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Angelina_Jolie [31]
3 years ago
8

What is the purpose of a blank in spectrophotometry.

Chemistry
1 answer:
wariber [46]3 years ago
7 0
Spectrophotometers are also calibrated by using a “blank” solution that we prepare containing all of the components of the solution to be analyzed except for the one compound we are testing for so that the instrument can zero out these background readings and only report values for the compound of interest.
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What is in a microscope the enlarges things so our eye can see them?
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3 years ago
How much potassium nitrate has to dissolve in water to absorb 105 kj of heat? there are 4.184 kj/kcal. also note the answer is t
joja [24]

Lattice energy of potassium nitrate (KNO3) = -163.8 kcal/mol

Hydration energy of KNO3 = -155.5 kcal/mole

Heat of solution is the amount of heat absorbed by water when 1 mole of KNO3 is dissolved in it

Heat of solution = Hydration energy - Lattice energy

                           = -155.5 -(-163.8) = 8.3 kcal/mol

1 kcal/mol = 4.184 kJ/mole

Therefore, 8.3 kcal/mole = 4.184 * 8.3 = 34.73 kJ/mol

Now,  34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved

The given 105 kJ of heat would correspond to : 105/34.73 = 3.023 moles of KNO3

Molar mass of KNO3 = 101.1 g/mole

Mass of KNO3 = Molar mass * moles

                       = 101.1 * 3.023 = 305.63 g = 0.3056 kg    

8 0
4 years ago
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Answer:

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6 0
3 years ago
A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
True [87]
First, in order to calculate the specific heat capacity of the metal in help in identifying it, we must find the heat absorbed by the calorimeter using:
Energy = mass * specific heat capacity * change in temperature
Q = 250 * 1.035 * (11.08 - 10)
Q = 279.45 cal/g

Next, we use the same formula for the metal as the heat absorbed by the calorimeter is equal to the heal released by the metal.

-279.45 = 50 * c * (11.08 - 45) [minus sign added as energy released]
c = 0.165

The specific heat capacity of the metal is 0.165 cal/gC
6 0
3 years ago
Read 2 more answers
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