01 x
eV
n = Energy Level = / E = -13.6 / 4.01 x
= 3.39 x
= 1.84 x
energy level...
Answer:
Energy= 2.7758 × 10^-11 J ;
71.112×10^-6 kJ.
Mass defect in Kilogram= 3.0885×10^-28 kg.
That is; 3.1×10^-28 kg(to two significant figure).
Explanation:
(Note: Check equation of reaction in the attached file/picture).
STEP ONE: we have to calculate the Mass defect.
Mass defect= Mass of reactants -- Mass of products.
Mass of the products: (140.9144+91.9262+3.060) u.
= 235.8666 u.
Mass of reactants: (1.0087+235.0439) u= 236.0526 u.
Therefore, the Mass defect= (236.0526 -- 235.8666) u
= 0.1860 u.
STEP TWO: Converting the Mass defect to energy;
0.18860 × 1.6605 × 10^27 kg
= 3.0885× 10^-28 kg
STEP THREE: Calculating energy released . Recall(from the question) c^2= 931.5 Mev/u. This is also equals to 9×10^16 m/s.
E=Mc^2.
Where E= energy released, c= speed of light, M= Mass.
Slotting in the values;
E= 3.0885×10^-28 kg × 9×10^16 m/s.
E=2.7758 × 10^-11 J.
Know that;( 1g of uranium × 1 mol of uranium ÷ 235.0439 g of uranium) × (6.002×10^23 atom of uranium/ 1 mol of uranium) × 2.7758× 10^-11.
=7.1112×10^-10 J
= 71.112×10^6 kJ.
