Na3X, Enter the group number of X?
2 answers:
Answer : The group number of X is 15.
Explanation :
In , 'Na' is a sodium metal and 'X' is an unknown element.
First we have to determine the oxidation state of element 'X'.
Let the oxidation state of element 'X' be, 'a'.
As we know that sodium is an alkali metal which belongs to group 1 and has (+1) charge.
That means, the oxidation state or charge on 'X' is (-3) and the group 15 elements (nitrogen, phosphorous, arsenic, antimony and bismuth) have five electrons in valence shell and thus require three more electrons to acquire the nearest noble gas configuration. Thus, the group 15 elements have the tendency to show (-3) oxidation state.
Hence, the group number of X is 15.
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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.
