The student dissolves the entire impure sample of CuSO4(s) in enough distilled water to make 100.ML of solution. Then the studen
t measures the absorbance of the solution and observes that it is 0.30 . Determine the concentration of CuSO4(aq) in the solution.
1 answer:
Answer:
Here's what I get
Explanation:
I assume you prepared a calibration curve — a plot of the absorbance vs the concentrations of known solutions.
It probably looked like the one below.
You drew the best straight line that went as close as possible to all the points without necessarily going through any of them.
Now you have a CuSO₄ solution with an absorbance of 0.30.
You start at 0.30 on the vertical axis and draw a line horizontally until it hits your best-fit line.
Then you draw a vertical line to the horizontal axis and read the corresponding concentration.
It hits the axis at 0.15 mol·L⁻¹.
You would say that the concentration of CuSO₄ is 0.15 mol·L⁻¹.
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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.
