Answer:
1. 9.57 × 10^-9 moles.
2. 7.38mol
Explanation:
1.) To find the number of moles there are in the number of particles in an atom, we divide the number of particles (nA) by Avagadro's constant (6.02 × 10^23)
Hence, to find the number of moles (n) of Manganese (Mn), we say:
5.76 x 10^15 atoms ÷ 6.02 × 10^23
5.76/6.02 × 10^(15-23)
= 0.957 × 10^-8
= 9.57 × 10^-9 moles.
2.) Mole = mass/molar mass
Molar mass of sodium chloride (NaCl) = 23 + 35.5
= 58.5g/mol
mole = 431.6 g ÷ 58.5g/mol
mole = 7.38mol
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!
13 - Periodic table
14 - Dimitri mandeleev
15 - groups
Mark me brainiest pls it right answer
Answer:
V₂ = 6.0 mL
Explanation:
Given data:
Initial volume = 9.0 mL
Initial pressure = 500 mmHg
Final volume = ?
Final pressure = 750 mmHg
Solution:
According to Boyle's Law
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = 500 mmHg × 9.0 mL / 750 mmHg
V₂ = 4500 mmHg .mL / 750 mmHg
V₂ = 6.0 mL