Answer:
LOD = 0,0177
LOQ = 0,0345
Explanation:
Detection limit (LOD) is defined as the lowest signal which, with a stated probability, can be distinguished from a suitable blank signal. In the same way, quantification limit (LOQ) is defined as the lowest analyte concentration that can be quantitatively detected with a stated accuracy and precision.
There are many formulas but the most used are:
LOD = X + 3σ
LOQ = X + 10σ
Where X is average and σ is standard desvation
For the blanks readings the average X is 0,0105 and σ is 0,0024
Thus:
<em>LOD = 0,0177</em>
<em>LOQ = 0,0345</em>
I hope it helps!
Answer:
c. 20.0332 g to 20,0 g
Explanation:
A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.
<em>Which of the following examples illustrates a number that is correctly rounded to three significant figures?
</em>
a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.
b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.
c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.
d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.
e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.
The reaction creates dihydrogen, hence if it's uncontrolled it could lead to potentially dangerous amounts of gas being released at once.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of 
= 
= 0.0458 M
Concentration of 
= 
= 0.0521 M
GIven that :
Ka = 
Thus; it's pKa = 4.72
pH ≅ 4.80
The most obvious answer for this would be a product i think.
