Which element is a solid at STP? A. H2 B. I2 C. N2 D. O2
1 answer:
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Answer : Option C) Atomic Size
Explanation : The atomic radius of the elements is found to be decreasing if we go from left to right in the modern periodic table. Accordingly, increases as the number of shielding electrons present in the atomic nucleus of the periodic elements which lies in the same row remains constant while the number of protons in each atomic shell increases.
The effective nuclear charge of an atom is defined as the net positive charge which is felt by the valence electron of the atomic element.
When is observed to decrease, it is seen that the atomic radius grows in size. So, it explains the inverse relationship between both. This phenomenon occurs, because there is more screening of the electrons from the nucleus taking place, which is observed due to decrease the attraction between the electron and the nucleus.
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.
