A 30.5 gram sample of glucose (c6h12o6) contains ________ mol of glucose.
1 answer:
You might be interested in
Answer:
(a) A strong acid
Explanation:
We have given the pH of the solution is 2.46
pH=2.46
So the concentration of
solution having H+ concentration more than is acidic
Since in the given solution, H+ concentration is 0.00346 M which is more than 10^{-7}[/tex] so this is an acidic solution
Note-The concentration of decide the behavior of the solution that is, it is acidic or basic
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.
